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Tpy6a [65]
2 years ago
15

Ashley invests $9,720 in a one-month money market account paying 3.16% simple annual interest and $8,140 in a two-year CD yieldi

ng 3.23% simple annual interest. Assuming Ashley does not reinvest or renew these investments, how much money will she have when both investments reach maturity, to the nearest dollar?

Mathematics
2 answers:
tester [92]2 years ago
5 0
For the first investment. A = P(1 + rt); where p = 9,720, r = 0.0316 and t = 1/12
A = 9720(1 + 0.0316/12) = 9720(1.0026) = $9,746

For the second investment,
A = 8140(1 + 0.0323 x 2) = 8140(1.0646) = $8,666

Total amount she had = $9,746 + $8,666 = $18,412
Elodia [21]2 years ago
3 0

Answer:

D. $18,411

Step-by-step explanation:

I just answered it correct

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2 years ago
12 less than m is no less than 132
kaheart [24]

Answer

M-12=-132

The less than moves the number to the back

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2 years ago
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"An ordinance requiring that a smoke detector be installed in all previously constructed houses has been in effect in a particul
Galina-37 [17]

Answer:

a) Probability that the claim is rejected when the actual value of p is 0.8 = P(X ≤ 15) = 0.0173

b) Probability of not rejecting the claim when p = 0.7, P(X > 15) = 0.8106

when p = 0.6, P(X > 15) = 0.4246

c) Check Explanation

The error probabilities are evidently lower when 15 is replaced with 14 in the calculations.

Step-by-step explanation:

p is the true proportion of houses with smoke detectors and p = 0.80

The claim that 80% of houses have smoke detectors is rejected if in a sample of 25 houses, not more than 15 houses have smoke detectors.

If X is the number of homes with detectors among the 25 sampled

a) Probability that the claim is rejected when the actual value of p is 0.8 = P(X ≤ 15)

This is a binomial distribution problem

A binomial experiment is one in which the probability of success doesn't change with every run or number of trials (probability that each house has a detector is 0.80)

It usually consists of a number of runs/trials with only two possible outcomes, a success or a failure (we are sampling 25 houses with each of them either having or not having a detector)

The outcome of each trial/run of a binomial experiment is independent of one another.

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = 25 houses sampled

x = Number of successes required = less than or equal to 15

p = probability of success = probability that a house has smoke detectors = 0.80

q = probability of failure = probability that a house does NOT have smoke detectors = 1 - p = 1 - 0.80 = 0.20

P(X ≤ 15) = Sum of probabilities from P(X = 0) to P(X = 15) = 0.01733186954 = 0.01733

b) Probability of not rejecting the claim when p= 0.7 when p= 0.6

For us not to reject the claim, we need more than 15 houses with detectors, hence, th is probability = P(X > 15), but p = 0.7 and 0.6 respectively for this question.

n = total number of sample spaces = 25 houses sampled

x = Number of successes required = more than 15

p = probability that a house has smoke detectors = 0.70, then 0.60

q = probability of failure = probability that a house does NOT have smoke detectors = 1 - p = 1 - 0.70 = 0.30

And 1 - 0.60 = 0.40

P(X > 15) = sum of probabilities from P(X = 15) to P(X = 25)

When p = 0.70, P(X > 15) = 0.8105639765 = 0.8106

When p = 0.60, P(X > 15) = 0.42461701767 = 0.4246

c) How do the "error probabilities" of parts (a) and (b) change if the value 15 in the decision rule is replaced by 14.

The error probabilities include the probability of the claim being false.

When X = 15

(Error probability when p = 0.80) = 0.0173

when p = 0.70, error probability = P(X ≤ 15) = 1 - P(X > 15) = 1 - 0.8106 = 0.1894

when p = 0.60, error probability = 1 - 0.4246 = 0.5754

When X = 14

(Error probability when p = 0.80) = P(X ≤ 14) = 0.00555

when p = 0.70, error probability = P(X ≤ 14) = 0.0978

when p = 0.60, error probability = P(X ≤ 14) = 0.4142

The error probabilities are evidently lower when 15 is replaced with 14 in the calculations.

Hope this Helps!!!

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Thirty-two percent of fish in a large lake are bass. Imagine scooping out a simple random sample of 15 fish from the lake and ob
klio [65]

Answer:

Thirty-two percent of fish in a large lake are bass. Imagine scooping out a simple random sample of 15 fish from the lake and observing the sample proportion of bass. What is the standard deviation of the sampling distribution? Determine whether the 10% condition is met.

A.  The standard deviation is 0.8795. The 10% condition is met because it is very likely there are more than 150 bass in the lake.

B. The standard deviation is 0.8795. The 10% condition is not met because there are less than 150 bass in the lake.

C. The standard deviation is 0.1204. The 10% condition is met because it is very likely there are more than 150 bass in the lake.

D. The standard deviation is 0.1204. The 10% condition is not met because there are less than 150 bass in the lake.

E. We are unable to determine the standard deviation because we do not know the sample mean. The 10% condition is met because it is very likely there are more than 150 bass in the lake

The answer is E.

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2 years ago
Evaluate u + xy, for u = 20, x = 9, and y = 8. 188 37 232 92
givi [52]

let's see

u + xy

Plug in numbers that you know:

20 + 9 * 8

20 + 72

92

8 0
2 years ago
Read 2 more answers
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