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4 years ago

9

Sheila deposited $1,050.00 into a saving account at her local bank. if the interest rate is 1.5%, then how much will he have aft

er 18 months (round your answer to the nearest cent)?

1 answer:

kogti [31]4 years ago

6 0

It is $1,073.63 because if you multiply the 1.5% to the amount saved in the account it gives you $1,073.63

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38. A baseball park's average game attendance is 2035 people. This is 65

marta [7]

Answer:

<u><em>1970</em></u> is the answer for the question "A baseball park's average game attendance is 2035 people. This is 65 more than the average attendance for last year. What was the average game attendance for last year?"

<u><em>- hope this helps!:)</em></u>

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3 years ago

The sum of two consecutive even integers is -134 . What are the two numbers?

Mamont248 [21]

2x + 2x+2 = -134
4x + 2 = -134
4x = -136 |: 4
x = -34

2x = 2 * (-34) = -68
2x + 2 = -68 + 2 = -66

This numbers are -66 and -68.

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3 years ago

Read 2 more answers

(cos 2theta)/(1 + sin 2theta) = (1 - tan theta)/(1 + tan theta)

andre [41]

Step-by-step explanation:

Here is the solution...hope it helps:)

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2 years ago

Use the information provided to determine a 95% confidence interval for the population variance. A researcher was interested in

Leno4ka [110]

Answer:

The 95% confidence interval for the population variance is (8.80, 32.45).

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for the population variance is given as follows:

$\frac{(n-1)\cdot s^{2}}{\chi^{2}_{\alpha/2}}\leq \sigma^{2}\leq \frac{(n-1)\cdot s^{2}}{\chi^{2}_{1-\alpha/2}}$

It is provided that:

<em>n</em> = 20

<em>s</em> = 3.9

Confidence level = 95%

⇒ <em>α</em> = 0.05

Compute the critical values of Chi-square:

$\chi^{2}_{\alpha/2, (n-1)}=\chi^{2}_{0.05/2, (20-1)}=\chi^{2}_{0.025,19}=32.852\\\\\chi^{2}_{1-\alpha/2, (n-1)}=\chi^{2}_{1-0.05/2, (20-1)}=\chi^{2}_{0.975,19}=8.907$

*Use a Chi-square table.

Compute the 95% confidence interval for the population variance as follows:

$\frac{(n-1)\cdot s^{2}}{\chi^{2}_{\alpha/2}}\leq \sigma^{2}\leq \frac{(n-1)\cdot s^{2}}{\chi^{2}_{1-\alpha/2}}$

$\frac{(20-1)\cdot (3.9)^{2}}{32.852}\leq \sigma^{2}\leq \frac{(20-1)\cdot (3.9)^{2}}{8.907}\\\\8.7967\leq \sigma^{2}\leq 32.4453\\\\8.80\leq \sigma^{2}\leq 32.45$

Thus, the 95% confidence interval for the population variance is (8.80, 32.45).

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3 years ago

Find the distance between the points with the given coordinates (-2, -5) and (-2, 4).

bija089 [108]

I hope this helps you

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4 years ago