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Julli [10]
3 years ago
9

744 ÷ 40 in long division

Mathematics
1 answer:
wariber [46]3 years ago
7 0
18.5 would be the answer and here’s my work

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what is this have no clue of how to do it ? ? n+31.53 =62.4 and this one to 9.2+n+8.4=20.8 please help !
mihalych1998 [28]
Ok so to solve to first one u do this:

62.4 - 31.53, which gives u 30.87. and then u add 30.87 and 31.33, and u get 62.4

for the second one u do the same thing.
4 0
3 years ago
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3(x + 4) = 2x + 4x - 6 Please help i'll give 90 pts and I only have 113.
dimaraw [331]

Answer:

x= 6

Step-by-step explanation:

3(x + 4) = 2x + 4x - 6

3x + 12 = 2x + 4x - 6

3x + 12 = 6x - 6

-3x = -18

x= 6

hope it helps

if it helps you, pls give me the brainliest.

5 0
4 years ago
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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
PRE-ALGEBRA<br> DUE NOW! PLEASE HELP!
Nata [24]
\bold{FULL SOLUTION:}

4 0
2 years ago
Find the domain and range of the relation, and state whether or not the relation is a function.
Brilliant_brown [7]
Domain is {1 ,2,3,4}
range is {3}
it's a function because each element in x has only one image in y
6 0
3 years ago
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