To solve this problem, we need to use related rates. Let's start by drawing a diagram:
```
/\
/ \
/ \
/ \
/ \
/__________\
```
We know that the radius of the conical pile is always two times its height, so we can label the diagram as follows:
```
/\
/ \
/ \
/ \
/ \
/__________\
/| r=2h \
/ |___________\
```
Now we need to find an equation that relates the height of the pile to its radius. We can use the formula for the volume of a cone:
```
V = (1/3)πr^2h
```
We want to solve for h in terms of r:
```
V = (1/3)πr^2h
3V/πr^2 = h
```
Now we can differentiate both sides of this equation with respect to time:
```
d/dt (3V/πr^2) = d/dt h
0 = (3/πr^2) dV/dt - (2/πr^3) dr/dt
```
We're given that the height is increasing at a rate of 2 cm/s when the pile is 11 cm high, so we know that:
```
dh/dt = 2 cm/s
h = 11 cm
```
We want to find the rate at which sand is leaving the bin, which is given by `dV/dt`. We can solve for this using the equation we derived:
```
0 = (3/πr^2) dV/dt - (2/πr^3) dr/dt
dV/dt = (2/3)πr^2 (dh/dt) / r
```
Now we just need to plug in the values we know:
```
dh/dt = 2 cm/s
h = 11 cm
r = 2h = 22 cm
dV/dt = (2/3)π(22)^2 (2) / 22
dV/dt = 264π/3
```
So the rate at which sand is leaving the bin when the pile is 11 cm high is `264π/3 cm^3/s`.
To solve this problem, we can use the relationship between the radius and height of the conical pile, as well as the given rate of height increase.
Since the radius (r) is always two times the height (h), we have r = 2h. The volume (V) of a cone is given by the formula V = (1/3)πr^2h. We can substitute r with 2h, so V = (1/3)π(2h)^2h.
Now, let's differentiate both sides with respect to time (t):
dV/dt = (1/3)π(8h^2)dh/dt
When the height is 11 cm, the rate of height increase (dh/dt) is 2 cm/s. We can substitute these values into the equation:
dV/dt = (1/3)π(8(11)^2)(2)
Solving for dV/dt:
dV/dt ≈ 2046.92 cm³/s
At that instant, the sand is leaving the bin at a rate of approximately 2046.92 cm³/s.
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Find the prime factorization of each number. Record without and then with exponents. 75
Answer:
5*5*3
5²*3
Step-by-step explanation:
Assuming that we have to find prime factorization of 75:
75 = 25 * 3
75 = 5*5*3
In 2000, 1500 rabbits live in a warren in a certain area. the number of rabbits increases exponentially at a discrete rate of 7% per year. predict population in 2008&2022
The predicted populations of rabbits in the warren in 2008 and 2022 are 2909 and 9933, respectively.
To predict the population of rabbits in the warren in 2008 and 2022, we can use the formula for exponential growth:
P(t) = P₀ (1 + r)ᵗ
Where P(t) is the population at time t, P₀ is the initial population, r is the growth rate, and t is the time elapsed.
For this problem, we know that P₀ = 1500, r = 0.07 (since the growth rate is 7%), and we want to find P(8) and P(22) (since we're predicting the population in 2008 and 2022, respectively).
Using the formula, we get:
[tex]P(8) = 1500( 1 + 0.07)^{8} = 2909[/tex]
So we can predict that there will be about 2909 rabbits in the warren in 2008.
To find P(22), we simply plug in t = 22:
[tex]P(22) = 1500 (1 + 0.07)^{22} = 9933[/tex]
So we can predict that there will be about 9933 rabbits in the warren in 2022.
Therefore, the predicted populations of rabbits in the warren in 2008 and 2022 are 2909 and 9933, respectively.
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Thirty-three cities were researched to determine whether they had a professional sports team, a symphony, or a children's museum. Of these cities, 17 had a professional sports team, 15 had a symphony, 14 had a children's museum, 9 had a professional sports team and a symphony, 6 had a professional sports team and a children's museum, 6 had a symphony and a children's museum, and 3 had all three activities.
Complete parts a) through e) below.
a) How many of the cities surveyed had only a professional sports team?
b) How many of the cities surveyed had a professional sports team and a symphony, but not a children's museum?
c) How many of the cities surveyed had a professional sports team or a symphony?
d) How many of the cities surveyed had a professional sports team or a symphony, but not a children's museum?
e) How many of the cities surveyed had exactly two of the activities?
Simplify your answers.
a) The number of cities that had only a professional sports team can be found by subtracting the number of cities that had a professional sports team and a symphony, the number of cities that had a professional sports team and a children's museum, and the number of cities that had all three activities from the total number of cities:
33 - (9 + 6 + 3) = 15 cities had only a professional sports team.
b) The number of cities that had a professional sports team and a symphony, but not a children's museum can be found by subtracting the number of cities that had all three activities from the number of cities that had a professional sports team and a symphony:
9 - 3 = 6 cities had a professional sports team and a symphony, but not a children's museum.
c) The number of cities that had a professional sports team or a symphony can be found by adding the number of cities that had a professional sports team, the number of cities that had a symphony, and then subtracting the number of cities that had both:
17 + 15 - 9 + 14 - 6 + 3 = 34 cities had a professional sports team or a symphony.
d) The number of cities that had a professional sports team or a symphony, but not a children's museum can be found by subtracting the number of cities that had all three activities from the answer to part c:
34 - 3 = 31 cities had a professional sports team or a symphony, but not a children's museum.
e) The number of cities that had exactly two of the activities can be found by adding up the number of cities that had a professional sports team and a symphony, the number of cities that had a professional sports team and a children's museum, and the number of cities that had a symphony and a children's museum, and then subtracting twice the number of cities that had all three activities:
9 + 6 + 6 - 2(3) = 15 cities had exactly two of the activities.
(1 point) Evaluate the integral by reversing the order of integration. 7 STE dedy
the integral by reversing the order of integration. the result will be F(β) - F(α), which is the integral evaluated after reversing the order of integration.
First, let's rewrite your integral more clearly:
∫∫ R 7x dy dx, where R is the region of integration.
To reverse the order of integration, we first need to determine the limits of integration for R in terms of x and y. Let's assume the current limits are a to b for x and c(y) to d(y) for y.
Now, we need to express these limits in terms of y and x. Let's denote the new limits as α to β for y and γ(x) to δ(x) for x.
After finding the new limits, we can rewrite the integral as:
∫∫ R 7x dx dy
Now, evaluate the integral by integrating first with respect to x and then with respect to y:
1. Integrate 7x with respect to x: (7/2)x^2 + C₁(x)
2. Apply the limits of integration for x: [(7/2)δ(x)^2 + C₁(δ(x))] - [(7/2)γ(x)^2 + C₁(γ(x))]
3. Integrate the result with respect to y: ∫[α, β] [(7/2)(δ(y)^2 - γ(y)^2)] dy
4. Apply the limits of integration for y: F(β) - F(α)
The final result will be F(β) - F(α), which is the integral evaluated after reversing the order of integration.
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(−3m
5
)(−2m
4
)=left parenthesis, minus, 3, m, start superscript, 5, end superscript, right parenthesis, left parenthesis, minus, 2, m, start superscript, 4, end superscript, right parenthesis, equals
The solution to the equation is m = 0.
What is an algebraic expression?
An algebraic expression is a mathematical phrase that contains variables, constants, and mathematical operations. It may also include exponents and/or roots. Algebraic expressions are used to represent quantities and relationships between quantities in mathematical situations, often in the context of problem-solving.
To solve the equation:
(-3m^ {5}) (-2m^ {4}) = (-6m^ {9})
We can simplify the left side of the equation by multiplying the terms:
(-3m^ {5}) (-2m^ {4}) = (6m^ {9})
Now we have:
6m^ {9} = (-6m^ {9})
To solve for m, we can divide both sides by 6m^ {9}:
m^ {9} = -m^ {9}
Since the powers of m on both sides are equal, we can simplify to:
2m^ {9} =0
Dividing both sides by 2, we get:
m^ {9} =0
Taking the ninth root of both sides, we get:
m = 0
Therefore, the solution to the equation is m = 0.
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Complete Question:
Simplify the expression: $(-3m^ {5}) (-2m^ {4}) $.
1.
The capacity of 1 jug is the same as the total capacity of 6 similar glasses. 10. 36 litres of
water is needed to fill 3 jugs and 19 glasses. What is the capacity of one glass?
10
The capacity of one glass is 36/37 liters.
Let's begin by representing the capacity of one jug as "J" and the capacity of one glass as "G". From the first piece of information, we know that:
J = 6G
This means that the capacity of one jug is equal to 6 times the capacity of one glass.
Next, we are given that 36 liters of water is needed to fill 3 jugs and 19 glasses. We can use this information to form an equation in terms of J and G.
The total capacity of 3 jugs is 3J and the total capacity of 19 glasses is 19G. Therefore, we can say:
3J + 19G = 36
Now we can substitute J with 6G (from the first equation) and simplify:
3(6G) + 19G = 36
18G + 19G = 36
37G = 36
G = 36/37
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In a recent BMO survey, many Canadian university students said they expected to owe
$26, 500 after graduation. A group of n = 64 university students are randomly selected, and
the average student loan debt is found to be $26,000 with a standard deviation of 500.
(a) Construct a 98% confidence interval for the true average student loan debt for
university students in Canada.
(b) Does this contradict the reported average of $26,500? Explain
(a) With 98% confidence, the true average student loan debt for university students in Canada falls between $25,883.5 and $26,116.5.
(b) No, this does not contradict the reported average of $26,500.
(a) To construct a 98% confidence interval, we can use the formula:
CI = [tex]\bar{x}[/tex] ± z* (σ/√n)
where [tex]\bar{x}[/tex] is the sample mean, σ is the population standard deviation (unknown), n is the sample size, and z* is the z-value from the standard normal distribution that corresponds to the desired level of confidence (98% in this case).
Substituting the given values, we get:
CI = 26,000 ± 2.33 * (500/√64)
CI = 26,000 ± 116.5
CI = (25,883.5, 26,116.5)
Therefore, we can say with 98% confidence that the true average student loan debt for university students in Canada falls between $25,883.5 and $26,116.5.
(b) No, this does not contradict the reported average of $26,500. The reported average is within the confidence interval we calculated, which means that it is a plausible value for the true average. The sample mean of $26,000 is slightly lower than the reported average, but this could be due to random sampling error.
Overall, we cannot make a definitive conclusion about the true average based on this sample alone, but we can say with 98% confidence that it falls within the range of $25,883.5 to $26,116.5.
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How much will the monthly payment be for a new car priced at $24,530 if the current finance rate is 36 months at 3. 16%? You will finance the 8% TT&L and make a 20% down payment
The monthly payment for a new car priced at $24,530 with the given terms is $630.15.
Calculating the down payment:
20% of $24,530
= 0.20 × 24,530
= $4,906.
Subtracting the down payment from the car price:
= $24,530 - $4,906
= $19,624 (amount to finance).
Adding the 8% TT&L (tax, title, and license) to the amount to finance:
8% of $24,530
= 0.08 × 24,530
= $1,962.40.
So, the total amount to finance
= $19,624 + $1,962.40
= $21,586.40.
Converting the annual interest rate of 3.16% to a decimal:
3.16% / 100
= 0.0316.
Calculating the monthly interest rate:
0.0316 / 12
= 0.002633.
Calculating the total number of payments: 36 months.
Using the monthly payment formula:
P = (PV * r * (1 + r)^n) / ((1 + r)^n - 1),
where P is the monthly payment,
PV is the present value or amount to finance,
r is the monthly interest rate, and
n is the total number of payments.
Now, pluggin in the values:
P = ($21,586.40 × 0.002633 × (1 + 0.002633)³⁶) / ((1 + 0.002633)³⁶ - 1)
= $630.15.
The monthly payment for the new car will be approximately $630.15.
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(1 point) Consider the power series (-1)"x" Vn +5 Find the radius of convergence R. If it is infinite, type "infinity" or "inf". Answer: R= What is the interval of convergence? Answer (in interval not
R= 1 and the interval of convergence is (-1, 1].
To find the radius of convergence, we can use the ratio test:
lim n→∞ |(-1)^n x^(n+1) Vn+5| / |(-1)^n x^n Vn+5| = lim n→∞ |x|
This limit exists for all x, and it equals 1 when |x| = 1. Therefore, the radius of convergence is R = 1.
To determine the interval of convergence, we need to check the endpoints x = -1 and x = 1 separately.
When x = -1, the series becomes:
∑ (-1)^n (Vn+5)
This is an alternating series with decreasing terms, so it converges by the alternating series test.
When x = 1, the series becomes:
∑ (-1)^n (Vn+5)
This is again an alternating series with decreasing terms, so it also converges by the alternating series test.
Therefore, the interval of convergence is (-1, 1].
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(a)
The masses of two animals at a zoo are described, where band care integers.
•The mass of an African elephant is 6, 125,000 grams, or about 6 x 10 grams.
• The mass of a silverback gorilla is 185, 000 grams, or about 2 x 10 grams.
What are the values of b and c?
bu
CH
(b) Part B
Using these estimated values, the mass of the African elephant is about 3 x 10 times the mass of the silverback gorilla, where m is an integer.
What is the value of m?
m
With the masses, the value of a and b will be 6 and 5.
The value of m is 6.
How to calculate the valueThe mass of an African elephant is 6,125,000 grams, or about 6 x 10⁶grams. Thus, b = 6.
The mass of a silverback gorilla is 185,000 grams, or about 1.85 x 10⁵grams. Thus, c = 5.
We are told that the mass of the African elephant is about 10 times the mass of the silverback gorilla, where m is an integer.
Let's write this as an equation:
6 x 10ⁿ = 10(1.85 x 10⁵)
Simplifying this equation, we get:
6 x 10ⁿ = 1.85 x 10⁶
10ⁿ = 3.08 x 10⁵
Taking the logarithm (base 10) of both sides, we get:
m = log(3.08 x 10)
Using a calculator, we find that:
m ≈ 5.49
Since m must be an integer, we round up to the nearest integer and get:
m = 6.
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Use the information to answer the question.
A company had a profit of -$4,758 in January and a profit of $3,642 in February. The company's profits for the months of March through May
were the same in each of these months. By the end of May, the company's total profits for the year were -$1,275.
What were the company's profits each month from March through May? Enter the answer in the box.
The company's profits for each month from March through May were $658.33.
How to find the company profits for the months of March through May?
To solve the problem, we need to use the information given and set up an equation. Let's call the profits for the months of March through May "P" (since they are the same for each month).
The company's total profits for the year can be calculated by adding up the profits for each month:
January profit + February profit + March profit + April profit + May profit = total profit
Plugging in the numbers we know:
-$4,758 + $3,642 + 3P + 3P + 3P = -$1,275
Simplifying the equation:
9P = $5,925
Dividing both sides by 9:
P = $658.33
So the company's profits for each month from March through May were $658.33.
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A bowl contains four balls numbered 1 through 4. You select the four balls successively,
without replacement, until none remain in the bowl.
a) What is the probability that the two even numbered balls are selected before the odd
numbered balls?
b) What is the probability that the balls are not selected in size order (i. E. 1, 2, 3, 4 or 4, 3, 2,1) ?
a) The probability of selecting two even numbered balls before odd numbered balls is 1/6.
b) The probability of not selecting balls in size order is 22/24 or 11/12.
a) There are 4! (24) ways to arrange the four balls. There are 2! ways (2) to arrange the even balls and 2! ways (2) to arrange the odd balls, so there are 2x2=4 favorable ways (2,4,1,3 and 4,2,1,3). The probability is 4/24, which simplifies to 1/6.
b) There are only 2 ways to select balls in size order (1,2,3,4 and 4,3,2,1). Subtracting these from the total arrangements (24-2) results in 22 non-size ordered selections. The probability is 22/24, which simplifies to 11/12.
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Larry went to Home Depot and buck 32 ft.² of treated plywood for $50 and 40 ft.² a regular plywood for $64 how much more does the treated plywood cost in the regular plywood in dollars per foot 
If Larry went to Home Depot and buck 32 ft.² The amount the treated plywood cost in the regular plywood in dollars per foot is: -$1.80 per foot
What is the cost?Treated plywood cost per square foot:
50 / 32
= $1.5625 per square foot
Regular plywood cost per square foot:
64 / 40
= -$1.60 per square foot
Difference in cost per square foot
1.5625 - 1.60
= -$0.0375 per square foot
Difference in cost per foot is:
(-$0.0375 / 0.0208)
≈ $1.80 per foot
Therefore based on the above calculation it treated plywood costs $1.80 less per foot than the regular plywood.
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1,2 2.1 2.2 2.3 20 Tilly wants to deposit money into her banking account at FNB bank: R3,50 + R1,00 per R100 is charged for an over the counter deposit. When an own ATM is used for deposits, R1,00 per R100 (First 2 free per month) is charged and for ATM withdrawals at own bank R6,50 +R1,00 per R100 is charged Determine the service fee which Tilly will incur, when: R700 is deposited over the counter? R700 was deposited as follows: R100 on Monday, R200 on Tuesday, and (3) R400 on Saturday during the first week of the month. R500 is withdrawn at an ATM? (2) (2) [14]
Service fee for the over the counter deposit of R700: R10.50
Service fee for the ATM withdrawal of R500: R11.50
To determine the service fee that Tilly will incur for the given transactions,
Over the Counter Deposit of R700:
Tilly is depositing R700, the fee will be:
Fee = R3.50 + (R1.00 per R100) x (R700 / R100)
= R3.50 + R7.00
= R10.50
Therefore, the service fee for the over the counter deposit of R700 is R10.50.
ATM Withdrawal of R500:
The fee for an ATM withdrawal at own bank is calculated as R6.50 + R1.00 per R100.
Since Tilly is withdrawing R500, the fee will be:
Fee = R6.50 + (R1.00 per R100) x (R500 / R100)
= R6.50 + R5.00
= R11.50
Therefore, the service fee for the ATM withdrawal of R500 is R11.50.
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Evaluate (8 + 2)^3 - 6
Step-by-step explanation:
First you need to do of bracket
(8+2)
=10
Second you need to do of exponential sign ^
10^3=1000(note this sign is also called cube)
Now,
1000-6
=994
Answer:
994
Step-by-step explanation:
We need to use order of operations to solve this problem.
( 8 + 2 ) ^ 3 - 6
According to PEMDAS, Parenthesis must be resolved first.
So we get:
10 ^ 3 - 6
Secondly, PEMDAS states that Exponents go next
1000 - 6
Finally, we only have one operation left, subtraction, so we can go ahead and do that.
994 is your answer.
I put a lot of thought and effort into my answers, so a brainliest would be much appreciated!
PLEASE HELP ME WITH THIS MATH PROBLEM!!! WILL GIVE BRAINLIEST!!! 20 POINTS!!!
The average price of a gallon of milk in the following years, using the exponential growth function, are:
a) 2018 = $2.90
2021 = $3.55
b) Based on the exponential growth function, the cost of milk is inflating at 7% per year.
c) Based on the percentage of inflation, the predicted price of a gallon of milk in 2025 is $4.66.
What is an exponential growth function?An exponential growth function is a mathematical equation that describes the relationship between two variables (dependent and independent).
Under the function, there is a constant ratio of growth with the number of years between the initial value and the desired value as the exponent.
The given function for the price of average gallon of milk from 2008 to 2021 is 3.55 = 2.90 (1 + x)³.
Average price of milk in 2018 = $2.90
Average price of milk in 2021 = $3.55
Change in the average price of milk = $0.65 ($3.55 - $2.90)
The percentage change from 2018 to 2021 = 22.41% ($0.65 ÷ $2.90 x 100)
The cost of milk is inflating annually at (1 + x)^3
x = 7%
Cost of milk in 2025 = y
Number of years from 2018 to 2025 = 7 years
y = 2.90 (1 + 0.07)⁷
y = 2.90 (1.07)⁷
y = $4.66
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A speaker is in the shape of a cube with an edge length of 3. 5 inches. The speaker is sold in a package in the shape of a square prism with a base area of 16 square inches and a height of 4. 25 inches. How much empty space, in cubic inches, remains in the package after the speaker is placed in the package?
The volume of the cube-shaped speaker is 42.875 cubic inches. The volume of the package is 68 cubic inches. Subtracting the volume of the speaker from the volume of the package gives the empty space remaining, which is 25.125 cubic inches.
The volume of the cube-shaped speaker is given by V₁ = (edge length)³ = 3.5³ = 42.875 cubic inches.
The volume of the square prism-shaped package is given by V₂ = (base area) x (height) = 16 x 4.25 = 68 cubic inches.
Therefore, the empty space remaining in the package after the speaker is placed in it is V₂ - V₁ = 68 - 42.875 = 25.125 cubic inches.
So, the empty space remaining in the package is 25.125 cubic inches.
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Find all solutions of the equation in radians.
sin(2x)sin(x)+cos(x)=0
Answer:
[tex]x&=\dfrac{1}{2}\pi +2\pi n,\;\; \dfrac{3}{2}\pi + 2 \pi n[/tex]
Step-by-step explanation:
Given equation:
[tex]\sin(2x)\sin(x)+\cos(x)=0[/tex]
Rewrite sin(2x) using the trigonometric identity sin(2x) = 2sin(x)cos(x):
[tex]\implies 2\sin(x)\cos(x)\sin(x)+\cos(x)=0[/tex]
[tex]\implies 2\sin^2(x)\cos(x)+\cos(x)=0[/tex]
Factor out cos(x):
[tex]\implies \cos(x)\left[2\sin^2(x)+1\right]=0[/tex]
Applying the zero-product property:
[tex]\textsf{Equation 1:}\quad\cos(x)=0[/tex]
[tex]\textsf{Equation 2:}\quad2\sin^2(x)+1=0[/tex]
Solve each part separately.
[tex]\underline{\sf Equation \; 1}[/tex]
[tex]\begin{aligned}\cos(x)&=0\\x&=\arccos(0)\\x&=\dfrac{1}{2}\pi +2\pi n,\;\; \dfrac{3}{2}\pi + 2 \pi n\end{aligned}[/tex]
[tex]\underline{\sf Equation \; 2}[/tex]
[tex]\begin{aligned}2\sin^2(x)+1&=0\\\sin^2(x)&=-\dfrac{1}{2}\;\;\;\;\;\;\leftarrow\;\textsf{No solution}\end{aligned}[/tex]
Therefore, the solutions of the equation in radians are:
[tex]\boxed{x&=\dfrac{1}{2}\pi +2\pi n,\;\; \dfrac{3}{2}\pi + 2 \pi n}[/tex]
For the pair of similar figures, use the given areas to find the scale factor from the figure on the left to the figure on the right. Write your answer as a fraction, if necessary.
scale factor =
Find the value of x to the nearest tenth. X= m
The value of x in the similar figures is 2.33
When two figures have the same shape but their sizes are different, then such figures are called similar figures
The two polygons are similar
We have to find the value of x
Let us form a proportional equation
4590/21=510/x
4590x=21×510
4590x=10710
Divide both sides by 4540
x=10710/4590
x=2.33
Hence, the value of x in the similar figures is 2.33
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Jack is making $9.50 per hour at his job re-stocking grocery shelves. His supervisor is impressed with his good work and gives him a 10% raise. How much per hour is Jack making now?
PLS HELP I WILL MARK YOU BRAINLIEST
Answer:
Jack now makes $10.45
Step-by-step explanation:
First, we need to find 10% of 9.50. To do that we must divide 9.50 by 10 or simply move the decimal one place to the left to get 0.95. 0.95 is 10% of 9.50. Now we must ADD 0.95 to 9.50 because Jack is getting a 10% raise. 0.95 + 9.50 = 10.45. Therefore Jack makes $10.45 per hour at his job.
Meena is going to see a movie and is taking her 2 kids. each movie ticket costs $14 and there are an assortment of snacks available to purchase for $5 each. how much total money would meena have to pay for her family if she were to buy 3 snacks for everybody to share? how much would meena have to pay if she bought xx snacks for everybody to share?
Meena would have to pay a total of $57 for her family if she were to buy 3 snacks for everybody to share.
Meena would have to pay a total of $42 + 5xx for her family if she bought xx snacks for everybody to share.
To calculate the total money Meena would have to pay for her family, including movie tickets and snacks, we'll first look at the scenario with 3 snacks to share.
1. Calculate the cost of movie tickets: Meena + 2 kids = 3 tickets at $14 each.
3 tickets * $14 = $42
2. Calculate the cost of 3 snacks at $5 each.
3 snacks * $5 = $15
3. Add the cost of movie tickets and snacks.
$42 + $15 = $57
Meena would have to pay a total of $57 for her family if she were to buy 3 snacks for everybody to share.
For the scenario where she buys xx snacks:
1. Calculate the cost of movie tickets (same as before):
3 tickets * $14 = $42
2. Calculate the cost of xx snacks at $5 each.
xx snacks * $5 = 5xx
3. Add the cost of movie tickets and snacks.
$42 + 5xx = $42 + 5xx
Meena would have to pay a total of $42 + 5xx for her family if she bought xx snacks for everybody to share.
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stys
ACA
2. A square with one side length represented by an
expression is shown below.
6(3x + 8) + 32 + 12x
Use the properties of operations to write three
different equivalent expressions to represent the
lengths of the other three sides of the square. One
of your expressions should contain only two terms.
We want to use properties to write expressions for the length of the other sides of the square.
Remember that the length of all the sides in a square is the same, so we only need to rewrite the above expression in two different ways.
First, we can use the distribute property in the first term:
[tex]\sf 6\times(3x + 8) + 32 + 12\times x[/tex]
[tex]\sf = 6\times3x + 6\times8 + 32 +12\times x[/tex]
[tex]= \sf 18\times x + 48 + 32 + 12\times x[/tex]
So this can be the length of one of the sides.
Now we can keep simplifying the above equation:
[tex]= \sf 18\times x + 48 + 32 + 12\times x[/tex]
To do it, we can use the distributive and associative property in the next way:
[tex]\sf 18\times x + 48 + 32 + 12\times x[/tex]
[tex]= \sf 18\times x + 12\times x + 48 + 32[/tex]
[tex]= \sf (18\times x + 12\times x) + (48 + 32)[/tex]
[tex]= \sf (18 + 12)\times x + 80[/tex]
[tex]= \sf 30\times x + 80[/tex]
This can be the expression to the other side.
Do not answer 7 and 9
Answer:
[tex]32 \div 4 = 8[/tex]
Answer: 32 divided by 4 =8
Step-by-step explanation:
143°
(8x+55)° find the value of X
Answer:11
Step-by-step explanation:Vertically opposite angles are the same.
Alternate(Z)angles are the same.Therefore we have an angle of 143 vertically opposite to the equation.143-55=88
88/8=11
At a music store in New York City, 62 people entering the store were selected at random and were asked to choose their favorite type of music. Of the 62, 14 chose rock, 16 chose country, 8 chose classical and 24 chose something other than rock, country, or classical.
a) Determine the empinical probability that the next person entering the store favors rock music.
b) Determine the empirical probability that the next person entering the store favors country music.
c) Determine the empirical probability that the next person entering the store favors something other than rock, country, or classical music
(a) The empirical probability that the next person favors rock music is approximately 0.226.
b) The empirical probability that the next person favors country music is approximately 0.258.
c) The empirical probability that the next person favors something other than rock, country, or classical music is approximately 0.387.
a) To determine the empirical probability that the next person entering the store favors rock music, you need to divide the number of people who chose rock (14) by the total number of people surveyed (62).
Empirical Probability (Rock) = Number of Rock Fans / Total Surveyed = 14 / 62 ≈ 0.226
b) To determine the empirical probability that the next person entering the store favors country music, you need to divide the number of people who chose country (16) by the total number of people surveyed (62).
Empirical Probability (Country) = Number of Country Fans / Total Surveyed = 16 / 62 ≈ 0.258
c) To determine the empirical probability that the next person entering the store favors something other than rock, country, or classical music, you need to divide the number of people who chose something other (24) by the total number of people surveyed (62).
Empirical Probability (Other) = Number of Other Fans / Total Surveyed = 24 / 62 ≈ 0.387
In summary:
a) The empirical probability that the next person favors rock music is approximately 0.226.
b) The empirical probability that the next person favors country music is approximately 0.258.
c) The empirical probability that the next person favors something other than rock, country, or classical music is approximately 0.387.
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Find the nth Maclaurin polynomial for the fu f(x) = tan x, n = 3 3 = P(x) I
The 3rd Maclaurin polynomial for f(x) = tan x is P(x) = x + (1/3)x^3.
Recall that the Maclaurin series expansion for tan x is given by:
tan x = x + (1/3)x^3 + (2/15)x^5 + ...
To find the 3rd Maclaurin polynomial, we only need to take the first three terms of the series expansion, since n = 3.
Thus, the 3rd Maclaurin polynomial for f(x) = tan x is given by:
P(x) = x + (1/3)x^3.
Note that the first two terms of the polynomial, x and (1/3)x^3, correspond to the first two terms of the Maclaurin series expansion for tan x.
The third term of the polynomial would correspond to the next term in the series expansion, which is (2/15)x^5. However, since we are only finding the 3rd Maclaurin polynomial, we do not need to include this term.
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The physics department of a college has 10 male professors, 7 female professors, 13 male teaching assistants, and 9 female teaching assistants. If a person is selected at random from the group, find the probability that the selected person is a teaching assistant or a female. (Input you answer as a decimal to four decimal places. )
The probability that the selected person is a teaching assistant or a female is 29.
What is the probability?
A probability is a numerical representation of the likelihood or chance that a specific event will take place. Both proportions ranging from 0 to 1 and percentages ranging from 0% to 100% can be used to describe probabilities.
Here, we have
Given: The physics department of a college has 10 male professors, 7 female professors, 13 male teaching assistants, and 9 female teaching assistants.
We have to find the probability that the selected person is a teaching assistant or a female.
N(A OR B) = N(A) + N(B) - N(A AND B)
N(teaching assistant) = 13 + 9 = 22
N(females) = 7 + 9 = 16
N(teaching assistant and female) = 9
N(teaching assistant OR female) = N(teaching assistant) + N(females) - N(teaching assistant AND female)
N(teaching assistant OR female) = 22 + 16 - 9
N(teaching assistant OR female) = 29
Hence, the probability that the selected person is a teaching assistant or a female is 29.
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Compound Interest:
In March 2003, Natalie invested $800 in an account that earns 4. 8% interest compounded monthly. After 5 years, she withdrew all the money and reinvested it in a new account that earns 6% interest compounded semiannually. Assuming there were no other deposits or withdrawals, how much total interest will she have earned by March 2025?
I NEED HELP, CAN SOMEONE HELP ME, PLEASE?
Natalie will have earned a total of $488.97 in interest by March 2025.
"What is compound interest formula?To solve this problem, we can use the formula for compound interest:
A = [tex]P(1 + r/n)^(nt)[/tex]
where A is the total amount, P is the principal amount, r is the annual interest rate as a decimal, n is the number of times the interest is compounded per year, and t is the time in years.
First, let's find out how much money Natalie will have in her first account after 5 years:
P = $800
r = 4.8% per year = 0.048
n = 12 (compounded monthly)
t = 5 years
A = [tex]800(1 + 0.048/12)^(12*5)[/tex]
A = $995.08
So after 5 years, Natalie will have $995.08 in her first account.
Next, let's find out how much money Natalie will have in her second account:
P = $995.08
r = 6% per year = 0.06
n = 2 (compounded semiannually)
t = 5 years
A = [tex]995.08(1 + 0.06/2)^(2*5)[/tex]
A = $1,288.97
So after reinvesting her money in the second account, Natalie will have $1,288.97 after 5 years.
Finally, let's calculate the total interest earned:
Total interest = A - P
Total interest = $1,288.97 - $800
Total interest = $488.97
Therefore, Natalie will have earned a total of $488.97 in interest by March 2025.
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A paving company is paving the rectangular student parking lot at hamilton high school. the length of the parking lot can be represented as (5 − 7) and the width as (3 + 4) . which expression represents the area of the parking lot?
The expression that represents the area of the parking lot is (5 - 7) * (3 + 4).
How to express the area of the parking lot?The length of the parking lot can be represented as (5 - 7), which simplifies to -2. The width of the parking lot can be represented as (3 + 4), which simplifies to 7. To find the area of a rectangle, we multiply the length by the width.
Therefore, the expression that represents the area of the parking lot is (-2) * 7. Multiplying -2 by 7 gives us -14. So, the area of the parking lot is -14 square units.
However, it is important to note that negative areas do not have practical meaning in this context, as areas are typically positive values. Therefore, the area of the parking lot would be considered 14 square units.
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I absolutely hate IQR so can someone help
The interquartile range (IQR) of the given data set is 4.
Interquartile range (IQR) is a measure of variability in a data set that measures the spread of the middle 50% of the data. It is calculated as the difference between the third quartile (Q3) and the first quartile (Q1).
How to fine the interquartile range (IQR)?To find the interquartile range (IQR), we first need to find the median of the data set.
The median is the middle value of the dataset when the data is arranged in order. In this case, the data set is already in order, so the median is the middle value or the average of the two middle values:
Median = (26 + 28) / 2 = 27
Now, we need to find the first quartile (Q1) and the third quartile (Q3) of the data set.
Q1 is the median of the lower half of the data set, and Q3 is the median of the upper half of the data set.
Lower half: 22, 24, 26
Upper half: 28, 30
Q1 = median of the lower half = (24 + 26) / 2 = 25
Q3 = median of the upper half = (28 + 30) / 2 = 29
Now we can find the IQR:
IQR = Q3 - Q1 = 29 - 25 = 4
Therefore, the interquartile range (IQR) of the given data set is 4.
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