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3 years ago

Can somebody help me with this?

Mathematics

1 answer:

Bess [88]3 years ago

7 0

The answer is C, 40, angle c for triangle abc is 50 bc the total of it and 130 must equal 180. so if you subtract the angle you are given, 90, and the new angle, 50, from 180 (total a triangle should have) then your answer is 40.

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A vehicle is purchased for $18,000, with a down payment of $6,098. The balance in financed for three years at an annual rate of

mestny [16]

The Present value of an annuity is given by PV = P(1 - (1 + r/t)^-nt)/(r/t)
where: P is the monthly payment, r is the annual rate = 7% = 0.07, t is the number of periods in one year = 12 and n is the number of years = 3.

18,000 - 6,098 = P(1 - (1 + 0.07/12)^-(3 x 12)) / (0.07/12)
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2 years ago

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irga5000 [103]

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Step-by-step explanation:

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8 0

3 years ago

Read 2 more answers

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Firlakuza [10]

$\frac{13}{10}$

$1.3$

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2 years ago

Read 2 more answers

Sara travels often for her job. The miles she traveled for the last seven work days are shown. 88, 72, 91, 65, 86, 93, 97 Which

Troyanec [42]

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2 years ago

A candidate for a US Representative seat from Indiana hires a polling firm to gauge her percentage of support among voters in he

mojhsa [17]

Answer:

(A) The minimum sample size required achieve the margin of error of 0.04 is 601.

(B) The minimum sample size required achieve a margin of error of 0.02 is 2401.

Step-by-step explanation:

Let us assume that the percentage of support for the candidate, among voters in her district, is 50%.

(A)

The margin of error, MOE = 0.04.

The formula for margin of error is:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}$

The critical value of z for 95% confidence interval is: $z_{\alpha/2}=1.96$

Compute the minimum sample size required as follows:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}\\0.04=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\(\frac{0.04}{1.96})^{2} =\frac{0.50(1-0.50)}{n}\\n=600.25\approx 601$

Thus, the minimum sample size required achieve the margin of error of 0.04 is 601.

(B)

The margin of error, MOE = 0.02.

The formula for margin of error is:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}$

The critical value of z for 95% confidence interval is: $z_{\alpha/2}=1.96$

Compute the minimum sample size required as follows:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}\\0.02=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\(\frac{0.02}{1.96})^{2} =\frac{0.50(1-0.50)}{n}\\n=2401.00\approx 2401$

Thus, the minimum sample size required achieve a margin of error of 0.02 is 2401.

8 0

3 years ago