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lawyer [7]
3 years ago
9

What is 5(h-4)=8?????????????????????

Mathematics
1 answer:
IRINA_888 [86]3 years ago
7 0
5(h-4)=8\\
5h-20=8\\
5h=28\\
h=\dfrac{28}{5}
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Find the approximate value of the circumference of a circle with the given radius. Use pi = 3.14. Round your results to one more
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3 years ago
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Which of the following relations is a function? A. (0, 4), (-2, 6), (0, 3), (-7, 2) B. (8, 1), (-2, 4), (0, 1), (8, 2) C. (0, 4)
fomenos

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Step-by-step explanation:

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The other choices are not functions because at least one of the inputs corresponds to 2 or more outputs.

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3 years ago
A candy manufacturer claims that the average weight of their individually wrapped candy is 3 grams. Sixty-four candy pieces were
Anon25 [30]

Answer:

t=\frac{2.8-3}{\frac{0.5}{\sqrt{64}}}=-3.2    

p_v =P(t_{(63)}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is significantly lower than 3 grams at 5% of signficance.  

Step-by-step explanation:

Data given and notation  

\bar X=2.8 represent the sample mean

s=0.5 represent the sample standard deviation for the sample  

n=64 sample size  

\mu_o =3 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)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is lower than 3 grams, the system of hypothesis would be:  

Null hypothesis:\mu \geq 3  

Alternative hypothesis:\mu < 3  

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{2.8-3}{\frac{0.5}{\sqrt{64}}}=-3.2    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=64-1=63  

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

p_v =P(t_{(63)}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is significantly lower than 3 grams at 5% of signficance.  

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