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3 years ago

10

A photo 5 inches wide and 8 inches long is enlarged. How long is the new photo if it is (1)10 inches; (2)15 inches; (3)8 inches

wide?

2 answers:

sweet [91]3 years ago

5 0

1. 16 inches long, 2. 24 inches long, 3. 12.8 inches long

telo118 [61]3 years ago

4 0

Answer:

1.16 in

2.24 in

3.12.8 in

Step-by-step explanation:

We are given that

Width of photo=5 in

Length of photo=8 in

We have to find the length of new photo in each case.

1. Let x represent the width and y represents the length of photo.

When width increase then length of photo is also increases then it is direct proportion.

$\frac{x_1}{y_1}=\frac{x_2}{y_2}$

Substitute the values then we get

$\frac{5}{8}=\frac{10}{y}$

$y=\frac{10\times 8}{5}=16$

Hence, the length of photo=16 in

2.Width=x=15 in

Again using the formula

$\frac{5}{8}=\frac{15}{y}$

$y=\frac{15\times 8}{5}=24$

Hence, the length of new photo=24 in

3. Width =x=8 in

Substitute the values in the given formula

$\frac{5}{8}=\frac{8}{y}$

$y=\frac{8\times 8}{5}=\frac{64}{5}=12.8$

Hence, the length of new photo=12.8 in

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Answer:

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a.

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b.

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ŷ = 109.46

First we calculate the standard deviation

variance = SSE/(n-2)

v = 1541.4/(9-2)

v = 1541.4/7

v = 220.2

s = √v

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Then we calculate mean(x) and ∑(x - (mean(x))²

X --- Y -- Mean(x) --- ∑(x - (mean(x))²

148 -- 161 -- 43-- 1849

96 || 105|| -9 || 81

91 ||101 || -14 || 196

110 || 142 || 5 || 25

90 || 100 || -15 || 225

102 || ||120 ||-3|| 9

136 || 167 ||31 ||961

90 || 140 ||-15 ||225

82 || 98 ||-23 || 529

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tα,df = n − 2 = t0.025,7 =2.365

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c.

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ŷ = 149.77

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Interval = 149.77 ± 39.0817

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