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2 years ago

HELP PLEASE LIKE NOW!!’ I’ll mark you brainliest !!!!!!!!

Mathematics

2 answers:

gizmo_the_mogwai [7]2 years ago

6 0

I believe that the answer is c.) When it was purchased (year 0), The coin was so worth four dollars.

iren [92.7K]2 years ago

4 0

Its c your welcome!!!!!!

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Solve for m. y=7m+6 solve for b. A=6b

ryzh [129]

y=7m+6

subtract 6 from each side

y-6 = 7m

divide by 7

(y-6)/7 =m

A = 6b

divide by 6

A/6 = b

8 0

3 years ago

Read 2 more answers

Brittany has a gift box in the shape of a cube. Each side of the box measures 15 cm. what is the volume of the gift box?

miss Akunina [59]

The formula for the volume of a cube is:

v=s³

Where v=volume and s=the length of a side

Since we know that the length of a side of the gift box is 15cm, we can solve for the volume.

v=15³=3375cm³

Answer: V=3375cm³

7 0

2 years ago

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Need Help, Trying To Finish .

hammer [34]

Answer:

8

Steps:

Plug in 2 for x in h(x)

h(2) = -(2)^2 + 6 * 2

h(2) = -4 + 12

h(2) = 8

3 0

3 years ago

An certain brand of upright freezer is available in three different rated capacities: 16 ft3, 18 ft3, and 20 ft3. Let X = the ra

ruslelena [56]

Answer:

$E(X)=16\cdot 0.3+18\cdot 0.1+20\cdot 0.6=18.6$

$E(X^2)=16^2\cdot 0.3+18^2\cdot 0.1+20^2\cdot 0.6=349.2$

$V(X) = E(X^2)-[E(X)]^2=349.2-(18.6)^2=3.24$

The expected price paid by the next customer to buy a freezer is $466

Step-by-step explanation:

From the information given we know the probability mass function (pmf) of random variable X.

$\left|\begin{array}{c|ccc}x&16&18&20\\p(x)&0.3&0.1&0.6\end{array}\right|$

Point a:

The Expected value or the mean value of X with set of possible values D, denoted by E(X) or μ is

$E(X) = $\sum_{x\in D} x\cdot p(x)$

Therefore

$E(X)=16\cdot 0.3+18\cdot 0.1+20\cdot 0.6=18.6$

If the random variable X has a set of possible values D and a probability mass function, then the expected value of any function h(X), denoted by E[h(X)] is computed by

$E[h(X)] = $\sum_{D} h(x)\cdot p(x)$

So $h(X) = X^2$ and

$E[h(X)] = $\sum_{D} h(x)\cdot p(x)\\E[X^2]=$\sum_{D}x^2\cdot p(x)\\ E(X^2)=16^2\cdot 0.3+18^2\cdot 0.1+20^2\cdot 0.6\\E(X^2)=349.2$

The variance of X, denoted by V(X), is

$V(X) = $\sum_{D}E[(X-\mu)^2]=E(X^2)-[E(X)]^2$

Therefore

$V(X) = E(X^2)-[E(X)]^2\\V(X)=349.2-(18.6)^2\\V(X)=3.24$

Point b:

We know that the price of a freezer having capacity X is 60X − 650, to find the expected price paid by the next customer to buy a freezer you need to:

From the rules of expected value this proposition is true:

$E(aX+b)=a\cdot E(x)+b$