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3 years ago

Please help me !

Mathematics

2 answers:

Alik [6]3 years ago

8 0

Solve for x in the second equation first

kumpel [21]3 years ago

8 0

Answer:

Solve for x in the second equation.

Step-by-step explanation:

To solve a system of equations by substition you have to solve one of the equations for X or Y and then substitute that value on the X of the other equation, if you want to avoid fractions, you can use the x of the secon equation, like this:

$6x+18y=24\\6x=24-18y\\x=\frac{24-18y}{6} \\x=4-3y$

So by solving X for the second equation you avoided fractions when substituting the value for X on the first equation.

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Shay makes 24 liters of homemade vegetable stock. She distributes the stock

ANTONII [103]

Answer:

3 liters

Step-by-step explanation:

pretty simple first grade question, 24/8

6 0

3 years ago

Read 2 more answers

We're testing the hypothesis that the average boy walks at 18 months of age (H0: p = 18). We assume that the ages at which boys

marusya05 [52]

Answer:

II. This finding is significant for a two-tailed test at .01.

III. This finding is significant for a one-tailed test at .01.

d. II and III only

Step-by-step explanation:

1) Data given and notation

$\bar X=19.2$ represent the battery life sample mean

$\sigma=2.5$ represent the population standard deviation

$n=25$ sample size

$\mu_o =18$ represent the value that we want to test

$\alpha$ 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)

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean battery life is equal to 18 or not for parta I and II:

Null hypothesis: $\mu = 18$

Alternative hypothesis: $\mu \neq 18$

And for part III we have a one tailed test with the following hypothesis:

Null hypothesis: $\mu \leq 18$

Alternative hypothesis: $\mu > 18$

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-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".

3) Calculate the statistic

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

$z=\frac{19.2-18}{\frac{2.5}{\sqrt{25}}}=2.4$

4) P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=25-1=24$

Since is a two tailed test for parts I and II, the p value would be:

$p_v =2*P(t_{(24)}>2.4)=0.0245$

And for part III since we have a one right tailed test the p value is:

$p_v =P(t_{(24)}>2.4)=0.0122$

5) Conclusion

I. This finding is significant for a two-tailed test at .05.

Since the $p_v$ . We reject the null hypothesis so we don't have a significant result. FALSE

II. This finding is significant for a two-tailed test at .01.

Since the $p_v >\alpha$ . We FAIL to reject the null hypothesis so we have a significant result. TRUE.

III. This finding is significant for a one-tailed test at .01.

Since the $p_v >\alpha$ . We FAIL to reject the null hypothesis so we have a significant result. TRUE.

So then the correct options is:

d. II and III only

6 0

3 years ago

A physical therapist wants to determine the difference in the proportion of men and women who participate in regular sustained p

Alekssandra [29.7K]

Using the z-distribution and the formula for the margin of error, it is found that:

a) A sample size of 54 is needed.

b) A sample size of 752 is needed.

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $\alpha$ , we have the following confidence interval of proportions.