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dmitriy555 [2]
3 years ago
11

For one month Siera calculated her home town’s average high temperature in degrees Fahrenheit. She wants to convert that tempera

ture from degrees Fahrenheit to degrees Celsius using the function . What does C(F) represent?
Mathematics
1 answer:
SVETLANKA909090 [29]3 years ago
4 0
<span>We have to find how can someone convert the temperature from degrees Farhenheit to degrees Celsius using the function. The temperature in degrees Celsius is equal to the temperature in degrees Farhenheit minus 32 times 5/9. Or C ( F ) = ( F - 32 ) * 5/9. For example: the temperature is F = 50°; C ( 50 ) = ( 50 - 32 ) * 5/9 = 18 * 5/ 9 = 10 °C. Answer: The function is: C ( F ) = ( F - 32 ) * 5/9.</span>
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Answer:

g(f(2)) = 9

Step-by-step explanation:

Evaluate f(2) and substitute the value obtained into g(x)

f(2) = 2 + 1 = 3 , then

g(3) = 3² = 9

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Mark and his three Friends ate out at Applebees's . Their bill totaled % 52.35 . If they left a 20% tip how much would each pers
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Answer:

Each person would pay $15.71

Step-by-step explanation:

First we can calculate how much everyone ate by multiplying the bill amount by 120% (20% tip and 100% of the bill).

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Step-by-step explanation:

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In a sample of 170 students at an Australian university that introduced the use of plagiarism-detection software in a number of
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Answer:

p_v =P(Z>4.146)=0.0000169  

Based on the p value obtained and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of students who NOT belief that such software unfairly targets students is higher than 0.5 or 50% .  

Step-by-step explanation:

1) Data given and notation  

n=170 represent the random sample taken  

X=58 represent the student's who belief that such software unfairly targets students

\hat px=\frac{58}{170}=0.341 estimated proportion of students who belief that such software unfairly targets students

\hat p=\frac{112}{170}=0.659 estimated proportion of students who NOT belief that such software unfairly targets students

p_o=0.50 is the value that we want to test  

\alpha=0.05 represent the significance level (no given)  

z would represent the statistic (variable of interest)  

p_v represent the p value (variable of interest)  

p= proportion of student's who belief that such software unfairly targets students

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that that a majority of students at the university do not share this belief. :  

Null hypothesis:p\leq 0.5  

Alternative hypothesis:p>0.5

We assume that the proportion follows a normal distribution.  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}    (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly (different,higher or less) from a hypothesized value p_o.  

<em>Check for the assumptions that he sample must satisfy in order to apply the test </em>

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

np_o =170*0.5=85>10

n(1-p_o)=170*(1-0.5)=85>10

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.659 -0.5}{\sqrt{\frac{0.5(1-0.5)}{170}}}=4.146  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided is \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a one right side test the p value would be:  

p_v =P(Z>4.146)=0.0000169  

Based on the p value obtained and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of students who NOT belief that such software unfairly targets students is higher than 0.5 or 50% .  

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