A manager samples the receipts of every fifth person who goes through the line. Out of 50 people, 4 had a mispriced item. If 1,5
1 answer:
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Answer: The correct statements are
The GCF of the coefficients is correct.
The variable c is not common to all terms, so a power of c should not have been factored out.
David applied the distributive property.
Step-by-step explanation:
GCF = Greatest common factor
1) GCF of coefficients : (80,32,48)
80 = 2 × 2 × 2 × 2 × 5
32 = 2 × 2 × 2 × 2 × 2
48 = 2 × 2 × 2 × 2 × 3
GCF of coefficients : (80,32,48) is 16.
2) GCF of variables :()
= b × b × b × b
= b × b
=b × b × b × b
GCF of variables :() is
3) GCF of and c: c is not the GCF of the polynomial. The variable c is not common to all terms, so a power of c should not have been factored out.
4)
David applied the distributive property.
Steps?
A graph shows zeros to be ±3. Factoring those out leaves the quadratic
(x-2)² +1
which has complex roots 2±i.
The function has roots -3, 3, 2-i, 2+i.
A graph shows zeros to be ±3. Factoring those out leaves the quadratic
(x-2)² +1
which has complex roots 2±i.
The function has roots -3, 3, 2-i, 2+i.
Answer:
(a) NULL HYPOTHESIS, :
3.5%
ALTERNATE HYPOTHESIS, :
> 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, :
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, :
> 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. = ~
where, = sample mean debt ratio of Ohio banks = 6%
s = sample standard deviation = = 1.83%
n = sample of banks = 7
So, test statistics = ~
= 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.