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poizon [28]
3 years ago
13

Kathryn currently has an account balance of $2,248.95. She opened the account 7 years ago with a deposit of $1,762.14. If the in

terest compounds quarterly, what is the interest rate on the account?
Mathematics
1 answer:
melomori [17]3 years ago
4 0
Hope it helps ...............
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C+12=15 idk this pls help
Vedmedyk [2.9K]

Answer:

C = 3

Step-by-step explanation:

................

4 0
3 years ago
Read 2 more answers
Cody pays $657 for six months of guitar lessons. He pays the same amount for lessons each month. Which of the following is the b
stiv31 [10]
The answer is D. $110
4 0
3 years ago
A club has 86 members, and there are 14 more girls than boys. how many boys and how many girls are members of the club?
Rus_ich [418]
X = boy count
x + 14 = girl count

x +x +14 = 86
2x + 14 = 86
2x = 72
x = 36
x + 14 = 50

There are 36 boys and 50 girls. Hope this helps!
6 0
3 years ago
A recent study from the University of Virginia looked at the effectiveness of an online sleep therapy program in treating insomn
Ludmilka [50]

Answer:

1. The 99% confidence interval for the difference in average is -6.47377 < μ₁ - μ₂ < -11.34623

2. The possible issues in the calculations includes;

a. The confidence level used in the confidence interval can influence the result of the confidence interval observed

b. The sample size is small

Step-by-step explanation:

1. The number of adults with insomnia in the sample = 45

The number of adults that participated in the therapy, n₁ = 22

The number of candidates that served as control group, n₂ = 23

The average score for the for the 22 participants of the program, \overline x_1 = 6.59

The standard deviation for the 22 participants of the program, s₁ = 4.10

The average score for the for the 23 subjects in the control group, \overline x_2 = 15.50

The standard deviation for the 23 subjects in the control group, s₂ = 5.34

The confidence interval for unknown standard deviation, σ, is given by the following expression;

\left (\bar{x}_{1}- \bar{x}_{2}  \right )\pm t_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}

α = 1 - 0.99 = 0.01

α/2 = 0.005

The degrees of freedom, df = 22 - 1 = 21

t_{\alpha /2} = t_{0.005, \, 21} = 1.721

Therefore, we have;

\left (6.59- 15.5  \right )\pm1.721 \cdot \sqrt{\dfrac{4.10^{2}}{22}+\dfrac{5.34^{2}}{23}}

The 99% confidence interval for the difference in average is therefore given as follows;

-6.47377 < μ₁ - μ₂ < -11.34623

Therefore, there is considerable evidence that the participants in the survey  had lower average score than the subjects in the control group

2. The possible issues in the calculations are;

a. The confidence level used in the confidence interval can influence the result of the confidence interval observed

b. The sample size is small

5 0
2 years ago
Rick, Maya, and Tom work in a call center. On a particular day, Rick worked 5 and ½ hours, Maya worked 6.1 hours, and Tom worked
storchak [24]
5h 30m + 6h 6m + 6h 30m = 17 hours and 66 minutes or 18 hours and 6 minutes.

1,110 minutes total.

Hope I helped! :)

5 0
4 years ago
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