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neonofarm [45]
3 years ago
6

PLEASE HELP !!!

Mathematics
1 answer:
katen-ka-za [31]3 years ago
4 0

Answer:

final cost of 3 pounds: <u>  8 dollars  </u>

final cost of p pounds: <u>   3.75p - 3.25  dollars  </u>

Step-by-step explanation:

Each pound costs $3.75, so 3 pounds cost 3*3.75 = 11.25 dollars. Subtract off the $3.25 to get the final cost to be 11.25-3.25 = 8 dollars. This takes care of the first part.

For the second part, the expression for the final cost is 3.75p - 3.25; where the 3.75p is the cost before the coupon is applied. If you plugged p = 3 into that expression, you should get 8 as a result. The variable p is some positive whole number. It's a place holder for the number of pounds of apples.

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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.

7 0
3 years ago
Carly is a catcher on the school softball team. By the end of the season, she had played 45% of the 49 games the team played. In
Eduardwww [97]
Carly did not play 22 games
6 0
3 years ago
What is the place value for 5 in 35.052
snow_tiger [21]
The place value is the hundreths 
4 0
3 years ago
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Transformations of exponential functions
AURORKA [14]

Answer:

It's the last one. We know it's to the right because the -8 is in the exponent and also, it's -8 not +8.

6 0
3 years ago
Plz plz Plz help!
likoan [24]

Answer:

$16,666.67

Step-by-step explanation:

PMT= PV*i Where PMT is the withdrawals ,PV is present value and i is the dicounting rate

PMT = $1,000.00

PV= ?

i = 6%

hence  $1,000 = PV*6%

PV=1,000/6%

PV = 16,666.67

4 0
3 years ago
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