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3 years ago

7

The low temperatures for Anchorage, Alaska last week were: -17°F, 8°F, -9°F, -11°F, 12°F, -13°F, and 2°F. What was the average l

ow temperature for the week?

1 answer:

Mademuasel [1]3 years ago

4 0

Answer:

-4 degrees Fahrenheit

Step-by-step explanation:

(-17 +8 + -9 + -11 +12 + -13 +2) / 7 = -4

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An internet service provider changes an initial fee of $65 and $18 per hour of labor for office internet installation. Which equ

OlgaM077 [116]

<h2>Answer:</h2><h2>The total cost (C) charged by the internet service provider = $65 + $18(n)</h2>

Step-by-step explanation:

Initial fee charged by the servic eprovider = $65

Cost per hour of labor for office internet installation = $18

The cost per hour be $18

Cost per two hours = 18 * 2 = $36

Cost per n number of hours = $18 (h)

where h represents the number of hours

The total cost (C) charged by the internet service provider = $65 + $18(n)

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3 years ago

A resident of Bayport claims to the City Council that the proportion of Westside residents

Kazeer [188]

Answer:

The test statistics is $z = -1.56$

The p-value is $p-value = 0.05938$

Step-by-step explanation:

From the question we are told

The West side sample size is $n_1 = 578$

The number of residents on the West side with income below poverty level is $k = 76$

The East side sample size $n_2=688$

The number of residents on the East side with income below poverty level is $u = 112$

The null hypothesis is $H_o : p_1 = p_2$

The alternative hypothesis is $H_a : p_1 < p_2$

Generally the sample proportion of West side is

$\^{p} _1 = \frac{k}{n_1}$

=> $\^{p} _1 = \frac{76}{578}$

=> $\^{p} _1 = 0.1315$

Generally the sample proportion of West side is

$\^{p} _2 = \frac{u}{n_2}$

=> $\^{p} _2 = \frac{112}{688}$

=> $\^{p} _2 = 0.1628$

Generally the pooled sample proportion is mathematically represented as

$p = \frac{k + u}{ n_1 + n_2 }$

=> $p = \frac{76 + 112}{ 578 + 688 }$

=> $p =0.1485$

Generally the test statistics is mathematically represented as

$z = \frac{\^ {p}_1 - \^{p}_2}{\sqrt{p(1- p) [\frac{1}{n_1 } + \frac{1}{n_2} ]} }$

=> $z = \frac{ 0.1315 - 0.1628 }{\sqrt{0.1485(1-0.1485) [\frac{1}{578} + \frac{1}{688} ]} }$

=> $z = -1.56$

Generally the p-value is mathematically represented as

$p-value = P(z < -1.56 )$

From z-table

$P(z < -1.56 ) = 0.05938$

So

$p-value = 0.05938$

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3 years ago

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3 years ago

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3 years ago

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