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iris [78.8K]
2 years ago
6

In 1925, airplane engines were cooled with water. Water freezes at a temperature of 32º Fahrenheit. Average winter low temperatu

res in Alaska are 50º colder than that. What is Average winter low temperature in Alaska?
Mathematics
1 answer:
Snezhnost [94]2 years ago
5 0

Answer:

From November to March, the temperature in Alaska is between 0°F/-18°C and -30°F/-35°C, which is very cold. May and September are both often the driest and the wettest months in Alaska's summer.

I hope this helps you

:)

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Write the number 5 million 4 thousand three hindered in standard form
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Answer:

The ans is 5004300

In standard form,

5004300. Five Millon four thousand and three hundred.

(hope that helped can i plz have brainlist it wil make my day :D hehe)

Step-by-step explanation:

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A manufacturer produces light bulbs that have a mean life of at least 500 hours when the production process is working properly.
denpristay [2]

Answer:

We conclude that the population mean light bulb life is at least 500 hours at the significance level of 0.01.

Step-by-step explanation:

We are given that a manufacturer produces light bulbs that have a mean life of at least 500 hours when the production process is working properly. The population standard deviation is 50 hours and the light bulb life is normally distributed.

You select a sample of 100 light bulbs and find mean bulb life is 490 hours.

Let \mu = <u><em>population mean light bulb life.</em></u>

So, Null Hypothesis, H_0 : \mu \geq 500 hours      {means that the population mean light bulb life is at least 500 hours}

Alternate Hypothesis, H_A : \mu < 500 hours     {means that the population mean light bulb life is below 500 hours}

The test statistics that would be used here <u>One-sample z test statistics</u> as we know about the population standard deviation;

                           T.S. =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean bulb life = 490 hours

           σ = population standard deviation = 50 hours

           n = sample of light bulbs = 100

So, <u><em>the test statistics</em></u>  =  \frac{490-500}{\frac{50}{\sqrt{100} } }

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The value of z test statistics is -2.

<u>Now, at 0.01 significance level the z table gives critical value of -2.33 for left-tailed test.</u>

Since our test statistic is higher than the critical value of z as -2 > -2.33, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis.</u>

Therefore, we conclude that the population mean light bulb life is at least 500 hours.

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