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MA_775_DIABLO [31]

3 years ago

6

Ralph deposited $3,000 into a bank account that earn simple interest each year after 3.5years he had earn $226.50 and interest i

f no money was deposited into or withdrawn from the account what was the annual interest rate

1 answer:

PilotLPTM [1.2K]3 years ago

4 0

Answer:

r=0.02147=2.147%

Step-by-step explanation:

Interest=principal X interest rate X time(yearly)

$226.50=$3000 * R * 3.5year

r=0.02147=2.147%

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2 years ago

Exponetial funtion is

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4 0

3 years ago

the variables x and y vary inversely with a constant variation of 7. Use the variation formula, k=xy, to find why when x=9

dolphi86 [110]

1.Make y the subject

k = xy

k/x = xy/x

y = k/x

2.Substitute

y = k/x

y = 7/9

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To confirm

k = xy

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Hope it helps☺

7 0

2 years ago

The weight of an adult swan is normally distributed with a mean of 26 pounds and a standard deviation of 7.2 pounds. A farmer ra

Snezhnost [94]

Let $X$ denote the random variable for the weight of a swan. Then each swan in the sample of 36 selected by the farmer can be assigned a weight denoted by $X_1,\ldots,X_{36}$ , each independently and identically distributed with distribution $X_i\sim\mathcal N(26,7.2)$ .

You want to find

$\mathbb P(X_1+\cdots+X_{36}>1000)=\mathbb P\left(\displaystyle\sum_{i=1}^{36}X_i>1000\right)$

Note that the left side is 36 times the average of the weights of the swans in the sample, i.e. the probability above is equivalent to

$\mathbb P\left(36\displaystyle\sum_{i=1}^{36}\frac{X_i}{36}>1000\right)=\mathbb P\left(\overline X>\dfrac{1000}{36}\right)$

Recall that if $X\sim\mathcal N(\mu,\sigma)$ , then the sampling distribution $\overline X=\displaystyle\sum_{i=1}^n\frac{X_i}n\sim\mathcal N\left(\mu,\dfrac\sigma{\sqrt n}\right)$ with $n$ being the size of the sample.

Transforming to the standard normal distribution, you have

$Z=\dfrac{\overline X-\mu_{\overline X}}{\sigma_{\overline X}}=\sqrt n\dfrac{\overline X-\mu}{\sigma}$

so that in this case,

$Z=6\dfrac{\overline X-26}{7.2}$

and the probability is equivalent to

$\mathbb P\left(\overline X>\dfrac{1000}{36}\right)=\mathbb P\left(6\dfrac{\overline X-26}{7.2}>6\dfrac{\frac{1000}{36}-26}{7.2}\right)$
$=\mathbb P(Z>1.481)\approx0.0693$

5 0

2 years ago