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Elenna [48]
3 years ago
10

6.5.5

Mathematics
1 answer:
KATRIN_1 [288]3 years ago
3 0
First number is 19 and second is 3
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Jina wanted to study how the area of a rectangle changes with the length if it’s width is fixed. She computed the areas of sever
olganol [36]

Answer:

The domain and the range of the function are, respectively:

Dom\{f\} = [0\,m,5\,m]

Ran\{f\} = [0\,m^{2}, 10\,m^{2}]

Step-by-step explanation:

Jina represented a function by a graphic approach, where the length, measured in meters, is the domain of the function, whereas the area, measured in square meters, is its range.

Dom\{f\} = [0\,m,5\,m]

Ran\{f\} = [0\,m^{2}, 10\,m^{2}]

8 0
3 years ago
What is |3.5| simplified?
erastovalidia [21]
|3.5| = 3.5

Any negative or positive number will always have an absolute value that is positive.

For example, | -3.5| = 3.5

3.5 can also be expressed a fraction.

3.5 * 2/2 = 7/2
8 0
3 years ago
The probability that a person who has made a reservation for a trip on a twelve-vehicle ferry will actually arrive and make the
skad [1K]

Answer:

D

Step-by-step explanation:

5 0
3 years ago
An NBA fan named Mark claims that there are more fouls called on his team 1 point
inn [45]

Answer:

t=\frac{12.2-11.5}{\frac{1.6}{\sqrt{34}}}=2.551    

df = n-1=34-1=33

p_v =P(t_{(33)}>2.551)=0.008  

Since the p value is less than the significance level of 0.05 we have enough evidence to reject the null hypothesis in favor of the claim

And the best conclusion for this case would be:

b)The p-value is 0.008, indicating sufficient evidence for his claim.

Step-by-step explanation:

Information provided

\bar X=12.2 represent the sample mean fould against

s=1.6 represent the sample standard deviation

n=34 sample size  

represent the value that we want to test

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

System of hypothesis

We need to conduct a hypothesis in order to check if the true mean is higher than 11.5 fouls per game:  

Null hypothesis:\mu \leq 11.5  

Alternative hypothesis:\mu > 11.5  

The statistic is given by:

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

The statistic is given by:

t=\frac{12.2-11.5}{\frac{1.6}{\sqrt{34}}}=2.551    

P value

The degreed of freedom are given by:

df = n-1=34-1=33

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

p_v =P(t_{(33)}>2.551)=0.008  

Since the p value is less than the significance level of 0.05 we have enough evidence to reject the null hypothesis in favor of the claim

And the best conclusion for this case would be:

b)The p-value is 0.008, indicating sufficient evidence for his claim.

7 0
3 years ago
Name:
Vlad [161]

Answer:

115

Step-by-step explanation:

59.50-30

=29.5

29.5/0.25=115

4 0
3 years ago
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