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Marianna [84]
3 years ago
5

Solve for v 2v+18=16-4(v+7)

Mathematics
2 answers:
Rina8888 [55]3 years ago
7 0
2v+18=16-4(v+7). Take away the parenthesis.
2v+18=16-4v-28. Combine your like terms.
2v+18=-12-4v. Add 12
2v+30=-4v. Subtract 2v
30=-6v. Divide by -6
v=-5. This is your final answer.
Jlenok [28]3 years ago
4 0

Answer: v= -5

I hope I'm correct. Also double check. :)

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Answer:

The 95% confidence interval for the difference between means is (-2164.21, -299.13).

The lower limit on the confidence interval is -$2164.21.

The upper limit on the confidence interval is -$299.13.

Step-by-step explanation:

The sample data is:

Gender   Mean          Std. dev.     n

Female    2577.75      1916.29     67

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The sample 1, of size n1=67 has a mean of 2577.75 and a standard deviation of 1916.29.

The sample 2, of size n2=33 has a mean of 3809.42 and a standard deviation of 2379.47.

The difference between sample means is Md=-1231.67.

M_d=M_1-M_2=2577.75-3809.42=-1231.67

The estimated standard error of the difference between means is computed using the formula:

s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{1916.29^2}{67}+\dfrac{2379.47^2}{33}}\\\\\\s_{M_d}=\sqrt{54808.468+171572.045}=\sqrt{226380.513}=475.795

The t-value for a 95% confidence interval is t=1.96.

The margin of error (MOE) can be calculated as:

MOE=t\cdot s_M=1.96 \cdot 475.795=932.54

Then, the lower and upper bounds of the confidence interval are:

LL=M_d-t \cdot s_{M_d} = -1231.67-932.54=-2164.21\\\\UL=M_d+t \cdot s_{M_d} = -1231.67+932.54=-299.13

The 95% confidence interval for the difference between means is (-2164.21, -299.13).

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Step-by-step explanation:

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