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Yanka [14]
3 years ago
14

What is the domain and range of Y=3X-7.25? can someone please help?

Mathematics
1 answer:
Tems11 [23]3 years ago
8 0

Answer:

Domain and range of the equation

Y = 3X-7.25

will be Set of all Real numbers.

Step-by-step explanation:

Because at any value of x, you will get a respective value of y. For example if you put X=0, then y=7.25. You can put any value of X in the equation and for each value equation will be true and will give respective value of y.

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A rectangle is drawn so that the width is 3 feet shorter than the length. the area of the rectangle is 40 square feet. find the
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A recipe submitted to a magazine by one of its subscribers’ states that the mean baking time for a cheesecake is 55 minutes. A t
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Answer:

t=\frac{59.4-55}{\frac{3.239}{\sqrt{10}}}=4.296    

The degrees of freedom are given by:

df=n-1=10-1=9  

The p value for this case is given by:

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And for this case the p value is lower than the significance level so we have enough evidence to reject the null hypothesis and then we can conclude that true mean is higher than 55.

Step-by-step explanation:

Information given

We have the following data: 54 55 58 59 59 60 61 61 62 65

The sample mean and deviation can be calculated with the following formulas:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

s=\sqrt{\frac{\sum_{i=1}^n (X-i -\bar x)^2}{n-1}}

\bar X=59.4 represent the sample mean

s=3.239 represent the sample standard deviation

n=10 sample size  

\alpha=0.05 represent the significance level

t would represent the statistic  

p_v represent the p value

System of hypothesis

We want to test if the true mean is higher than 55, the system of hypothesis would be:  

Null hypothesis:\mu \leq 55  

Alternative hypothesis:\mu > 55  

Replacing the info given we got:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

And replacing the info given we got:

t=\frac{59.4-55}{\frac{3.239}{\sqrt{10}}}=4.296    

The degrees of freedom are given by:

df=n-1=10-1=9  

The p value for this case is given by:

p_v =P(t_{(9)}>4.296)=0.001  

And for this case the p value is lower than the significance level so we have enough evidence to reject the null hypothesis and then we can conclude that true mean is higher than 55

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3 years ago
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svlad2 [7]

Answer:

3/4

Step-by-step explanation:

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