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3 years ago

Suppose that $1200 is invested at 612%, compounded quarterly. How much is in the account at the end of 5 years?

Mathematics

1 answer:

asambeis [7]3 years ago

5 0

Answer:

$\$1,656.50$

Step-by-step explanation:

we know that

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

in this problem we have

$t=5\ years\\ P=\$1,200\\ r=6\frac{1}{2}\%=6.5\%=6.5/100=0.065\\n=4$

substitute in the formula above

$A=1,200(1+\frac{0.065}{4})^{4*5}$

$A=1,200(1.01625)^{20}$

$A=\$1,656.50$

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bekas [8.4K]

If one inch represents 15 feet, then:

2 inches represent:
2 * 15 = 30 feet

1 2/3 inches represent:
1 2/3 * 15 =
= 5/3 * 15 =
= 75/3 =
= 25 feet

So the real area of the court is 30 * 25 = 750 square feet.

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3 years ago

Your cousin gave you 27 dollars with which to buy a present. this covered 9/10 of the cost. how much did the present cost

yaroslaw [1]

27/10 is 2.7
2.7*9 gets you $24.3

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3 0

3 years ago

Read 2 more answers

Suppose that in a random selection of 100 colored candies, 24% of them are blue. The candy company claims that the percentage

LuckyWell [14K]

Answer:

We conclude that the percentage of blue candies is equal to 23% at 0.01 significance level.

Step-by-step explanation:

We are given that in a random selection of 100 colored candies, 24% of them are blue.

The candy company claims that the percentage of blue candies is equal to 23%.

Let p = percentage of blue candies.

So, Null Hypothesis, $H_0$ : p = 23% {means that the percentage of blue candies is equal to 23%}

Alternate Hypothesis, $H_A$ : p $\neq$ 23% {means that the percentage of blue candies is different from 23%}

The test statistics that would be used here One-sample z proportion test statistics;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n}} }$ ~ N(0,1)

where, $\hat p$ = sample proportion of blue colored candies = 24%

n = sample of colored candies = 100

So, test statistics = $\frac{0.24-0.23}{\sqrt{\frac{0.24(1-0.24)}{100}} }$

= 0.234

The value of z test statistics is 0.234.

Now, at 0.01 significance level the z table gives critical values of -2.5758 and 2.5758 for two-tailed test.

Since our test statistics lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the percentage of blue candies is equal to 23%.

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3 years ago

Find the slope of the line that passes through (6, 68) and (54,-7)

soldi70 [24.7K]

Answer:

-25/16

Step-by-step explanation:

m=(y2-y1)/(x2-x1)

m=(-7-68)/(54-6)

m=-75/48

m=-25/16

7 0

2 years ago

Read 2 more answers

A grocery store’s receipts show that Sunday customer purchases have a skewed distribution with a mean of 27$ and a standard devi

34kurt

Answer:

(a) The probability that the store’s revenues were at least $9,000 is 0.0233.

(b) The revenue of the store on the worst 1% of such days is $7,631.57.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sum of values of X, i.e ∑X, will be approximately normally distributed.

Then, the mean of the distribution of the sum of values of X is given by,

$\mu_{X}=n\mu$

And the standard deviation of the distribution of the sum of values of X is given by,

$\sigma_{X}=\sqrt{n}\sigma$

It is provided that:

$\mu=\$27\\\sigma=\$18\\n=310$

As the sample size is quite large, i.e. n = 310 > 30, the central limit theorem can be applied to approximate the sampling distribution of the store’s revenues for Sundays by a normal distribution.

(a)

Compute the probability that the store’s revenues were at least $9,000 as follows:

$P(S\geq 9000)=P(\frac{S-\mu_{X}}{\sigma_{X}}\geq \frac{9000-(27\times310)}{\sqrt{310}\times 18})\\\\=P(Z\geq 1.99)\\\\=1-P(Z$

Thus, the probability that the store’s revenues were at least $9,000 is 0.0233.

(b)

Let s denote the revenue of the store on the worst 1% of such days.

Then, P (S < s) = 0.01.

The corresponding z-value is, -2.33.

Compute the value of s as follows:

$z=\frac{s-\mu_{X}}{\sigma_{X}}\\\\-2.33=\frac{s-8370}{316.923}\\\\s=8370-(2.33\times 316.923)\\\\s=7631.56941\\\\s\approx \$7,631.57$

Thus, the revenue of the store on the worst 1% of such days is $7,631.57.

5 0

3 years ago