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lara31 [8.8K]
3 years ago
15

Charles is 19 years old. his grandfather is 67 years older than he. what was thee total age 9 years ago?

Mathematics
2 answers:
Kryger [21]3 years ago
6 0

Answer:

87 years

Step-by-step explanation:

To find the total age 9 years ago you would subtract 9 from Charles's age only, 19. His grandfather is still always going to be 67 years older than he.

19 - 9 = 10

Add 10 and 67 together to find his grandfather's age back then.

10 + 67 = 77

Now add 10 and 77 together.

10 + 77 = 87

The total age 9 years ago was 87 years.

almond37 [142]3 years ago
3 0
Today: x=19
y=67+19=86

9 years ago: x=19-9=10
y=86-9=77

The total age 9 years ago: x+y=10+77=87
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The square below has a side length of 14 inches. What is the area of the shaded region?
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Step-by-step explanation:

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2 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]

Answer:

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

    ALTERNATE HYPOTHESIS, H_1 : \mu > 3.5%

(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
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I hope this helps!
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kompoz [17]
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3 years ago
HELP NEED IT RN!
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Answer:

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