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2 years ago

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you are buying a new printer and a new scanner for your computer and you cannot spend over $150. The printer you want costs $80.

Write an inequality that describes the most that you can spend on the scanner and still stay within your budget. If you can buy a scanner that costs $75 will you remain within your budget?

1 answer:

ira [324]2 years ago

6 0

It is given in the question that

You are buying a new printer and a new scanner for your computer and you cannot spend over $150. The printer you want costs $80.

Let the price of scanner be $ x .

So we have

$x+ 80\leq 150$

Subtracting 80 from both sides

$x\leq 70$

It means , the price should be not more then $70 .

Therefore in case, the price of scanner is $75, then you will not remain within your budget .

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2 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.

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

<em>Point a:</em>

The Expected value or the mean value of X with set of possible values D, denoted by <em>E(X)</em> or <em>μ </em>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 <em>E[h(X)]</em> 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

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

<em>Point b:</em>

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:

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

We have a = 60, b = -650, and <em>E(X)</em> = 18.6. Therefore

The expected price paid by the next customer is

$60\cdot E(X)-650=60\cdot 18.6-650=466$

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2 years ago

a square painting has an area of 81x^2-90x-25. A second square painting has an area of 25x^2+30x+9. What is an expression that r

galina1969 [7]

Answer:

The answer in the procedure

Step-by-step explanation:

Let

A1 ------> the area of the first square painting

A2 ----> the area of the second square painting

D -----> the difference of the areas

we have

$A1=81x^{2}-90x-25$

$A2=25x^{2}+30x+9$

case 1) The area of the second square painting is greater than the area of the first square painting

The difference of the area of the paintings is equal to subtract the area of the first square painting from the area of the second square painting

D=A2-A1

$D=(25x^{2}+30x+9)-(81x^{2}-90x-25)$

$D=(-56x^{2}+120x+34)$

case 2) The area of the first square painting is greater than the area of the second square painting

The difference of the area of the paintings is equal to subtract the area of the second square painting from the area of the first square painting

D=A1-A2

$D=(81x^{2}-90x-25)-(25x^{2}+30x+9)$

$D=(56x^{2}-120x-34)$

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2 years ago