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3 years ago

5

A group of students are planning a mural at a wall the rectangular wall has dimensions of (6x+7) by (8x+5) and they are planning

the mural to be (x+4) by(2x+5) what is the area of the remaining wall after the mural has been Painted

1 answer:

zvonat [6]3 years ago

8 0

Answer: $46x^2+73x+15$

Step-by-step explanation:

The area of a rectangle can be calculated with the formula:

$A=lw$

l: the length of the rectangle.

w: the width of the rectangle.

The area of the remaning wall after the mural has been painted, will be the difference of the area of the wall and the area of the mural.

Knowing that the dimensions of the wall are $(6x+7)$ by $(8x+5)$ , its area is:

$A_w=(6x+7)(8x+5)\\\\A_w=48x^2+30x+56x+35\\\\A_w=48x^2+86x+35$

As they are planning that the dimensions of the mural be $(x+4)$ by $(2x+5)$ , its area is:

$A_m=(x+4)(2x+5)\\\\A_m=2x^2+5x+8x+20\\\\A_m=2x^2+13x+20$

Then the area of the remaining wall after the mural has been painted is:

$A_{(remaining)}=A_w-A_m\\\\A_{(remaining)}=48x^2+86x+35-(2x^2+13x+20)\\\\A_{(remaining)}=48x^2+86x+35-2x^2-13x-20\\\\A_{(remaining)}=46x^2+73x+15$

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The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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Normal probability distribution

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In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

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$p = 0.71$

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Here $n = 16$ , we want $P(X \geq 13)$ . So

$P(X \geq 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 13) = C_{16,13}.(0.71)^{13}.(0.29)^{3} = 0.1591$

$P(X = 14) = C_{16,14}.(0.71)^{14}.(0.29)^{2} = 0.0835$

$P(X = 15) = C_{16,15}.(0.71)^{15}.(0.29)^{1} = 0.0273$

$P(X = 16) = C_{16,16}.(0.71)^{16}.(0.29)^{0} = 0.0042$

$P(X \geq 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16) = 0.1591 + 0.0835 + 0.0273 + 0.0042 = 0.2741$

0.2741 = 27.41% probability that at least 13 believe global warming is occurring

(b) For a sample of 160 Americans, what is the probability that at least 110 believe global warming is occurring?

Now $n = 160$ . So

$\mu = E(X) = np = 160*0.71 = 113.6$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{160*0.71*0.29} = 5.74$

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$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{109.5 - 113.6}{5.74}$

$Z = -0.71$

$Z = -0.71$ has a pvalue of 0.2389

1 - 0.2389 = 0.7611

0.7611 = 76.11% probability that at least 110 believe global warming is occurring

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