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Hunter-Best [27]
2 years ago
14

Marissa deposited $900 in her savings account. the rate of simple interest is 5% per year. find the balance at the end of 4 year

s
Mathematics
1 answer:
Anvisha [2.4K]2 years ago
5 0

Answer:

1080

Step-by-step explanation:

$900.00 per year at 5% is 45.00

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A carpenter is making doors that are 20582058 millimeters tall. If the doors are too long they must be trimmed, and if they are
melamori03 [73]

Answer:

Z= 0.253

Z∝/2 = ± 1.96

Step-by-step explanation:

Formulate the null and alternative hypotheses as

H0 : u1= u2 against Ha : u1≠ u2 This is a two sided test

Here ∝= 0.005

For alpha by 2 for a two tailed test Z∝/2 = ± 1.96

Standard deviation = s= 15

n= 10

The test statistic used here is

Z = x- x`/ s/√n

Z= 2058- 2046 / 15 / √10

Z= 0.253

Since the calculated value of Z= 0.253 falls in the critical region we reject the null hypothesis.

There is  evidence at the 0.05 level that the doors are too short and unusable.

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3 years ago
A financial talk show host claims to have a 55.3 % success rate in his investment recommendations. You collect some data over th
baherus [9]

Answer:

There is a 25.52% probability of observating 4 our fewer succesful recommendations.

Step-by-step explanation:

For each recommendation, there are only two possible outcomes. Either it was a success, or it was a failure. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

In this problem we have that:

p = 0.553, n = 10

If the claim is correct and the performance of recommendations is independent, what is the probability that you would have observed 4 or fewer successful:

This is

P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{10,0}.(0.553)^{0}.(0.447)^{10} = 0.0003

P(X = 1) = C_{10,1}.(0.553)^{1}.(0.447)^{9} = 0.0039

P(X = 2) = C_{10,2}.(0.553)^{2}.(0.447)^{8} = 0.0219

P(X = 3) = C_{10,3}.(0.553)^{3}.(0.447)^{7} = 0.0724

P(X = 4) = C_{10,4}.(0.553)^{4}.(0.447)^{6} = 0.1567

P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0003 + 0.0039 + 0.0219 + 0.0724 + 0.1567 = 0.2552

There is a 25.52% probability of observating 4 our fewer succesful recommendations.

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