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Naddika [18.5K]
3 years ago
7

Ernie Rolph deposits $4500 at 2% annual simple interest for 1 year. How much interest will he earn?

Mathematics
1 answer:
vaieri [72.5K]3 years ago
8 0

Answer: the amount of interest that he would earn is $90.

Step-by-step explanation:

The formula for determining simple interest is expressed as

I = PRT/100

Where

I represents interest paid on the amount of money deposited.

P represents the principal or amount of money deposited.

R represents interest rate on the deposit.

T represents the duration of the deposit in years.

From the information given,

P = $4500

R = 2%

T = 1 year

Therefore,

I = (4500 × 2 × 1)/100

I = $90

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Luis incurred 107 extra text messages.

Step-by-step explanation:

Giving the following information:

Luis's cell phone plan costs $39 per month. Text messages cost an additional $0.15 each.

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The following data represent the pH of rain for a random sample of 12 rain dates in Tucker County, West Virginia. A normal proba
Sphinxa [80]

Answer:

Step-by-step explanation:

n = 12

Mean = (4.58 + 5.72 + 4.77 + 4.76 + 5.19 + 5.05 + 4.80 + 4.77 + 4.75 + 5.02 + 4.74 + 4.56)/12 = 4.8925

Standard deviation = √(summation(x - mean)²/n

Summation(x - mean)² = (4.58 - 4.8925)^2 + (5.72 - 4.8925)^2 + (4.77 - 4.8925)^2 + (4.76 - 4.8925)^2 + (5.19 - 4.8925)^2 + (5.05 - 4.8925)^2 + (4.80 - 4.8925)^2 + (4.77 - 4.8925)^2 + (4.75 - 4.8925)^2 + (5.02 - 4.8925)^2 + (4.74 - 4.8925)^2 + (4.56 - 4.8925)^2 = 1.122225

Standard deviation = √(1.122225/12

s = 0.31

a) Point estimate = sample mean = 4.8925

Confidence interval is written in the form,

(Sample mean - margin of error, sample mean + margin of error)

Margin of error = z × s/√n

Where

From the information given, the population standard deviation is unknown and the sample size is small, hence, we would use the t distribution to find the z score

In order to use the t distribution, we would determine the degree of freedom, df for the sample.

df = n - 1 = 12 - 1 = 11

b) Since confidence level = 95% = 0.95, α = 1 - CL = 1 – 0.95 = 0.05

α/2 = 0.05/2 = 0.025

the area to the right of z0.025 is 0.025 and the area to the left of z0.025 is 1 - 0.025 = 0.975

Looking at the t distribution table,

z = 2.201

Margin of error = 2.201 × 0.31/√12

= 0.197

95% confidence interval = 4.8925 ± 0.197

Upper limit = 4.8925 + 0.197 = 5.0895

Lower limit = 4.8925 - 0.197 = 4.6955

We are 95% confident that the population mean of the rain water ph lies between 4.6955 and 5.0895

c) For 99% confidence level, z = 3.106

Margin of error = 3.106 × 0.31/√12

= 0.278

99% confidence interval = 4.8925 ± 0.278

Upper limit = 4.8925 + 0.278 = 5.1705

Lower limit = 4.8925 - 0.278 = 4.6145

We are 99% confident that the population mean of the rain water ph lies between 4.6145 and 5.1705

d) The interval gets wider as the confidence level is increased. This is logical since the test score is higher for 99% and therefore, increases the range of values. Since we want to be more confident, the range of values must be extended.

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