Samantha's age is 3 years more than half her uncle's age. If her uncle's age is z years, which of the following expressions best shows Samantha's age?

Answer:

The answer is 68

Step-by-step explanation:

Plug -5 into each x value then solve.

Answer:

R = 4cm

Step-by-step explanation:

V = π(R)²h = 480π

Divide by π

(R)²h = 480

Substitute h with 30

30*(R)² = 480

Divide by 30

(R)² = 16

Square root

R = 4cm

We do not accept R= -4 because it's a length, so it must be positive.

Using the Empirical Rule, it is found that the desired probabilities are given as follows.

a) P(x > 158)  = 0.16.

b) P(149 < x < 152) = 0.135.

It states that, for a normally distributed random variable:

Additionally, considering the symmetry of the normal distribution, 50% of the measures are below the mean and 50% are above.

Item a:

158 is one standard deviation above the mean, hence the probability is given by, considering that 32% of the measures are more than 1 standard deviation from the mean:

P(x > 158) = 0.5 x 0.32 = 0.16.

Item b:

Between one and two standard deviations below the mean, hence:

P(149 < x < 152) = 0.5 x (0.95 - 0.68) = 0.5 x 0.27 = 0.135.

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Answer:

24

Step-by-step explanation:

Lets start by working out 80% of 30

80% x 30=24

so the student answered 24 questions correctly

Answer:

12

Step-by-step explanation:

you add 92 with the 4 and get 96 then you divide the 8x with the 96 and get 12

Hope it helps <333

x = 12

Our goal is to find the value of x .

To find the value of x we need to get x by itself .

The first step is to move all numbers to the right . Luckily , there's only one number here : 4.

So we  add 4 on both sides :

8x=96

Now divide by 8 on both sides :

x = 12

[tex]\footnotesize\text{hope\:helpful~}[/tex]

Answer:

d

Step-by-step explanation:

hahaha

[tex]\qquad\qquad\huge\underline{{\sf Answer}}[/tex]

[tex] \textbf{Let's calculate its volume :} [/tex]

[tex] \textbf{Volume if the prism is 52.5 ft³} [/tex]

Answer:

I'm not sure if you are allowed to use improper fractions but that would be the only way to have an answer be greater than both of its factors if one of the factors has to be a fraction.

I would do something simple like:

[tex]\frac{2}{1} * 2 = 4[/tex]

Step-by-step explanation:

Since both factors are greater than 1 the product will be greater than both the factors.

Answer:

Equation is below. See model attached.

6/4 x 3 = 18/4

Answer:

$31,500

Step-by-step explanation:

10% = 0.1

100% = 1

1 - 0.1 = 0.9

35,000 x 0.9 = 31,500

Answer:

x=3,-2

Step-by-step explanation:

take the first part ;(x+2)=0. x=-2 x_3=0 so x=3

The worth of the boat of Maya which goes for 15% deprication this year at the end of the first year of purchase, with the considered purchase price, is $15,589

Suppose the value of which a thing is expressed in percentage is "a'

Suppose the percent that considered thing is of "a" is b%

Then since percent shows per 100 (since cent means 100), thus we will first divide the whole part in 100 parts and then we multiply it with b so that we collect b items per 100 items(that is exactly what b per cent means).

Thus, that thing in number is

[tex]\dfrac{a}{100} \times b[/tex]

Deprecated price = Initial price- amount of deprication

We have:

Initial price = $18,340

And the amount of deprication = 15% of $18,340

= [tex]\dfrac{18340}{100} \times 15 = 2751 \: \rm dollars[/tex]

(as deprication on first year was by 15% of the original price, and we want to know the deprication for the first year only)

Thus, we get:

Depricated price of Maya's boat = $18,340 - $2,751 = $15,589

Thus, the worth of the boat of Maya which goes for 15% deprication this year at the end of the first year of purchase, with the considered purchase price, is $15,589

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Answer: 31.4

Step-by-step explanation:

Circumference = πd  or 2πr

π = 3.14

Since the diameter is given we will use Circumference = πd

So

3.14 * 10 = 31.4

Answer:

C ≈ 31.4

Step-by-step explanation:

The circumference of a circle can be calculated using either of the following formulas: C=d or C=2r.

The circumference of a circle is the distance around the outside of the circle. It is like the perimeter of other shapes like squares. You can think of it as the line that defines the shape. For shapes made of straight edges this line is called the perimeter but for circles this defining line is called the circumference.

The radius (r) and the diameter (d) are two more crucial distances on a circle (d). Every circle has three distinguishing features: a radius, a diameter, and a circumference. The circumference may be calculated using the radius or diameter and pi. The diameter of a circle is the distance between one side and the other at its widest points. The circumference of a circle will always pass through its center. This distance is divided in half by the radius. The radius may alternatively be thought of as the distance between the circle's center and its edge.

You can calculate the circumference of a circle if you know its diameter or radius. To begin, keep in mind that pi is an irrational number represented by the symbol. 3.14 is a close approximation.

The formula for calculating a circle's circumference is:

The circumference of a circle is equal to its diameter multiplied by its circumference.

C = d is a common notation for this. This indicates that the circle's circumference is three "and a half" times its diameter. 

Given radius (R) = 5

⇒ Diameter =  2R =  10

⇒ Circumference =  2πR

=  10π

=  31.415926535898

≈     31. 4

Answer:

B. 1 :)

Step-by-step explanation:

Every time the x value gets bigger, the y laue does too. Positive numbers are larger than negative ones so every time we get closer to 0, our values get bigger :)

Have an amazing day!!

Please rate and mark as brainliest!!

Answer:

There were 29 adult admissions; and 72 child's.

Answer:

The perimeter of an equilateral traingle with height 42cm is 126cm.

Step-by-step explanation:

Given that the height or edge of an equilateral traingle is 42cm. With this information, we are asked to find the perimeter of an equilateral traingle.

The perimeter of equilateral traingle is defined as the three times of the length of edge. Mathematically;

→ Perimeter = 3a

By substituting the given values in the formula, we get the following results:

→ Perimeter = 3(42)

→ Perimeter = 126

Hence, the perimeter of an equilateral traingle with height 42cm is 126cm.

Additional information:

Answer:

a = -4, b = 5, c = -2

(the third option)

The values are a = -4, b = 5  and  c = -2 in the quadratic equation.

A quadratic equation is the second-order degree algebraic expression in a variable. the standard form of this expression is  ax² + bx + c = 0 where a. b are coefficients and x is the variable and c is a constant.

We are given to determine the values of a, b and c in the following quadratic equation :

0 = 5x - 4x² - 2

We know that a general quadratic equation is of the following form :

ax² + bx + c = 0

where a is the coefficient of x², b is the coefficient of x and c is the constant term.

Comparing the general equation with equation (i), we have that;

coefficient of x², a = -4,

coefficient of x, b = 5

and the constant term, c = -2.

Thus, the values are; a = -4, b = 5  and  c = -2.

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[tex]\text{Given that,} ~f(x) = 2x-4~~ \text{and}~~ g(x) = 2x+8\\\\ g(f(3))\\\\=g(2\cdot 3 -4)\\\\=g(2)\\\\=2(2)+8\\\\=4+8\\\\=12[/tex]

Answer:

7

Step-by-step explanation:

I took the k12 quiz

Answer:

7

Step-by-step explanation:

I  took the quiz on K-12 the quiz name is 6.04 Quiz: Mean Absolute Deviation (MAD)

Answer:

The answer is six I think

7:13

hope it helps...!!

Answer:

51

Step-by-step explanation:

you plug in 3 for x.

5(3)^2 - 3 + 9

5(9) - 3 + 9

45-3+9

51

Answer: 5x^2+6

Step-by-step explanation: sorry if wrong

Answer:

Step-by-step explanation:

In this question we have provided an equation that is 9 ( 3x - 16 ) + 15 = 6x - 24 . And we are asked to write the steps to solve the equation with explanation and to find the value of X .

Solution : -

[tex] \longmapsto \quad \: 9(3x - 16) + 15 = 6x - 24[/tex]

Step 1 : Solving parenthesis on left side using distributive property which means multiplying 9 with 3x as well as -16 :

[tex] \longmapsto \quad \:27x - \bold{144 }+ \bold{15 }= 6x - 24[/tex]

Step 2 : Solving like terms on left side that are -144 and 15 :

[tex] \longmapsto \quad \:27x -129 = 6x - 24[/tex]

Step 3 : Adding 129 on both sides :

[tex] \longmapsto \quad \:27x - \cancel{129} - \cancel{129} = 6x \bold{ - 24 } + \bold{129}[/tex]

Now on cancelling -129 with 129 on left side and solving the terms that are -24 and 129 on right side , We get :

[tex] \longmapsto \quad \:27x = 6x + 105[/tex]

Step 4 : Subtracting with 6x on both sides :

[tex] \longmapsto \quad \: \bold{27x} - \bold{6x} = \cancel{6x} +105 - \cancel{ 6x}[/tex]

On calculating further, We get :

[tex] \longmapsto \quad \:21x = 105[/tex]

Step 5 : Now we are Dividing with 21 on both sides so that we can isolate the variable that is x :

[tex] \longmapsto \quad \: \dfrac{ \cancel{21}x}{ \cancel{21}} = \dfrac{105}{ 21} [/tex]

Now , by cancelling 21 with 21 on left side , We get :

[tex] \longmapsto \quad \:x = \cancel{\dfrac{105}{21}} [/tex]

Step 6 : Now our final step is to simplify the value of x that is 105/21 . We know that 21 × 5 is equal to 105 . So :

[tex] \longmapsto \quad \: \purple{\underline{\boxed{\frak{ x = 5 }}}}[/tex]

Verifying : -

Now we are verifying our answer by substituting value of x in the given equation . So ,

Therefore, our value for x is correct that means it'll makes the equation true .

We can express the forcing function (the piecewise expression on the right side) in terms of the step function as [tex]e^{-t}(u(t) - u(t-1))[/tex] where

[tex]u(t) = \begin{cases}1&\text{for }t\ge0\\0&\text{otherwise}\end{cases}[/tex]

Let F(s) be the Laplace transform of a function f(t). Now recall the transform pair

[tex]f(t-c) u(t-c) \mapsto e^{-cs} F(s)[/tex]

This means

[tex]e^{-t} u(t) \mapsto \dfrac1{s+1}[/tex]

[tex]e^{-t} u(t-1) = \dfrac1e \times e^{-(t-1)} u(t-1) \mapsto  \dfrac{e^{-(s+1)}}{s+1}[/tex]

I assume you're familiar with the transform rule for derivatives of y(t). Now we're ready to take the transform of both sides of the ODE:

[tex]y'' + 3y' + 2y = e^{-t}(u(t) - u(t-1))[/tex]

[tex]\implies \left(s^2 Y(s) - s y(0) - y'(0)\right) + 3 \left(s Y(s) - y(0)\right) + 2 Y(s) = \dfrac{1 - e^{-(s+1)}}{s+1}[/tex]

Plug in the initial values and solve for Y(s) :

[tex]\left(s^2 Y(s) - s + 1\right) + 3 \left(s Y(s) + 1\right) + 2 Y(s) = \dfrac{1 - e^{-(s+1)}}{s+1}[/tex]

[tex](s^2 + 3s + 2) Y(s) - s + 4 = \dfrac{1 - e^{-(s+1)}}{s+1}[/tex]

[tex]Y(s) = \dfrac{1 - e^{-(s+1)} + (s-4)(s+1)}{(s+1)(s^2 + 3s + 2)}[/tex]

[tex]Y(s) = \dfrac{1 - e^{-(s+1)} + (s-4)(s+1)}{(s+1)^2 (s+2)}[/tex]

Consider the partial fraction expansion

[tex]\dfrac1{(s+1)^2(s+2)} = \dfrac a{s+1} + \dfrac b{(s+1)^2} + \dfrac c{s+2}[/tex]

Solve for the coefficients:

[tex]1 = a(s+1)(s+2) + b(s+2) + c(s+1)^2[/tex]

[tex]s = -1 \implies b = 1[/tex]

[tex]s = -2 \implies c = 1[/tex]

[tex]1 = (a+c)s^2 + \cdots \implies a+c = 0 \implies a = -1[/tex]

Hence we can expand Y(s) as

[tex]Y(s) = \dfrac1{(s+1)^2} + \dfrac1{s+2}  + \dfrac{e^{-(s+1)}}{s+1} - \dfrac{e^{-(s+1)}}{(s+1)^2} - \dfrac{e \times e^{-(s+2)}}{s+2}[/tex]

The last transform pair we need is

[tex]e^{ct} f(t) \mapsto F(s - c)[/tex]

Now, taking inverse transforms of everything yields

[tex]\dfrac1{(s+1)^2} \mapsto te^{-t}[/tex]

[tex]\dfrac1{s+2} \mapsto e^{-2t}[/tex]

[tex]\dfrac{e^{-(s+1)}}{s+1} \mapsto e^{-t} u(t-1)[/tex]

[tex]\dfrac{e^{-(s+1)}}{(s+1)^2} \mapsto e^{-t} (t-1) u(t-1)[/tex]

[tex]\dfrac{e \times e^{-(s+2)}}{s+2} \mapsto e^{-(2t-1)} u(t-1)[/tex]

and putting everything together gives the same solution as the one provided.

Answer:

68

Step-by-step explanation:

62+50=112

180-112=68

[tex]sin(2x) = 2sin(x) \\ 2sin(x)cos(x) = 2sin(x) \\ sin(x)cos(x) = sin(x) \\ sin(x)cos(x) - sin(x) = 0 \\ \sin(x) (cosx - 1) = 0 \\ \sin(x) = 0 \\ x = \pi \: or \: x = 0 \\ cos(x) = 1 \\ x = 0 \: or \: x = \pi \\ \\ s[/tex]

[tex]s = > {0 < = > \pi}[/tex]

Answer:

Domain is all real numbers

Range is (3, ∞) or y > 3

Step-by-step explanation: