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3 years ago

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Andy brought a computer. The length of the computer is 32 inches and the width is 14 inches. Find the area of the computer.

2 answers:

Lady_Fox [76]3 years ago

8 0

Ah, no problem! All area is the width times the length.
All you got to do is 32 x 14!

32 x 14 = 448

Keep in mind Width x Length & you'll be fine!

Semenov [28]3 years ago

7 0

The answer to your problem is 448 sq.inches

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Solid fats are more likely to raise blood cholesterol levels than liquid fats. Suppose a nutritionist analyzed the percentage of

Andrews [41]

Answer:

$t = 31.29$

Step-by-step explanation:

Given

$\begin{array}{ccccccc}{Stick} & {25.8} & {26.9} & {26.2} & {25.3} & {26.7}& {26.1} \ \\ {Liquid} & {16.9} & {17.4} & {16.8} & {16.2} & {17.3}& {16.8} \ \end{array}$

Required

Determine the test statistic

Let the dataset of stick be A and Liquid be B.

We start by calculating the mean of each dataset;

$\bar x =\frac{\sum x}{n}$

n, in both datasets in 6

For A

$\bar x_A =\frac{25.8+26.9+26.2+25.3+26.7+26.1}{6}$

$\bar x_A =\frac{157}{6}$

$\bar x_A =26.17$

For B

$\bar x_B =\frac{16.9+17.4+16.8+16.2+17.3+16.8}{6}$

$\bar x_B =\frac{101.4}{6}$

$\bar x_B =16.9$

Next, calculate the sample standard deviation

This is calculated using:

$s = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}$

For A

$s_A = \sqrt{\frac{\sum(x - \bar x_A)^2}{n-1}}$

$s_A = \sqrt{\frac{(25.8-26.17)^2+(26.9-26.17)^2+(26.2-26.17)^2+(25.3-26.17)^2+(26.7-26.17)^2+(26.1-26.17)^2}{6-1}}$

$s_A = \sqrt{\frac{1.7134}{5}}$

$s_A = \sqrt{0.34268}$

$s_A = 0.5854$

For B

$s_B = \sqrt{\frac{\sum(x - \bar x_B)^2}{n-1}}$

$s_B = \sqrt{\frac{(16.9 - 16.9)^2+(17.4- 16.9)^2+(16.8- 16.9)^2+(16.2- 16.9)^2+(17.3- 16.9)^2+(16.8- 16.9)^2}{6-1}}$

$s_B = \sqrt{\frac{0.92}{5}}$

$s_B = \sqrt{0.184}$

$s_B = 0.4290$

Calculate the pooled variance

$S_p^2 = \frac{(n_A - 1)*s_A^2 + (n_B - 1)*s_B^2}{(n_A+n_B-2)}$

$S_p^2 = \frac{(6 - 1)*0.5854^2 + (6 - 1)*0.4290^2}{(6+6-2)}$

$S_p^2 = \frac{2.6336708}{10}$

$S_p^2 = 0.2634$

Lastly, calculate the test statistic using:

$t = \frac{(\bar x_A - \bar x_B) - (\mu_A - \mu_B)}{\sqrt{S_p^2/n_A +S_p^2/n_B}}$

We set

$\mu_A = \mu_B$

So, we have:

$t = \frac{(\bar x_A - \bar x_B) - (\mu_A - \mu_A)}{\sqrt{S_p^2/n_A +S_p^2/n_B}}$

$t = \frac{(\bar x_A - \bar x_B) }{\sqrt{S_p^2/n_A +S_p^2/n_B}}$

The equation becomes

$t = \frac{(26.17 - 16.9) }{\sqrt{0.2634/6 +0.2634/6}}$

$t = \frac{9.27}{\sqrt{0.0878}}$

$t = \frac{9.27}{0.2963}$

$t = 31.29$

<em>The test statistic is 31.29</em>

3 0

3 years ago

Canadians who visit the United States often buy liquor and cigarettes, which are much cheaper in the United States. However, the

fenix001 [56]

Answer:

(a): Marginal pmf of x

$P(0) = 0.72$

$P(1) = 0.28$

(b): Marginal pmf of y

$P(0) = 0.81$

$P(1) = 0.19$

(c): Mean and Variance of x

$E(x) = 0.28$

$Var(x) = 0.2016$

(d): Mean and Variance of y

$E(y) = 0.19$

$Var(y) = 0.1539$

(e): The covariance and the coefficient of correlation

$Cov(x,y) = 0.0468$

$r \approx 0.2657$

Step-by-step explanation:

Given

<em>x = bottles</em>

<em>y = carton</em>

<em>See attachment for complete question</em>

<em />

Solving (a): Marginal pmf of x

This is calculated as:

$P(x) = \sum\limits^{}_y\ P(x,y)$

So:

$P(0) = P(0,0) + P(0,1)$

$P(0) = 0.63 + 0.09$

$P(0) = 0.72$

$P(1) = P(1,0) + P(1,1)$

$P(1) = 0.18 + 0.10$

$P(1) = 0.28$

Solving (b): Marginal pmf of y

This is calculated as:

$P(y) = \sum\limits^{}_x\ P(x,y)$

So:

$P(0) = P(0,0) + P(1,0)$

$P(0) = 0.63 + 0.18$

$P(0) = 0.81$

$P(1) = P(0,1) + P(1,1)$

$P(1) = 0.09 + 0.10$

$P(1) = 0.19$

Solving (c): Mean and Variance of x

Mean is calculated as:

$E(x) = \sum( x * P(x))$

So, we have:

$E(x) = 0 * P(0) + 1 * P(1)$

$E(x) = 0 * 0.72 + 1 * 0.28$

$E(x) = 0 + 0.28$

$E(x) = 0.28$

Variance is calculated as:

$Var(x) = E(x^2) - (E(x))^2$

Calculate $E(x^2)$

$E(x^2) = \sum( x^2 * P(x))$

$E(x^2) = 0^2 * 0.72 + 1^2 * 0.28$

$E(x^2) = 0 + 0.28$

$E(x^2) = 0.28$

So:

$Var(x) = E(x^2) - (E(x))^2$

$Var(x) = 0.28 - 0.28^2$

$Var(x) = 0.28 - 0.0784$

$Var(x) = 0.2016$

Solving (d): Mean and Variance of y

Mean is calculated as:

$E(y) = \sum(y * P(y))$

So, we have:

$E(y) = 0 * P(0) + 1 * P(1)$

$E(y) = 0 * 0.81 + 1 * 0.19$

$E(y) = 0+0.19$

$E(y) = 0.19$

Variance is calculated as:

$Var(y) = E(y^2) - (E(y))^2$

Calculate $E(y^2)$

$E(y^2) = \sum(y^2 * P(y))$

$E(y^2) = 0^2 * 0.81 + 1^2 * 0.19$

$E(y^2) = 0 + 0.19$

$E(y^2) = 0.19$

So:

$Var(y) = E(y^2) - (E(y))^2$

$Var(y) = 0.19 - 0.19^2$

$Var(y) = 0.19 - 0.0361$

$Var(y) = 0.1539$

Solving (e): The covariance and the coefficient of correlation

Covariance is calculated as:

$COV(x,y) = E(xy) - E(x) * E(y)$

Calculate E(xy)

$E(xy) = \sum (xy * P(xy))$

This gives:

$E(xy) = x_0y_0 * P(0,0) + x_1y_0 * P(1,0) +x_0y_1 * P(0,1) + x_1y_1 * P(1,1)$

$E(xy) = 0*0 * 0.63 + 1*0 * 0.18 +0*1 * 0.09 + 1*1 * 0.1$

$E(xy) = 0+0+0 + 0.1$

$E(xy) = 0.1$

So:

$COV(x,y) = E(xy) - E(x) * E(y)$

$Cov(x,y) = 0.1 - 0.28 * 0.19$

$Cov(x,y) = 0.1 - 0.0532$

$Cov(x,y) = 0.0468$

The coefficient of correlation is then calculated as:

$r = \frac{Cov(x,y)}{\sqrt{Var(x) * Var(y)}}$

$r = \frac{0.0468}{\sqrt{0.2016 * 0.1539}}$

$r = \frac{0.0468}{\sqrt{0.03102624}}$

$r = \frac{0.0468}{0.17614266944}$

$r = 0.26569371378$

$r \approx 0.2657$ --- approximated

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rosijanka [135]

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Sergeeva-Olga [200]

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