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Andreyy89
3 years ago
9

9100 dollard placed into an account with an annual intrest of 5% to the neareast year how long will it take for the account to v

alue 36900 dollars
Mathematics
1 answer:
shusha [124]3 years ago
4 0

Answer:

81 years

Step-by-step explanation:

The answer is 81 because 5% of 9100 is 455 and 36900 divided by 455 is rounded to 81.

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(4.23)(1.6) please multiple
Wewaii [24]

Answer:

The correct answer is 6.768.

Step-by-step explanation:

(4.23)(1.6) = 6.768

8 0
3 years ago
Read 2 more answers
If f(x)=-9x-3, find f(-5)
Andrews [41]

Answer:

42

Step-by-step explanation:

f(-5) = -9 × (-5) - 3 = 45 - 3 = 42

If you have any questions about the way I solved it, don't hesitate to ask me in the comments below =)

5 0
3 years ago
Please help, I am super stuck and I cant lose points<br><br> this homework is 16 points total
bekas [8.4K]
Hello,

7)  A∪C={1,2,3,4,5,7,9}

8) A∩B={2,4}

C'= complement of C ={2,4,6}

9) A∪B∩C'={1,2,3,4,6,8}∩{2,4,6}={2,4,6}

10) A∪(B∩C')={1,2,3,4}∩{2,4,6}={1,2,3,4,6}

Are you blind?
4 0
3 years ago
Mika is a local farmer who is interested in how often residents in her town go to a farmer's market each month. She surveys 124
TEA [102]

Answer:

The margin of error is of 0.3012, and it means that we should be 99% confident that the population mean would be within 0.3012 of the sample mean.

Step-by-step explanation:

Margin of error

M = z\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation and n is the size of the sample.

Standard deviation of 1.3

This means that \sigma = 1.3

She surveys 124 families

This means that n = 124

Margin of error and meaning:

M = z\frac{\sigma}{\sqrt{n}}

M = 2.58\frac{1.3}{\sqrt{124}}

M = 0.3012

The margin of error is of 0.3012, and it means that we should be 99% confident that the population mean would be within 0.3012 of the sample mean.

6 0
2 years ago
The United States Coast Guard assumes the mean weight of passengers in commercial boats is 185 pounds. The previous value was lo
Valentin [98]

Answer:

There is a 5.5% probability that a random sample of passengers will have a mean weight that is as extreme or more extreme (either above or below the mean) than was observed in this sample.

Step-by-step explanation:

To solve this problem, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

In this problem, we have that:

\mu = 185, \sigma = 26.7, n = 48, s = \frac{26.7}{\sqrt{48}} = 3.85

The weights of a random sample of 48 commercial boat passengers were recorded. The sample mean was determined to be 177.6 pounds. Find the probability that a random sample of passengers will have a mean weight that is as extreme or more extreme (either above or below the mean) than was observed in this sample.

The probability of an extreme value below the mean.

This is the pvalue of Z when X = 177.6.

So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{177.6 - 185}{3.85}

Z = -1.92

Z = -1.92 has a pvalue of 0.0274.

So there is a 2.74% of having a sample mean as extreme than that and lower than the mean.

The probability of an extrema value above the mean.

Measures above the mean have a positive z score.

So this probability is 1 subtracted by the pvalue of Z = 1.92

Z = 1.92 has a pvalue of 0.9726.

So there is a 1-0.9726 = 0.0274 = 2.74% of having a sample mean as extreme than that and above than the mean.

Total:

2*0.0274 = 0.0548 = 0.055

There is a 5.5% probability that a random sample of passengers will have a mean weight that is as extreme or more extreme (either above or below the mean) than was observed in this sample.

4 0
3 years ago
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