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3 years ago

10 - 4 (3-2y) = 8y -5

Mathematics

1 answer:

Lisa [10]3 years ago

8 0

Answer:

No Solutions

Step-by-step explanation:

10 - 4(3-2y) = 8y - 5

Distribute:

10 - 12 + 8y = 8y - 5

Simplify:

-2 + 8y = 8y - 5

Same to both sides

-2 + 5 = 8y - 8y

3 = 0y

No Solutions

Let me know if you have any questions

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3 years ago

The bad debt ratio for a financial institution is defined to be the dollar values of loans defaulted divided by the total dollar

Nimfa-mama [501]

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(a) NULL HYPOTHESIS, $H_0$ : $\mu \leq$ 3.5%

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(b) We conclude that the the mean bad debt ratio for Ohio banks is higher than the mean for all federally insured banks.

Step-by-step explanation:

We are given that a random sample of seven Ohio banks is selected.The bad debt ratios for these banks are 7, 4, 6, 7, 5, 4, and 9%.The mean bad debt ratio for all federally insured banks is 3.5%.

We have to test the claim of Federal banking officials that the mean bad debt ratio for Ohio banks is higher than the mean for all federally insured banks.

(a) Let, NULL HYPOTHESIS, $H_0$ : $\mu \leq$ 3.5% {means that the the mean bad debt ratio for Ohio banks is less than or equal to the mean for all federally insured banks}

ALTERNATE HYPOTHESIS, $H_1$ : $\mu$ > 3.5% {means that the the mean bad debt ratio for Ohio banks is higher than the mean for all federally insured banks}

The test statistics that will be used here is One-sample t-test;

T.S. = $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean debt ratio of Ohio banks = 6%

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 1.83%

n = sample of banks = 7

So, test statistics = $\frac{6-3.5}{\frac{1.83}{\sqrt{7} } }$ ~ $t_6$

= 3.614

(b) Now, at 1% significance level t table gives critical value of 3.143. Since our test statistics is more than the critical value of t so we have sufficient evidence to reject null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the the mean bad debt ratio for Ohio banks is higher than the mean for all federally insured banks.

Hence, Federal banking officials claim was correct.

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3 years ago