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joja [24]
3 years ago
11

Tanisha wants to have $1000 in her bank account in 5 years. How much money should she deposit if her account earns 6% interest w

hich is compounded 2 times per year?
Mathematics
2 answers:
Artist 52 [7]3 years ago
5 0

Answer:

747,26

Step-by-step explanation:

1000=x (1+0.06)^5

1000/((1+0.06)^5)=x

Lilit [14]3 years ago
3 0

Answer:

744,09

Step-by-step explanation:

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Brian saved $8 on a pair of sneakers with a 25% coupon at Kohls
Scilla [17]

Answer: 32$

Step-by-step explanation: 8 x 4 = 32

32 divided by 4 = 8

8 0
3 years ago
The Rocky Mountain News (January 24, 1994) indicated that the 20-year mean snowfall in the Denver/Boulder region is 28.76 inches
ycow [4]

Answer:

The p-value of the test is 0.0007 < 0.05, indicating that the the snowfall for the 1993-1994 winters was higher than the previous 20-year average.

Step-by-step explanation:

20-year mean snowfall in the Denver/Boulder region is 28.76 inches. Test if the snowfall for the 1993-1994 winters has higher than the previous 20-year average.

At the null hypothesis, we test if the average was the same, that is, of 28.76 inches. So

H_0: \mu = 28.76

At the alternate hypothesis, we test if the average incresaed, that is, it was higher than 28.76 inches. So

H_1: \mu > 28.76

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

28.76 is tested at the null hypothesis:

This means that \mu = 28.76

Standard deviation of 7.5 inches. However, for the winter of 1993-1994, the average snowfall for a sample of 32 different locations was 33 inches.

This means that \sigma = 7.5, X = 33, n = 32.

Value of the test statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{33 - 28.76}{\frac{7.5}{\sqrt{32}}}

z = 3.2

P-value of the test and decision:

The p-value of the test is the probability of finding a sample mean above 33, which is 1 subtracted by the p-value of z = 3.2. In this question, we consider the standard level \alpha = 0.05.

Looking at the z-table, z = 3.2 has a p-value of 0.9993.

1 - 0.9993 = 0.0007

The p-value of the test is 0.0007 < 0.05, indicating that the the snowfall for the 1993-1994 winters was higher than the previous 20-year average.

5 0
3 years ago
Clara pays $76 for 8 guitar lessons. How much would 10 guitar lessons cost? Enter your answer in the box.
WARRIOR [948]
You would divide 76 by 8. That would give you 9.5. Then you multiply by 10 and that gives you 95. Hope this helps (;
5 0
4 years ago
Multiply (2.4 ⋅ 1014) ⋅ (4 ⋅ 107). Express the answer in scientific notation
IRINA_888 [86]
9.6*10^21 is your answer
3 0
4 years ago
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According to the survey described here, what was the sample of the survey? “A recent telephone survey of 900 teenagers aged 16 t
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Answer:

i think its b

Step-by-step explanation:

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