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AfilCa [17]
2 years ago
3

The temperature in Coast City is 3 degrees below zero on the Fahrenheit scale . The temperature in Dodge City is 5F colder than

a temperature that is twice as cold as Coast City’s temperature. What is the temperature in Dodge City?
a -11F
b-1F
c 1F
d11F
Mathematics
2 answers:
Mumz [18]2 years ago
8 0
When answering word questions, change the words into numbers/math symbols.

If the temperature in Coast City is 3 degrees below zero, then it is "-3º F".

Because the next sentence is more complex, we should break it down into two parts since it is really asking two questions: "What is the temperature that is twice as cold as Coast City's temperature?" and "What is 5ºF colder than twice Coast City's temperature?"

To answer the first question, we change the phrase "twice as cold" to numerical symbols. When something is twice as big/cold/old/small/etc of something, it means you must multiply the original number by 2. Since the temperature of Coast City is -3ºF, then twice that temperature is -3 x 2 = -6. So twice the temperature of Coast City is -6ºF. Now we must answer the second question. What is 5ºF colder than -6ºF? Converting this to numbers and mathematical equations, we get "-6 – 5 = ?" This answer to this is "-11". So 5ºF colder than -6ºF is -11ºF. So A, -11ºF is the answer.
frozen [14]2 years ago
5 0

Answer:

A is correct answer

Step-by-step explanation:

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Answer:

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And that represent the 1.4%.

b) And the 95% confidence interval would be given (0.00370;0.0243).

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Step-by-step explanation:

Data given and notation  

n=500 represent the random sample taken    

X=7 represent the households with three or more large-screen TVs

\hat p=\frac{7}{500}=0.014 estimated proportion of households with three or more large-screen TVs

\alpha=0.05 represent the significance level (no given, but is assumed)    

z would represent the statistic (variable of interest)    

p= population proportion of households with three or more large-screen TVs

Part a

The percentage of households in the town with three or more largescreen TVs is estimated as :

The best estimation for the population proportion is :

\hat p=\frac{7}{500}=0.014

And that represent the 1.4%.

Part b

Yes is possible. We hav that np>10 and n(1-p)>10 so we have the assumption of normality to find the interval.

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=1.96

And replacing into the confidence interval formula we got:

0.014 - 1.96 \sqrt{\frac{0.014(1-0.014)}{500}}=0.00370

0.014 + 1.96 \sqrt{\frac{0.014(1-0.014)}{500}}=0.0243

And the 95% confidence interval would be given (0.00370;0.0243).

And the % would be between 0.37% and 2.43%.

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