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Sladkaya [172]
2 years ago
7

The temperature drops an average of 1 degree celsius for every 100 meters that a hot air ballon rises. If the temperature is 25

degrees celsius on the ground, what is the approximate temperature 1200 meters above ground.
Mathematics
1 answer:
marysya [2.9K]2 years ago
7 0

The required temperature at the height of 1200 meters is 13 degree Celcius.

Given that,
Temperature drops an average of 1 degree celsius for every 100 meters that a hot air balloon rises.
The temperature is 25 degrees celsius on the ground.
The approximate temperature is 1200 meters above the ground is to be determined.

<h3>What are functions?</h3>

Functions are the relationship between sets of values. e g y=f(x), for every value of x there is its exists in a set of y. x is the independent variable while Y is the dependent variable.

The expression of for temperature drops an average of 1-degree celsius for every 100 meters that a hot air balloon rises.
Let x be the height in meters and y be the temperature in degrees celcius
y = -x/100 + 25
For temperature 1200 meters above ground
x = 1200
y = -1200/100 + 25
y = -12 + 25
y = 13

Thus, the required temperature at the height of 1200 meters is 13 degrees Celcius.

learn more about function here:

brainly.com/question/21145944

#SPJ1





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Step-by-step explanation:

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0.32 = 32 hundredths + 32/100 = 32%

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Suppose 462 of 500 randomly selected college students said they would be embarrassed to truthfully admit they could not read at
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We conclude that there is a significant difference in the proportion of all college students who would be embarrassed by these two admissions.

Step-by-step explanation:

We are given that 462 of 500 randomly selected college students said they would be embarrassed to truthfully admit they could not read at a fourth grade level.

Further suppose 135 of 500 randomly selected college students said they would be embarrassed to truthfully admit they could not "do math" at a fourth grade level (like fractions).

<em>Let </em>p_1<em> = proportion of college students who would be embarrassed to truthfully admit they could not read at a fourth grade level.</em>

p_2<em> = proportion of college students who would be embarrassed to truthfully admit they could not "do math" at a fourth grade level.</em>

So, Null Hypothesis, H_0 : p_1-p_2 = 0  or  p_1= p_2     {means that there is not any significant difference in the proportion of all college students who would be embarrassed by these two admissions}

Alternate Hypothesis, H_A : p_1-p_2 \neq 0  or  p_1\neq p_2     {means that there is a significant difference in the proportion of all college students who would be embarrassed by these two admissions}

The test statistics that would be used here <u>Two-sample z proportion</u> <u>statistics</u>;

                       T.S. =  \frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}  } }  ~ N(0,1)

where, \hat p_1 = sample proportion of college students who would be embarrassed to admit they could not read at a fourth grade level = \frac{462}{500} = 0.924

\hat p_2 = sample proportion of college students who would be embarrassed to admit they could not "do math" at a fourth grade level = \frac{135}{500} = 0.27

n_1 = sample of college students = 500

n_2 = sample of college students = 500

So, <em><u>test statistics</u></em>  =  \frac{(0.924-0.27)-(0)}{\sqrt{\frac{0.924(1-0.924)}{500}+\frac{0.27(1-0.27)}{500}  } }

                              =  28.28

The value of z test statistics is 28.28.

<u>Also, P-value of the test statistics is given by;</u>

          P-value = P(Z > 28.28) = Less than 0.0005%

<u></u>

<u>Now, at 0.10 significance level the z table gives critical values of -1.645 and 1.645 for two-tailed test.</u>

Since our test statistics doesn't lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u>we reject our null hypothesis</u>.

Therefore, we conclude that there is a significant difference in the proportion of all college students who would be embarrassed by these two admissions.

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