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3 years ago

What is 14% of 39? (help)

Mathematics

2 answers:

pogonyaev3 years ago

7 0

The correct answer is 5.46.

Airida [17]3 years ago

6 0

The answer to this is 35.8% and heres why:

We assume, that the number 39 is 100% because it's the output value of the task and we all know that x is the value we are looking for. If 100% equals 39, so we can write it down as 100%=39 and x% equals 14 of the output value, so we can write it down as x%=14. And with that we have 2 equations we can solve.

Lets solve:

100%=39 and x%=14.

100%/x%=39/14

(100/x)*x=(39/14)*x - we multiply both sides of the equation by x

100=2.78*x - we divide both sides of the equation by (2.78) to get x

100/2.78=x

35.8=x

x=35.8

Hope this helped :)

Have a great day

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Answer:

$t=\frac{51.6-48}{\frac{1.1}{\sqrt{10}}}=10.349$

$p_v =P(t_9>10.349)=1.34x10^{-6}$

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the the actual mean is significantly higher than 48 MPa at 5% of significance.

Step-by-step explanation:

1) Data given and notation

$\bar X=51.6$ represent the sample mean

$s=1.1$ represent the standard deviation for the sample

$n=10$ sample size

$\mu_o =48$ 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 determine if the mean score is higher than 48, the system of hypothesis would be:

Null hypothesis: $\mu \geq 48$

Alternative hypothesis: $\mu > 48$

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (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:

$t=\frac{51.6-48}{\frac{1.1}{\sqrt{10}}}=10.349$

Calculate the P-value

First we find the degrees of freedom:

$df=n-1=10-1=9$

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

$p_v =P(t_9>10.349)=1.34x10^{-6}$

Conclusion

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3 years ago