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Brums [2.3K]
3 years ago
12

Jimmy invests $7,000 in an account that pays 3% interest compounded semi-annually. What is his balance after 8 years?

Mathematics
1 answer:
Bumek [7]3 years ago
7 0

Answer:

$8882.9

Step-by-step explanation:

A=p(1+(r/n))^nt

Given:P=7000, r=(3÷100%)=0.03 , n=2, t=8

A = 7000(1+(.03/2))^(2×8)

A = 7000 (1+0.015)^16

A = 7000 × 1.015^16

A = $8882.9

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I want to just get in with life.
Ahat [919]

Answer:

The equation would be y = -2x + 1

Step-by-step explanation:

All you have to do is put it in the form y = mx + b where m is the slope and b is the y-intercept (for example, the equation y= 2x + 1 has a slope of 2 and a y-intercept of 1.)

Since you have slope, or m, = -2 and the y-intercept, or b, = 1, the equation would be y = -2x + 1

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Ilya [14]

answer:

a. scarf costs today = $6

b. scarf costs 3 days from now = $3

step-by-step explanation:

  • 1/2 off means you would multiply the amount by 1/2 then subtract

a. original price of scarf: $12

12 X 1/2 = 6

12 - 6 = 6

— scarf costs today = $6

  • do the same thing but this time it is a different number

b. 12 X 3/4 = 9

12 - 9 = 3

— scarf costs 3 days from now = $3

3 0
3 years ago
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Viefleur [7K]
Hello!

There are only six was to to role a number less than four. 1 and 1. 1 and 2. 2 and 1. 1 and 3. 2 and 2. 3 and 3 and 1.

The first number on the die could be 1, and the second die could have a probability of rolling the second number on it. Giving us a probability of 1/6.

We can subtract the calculated probability from rolling less than four from 1 to find the probability of rolling more than 5:

1 - 1/6 = 5/6

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4 years ago
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A small town has 2000 families. The average number of children per family is mu = 2.5, with a standard deviation sigma = 1.7. A
Nikitich [7]

Answer:

Let X the random variable that represent the number of children per fammili of a population, and for this case we know the following info:

Where \mu=2.5 and \sigma=1.7

We select a sample of n =64 >30 and we can apply the central limit theorem. From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And for this case the standard error would be:

\sigma_{\bar X} = \frac{1.7}{\sqrt{64}}= 0.2125

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

Let X the random variable that represent the number of children per fammili of a population, and for this case we know the following info:

Where \mu=2.5 and \sigma=1.7

We select a sample of n =64 >30 and we can apply the central limit theorem. From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And for this case the standard error would be:

\sigma_{\bar X} = \frac{1.7}{\sqrt{64}}= 0.2125

6 0
4 years ago
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