Round 565.948063818 to 1 decimal place.

The annual interest rate of Marta's savings account is 7.2% in a simple interest is calculated quarterly. What is the periodic i

kkurt [141]

1. You have the following information:

-The annual interest rate is 7.2%.
- The simple interest is calculated quarterly.

2. Then, to solve this exercise and calculate the periodic interest rate of Marta's Account, you only have to divide the annual interest rate (7.2%) by 1/4 year. So, you have:

=(7.2%)(1/4)
=(7.2%)/4
=1.8%

What is the periodic interest rate of Marta's Account?

The answer is: 1.8%

7 0

2 years ago

What the number is 1/10 the value of 237

Troyanec [42]

1/10 the value of 237 means we must multiply 1/10 by 237

1/10 * 237 = 23.7

Hope this helps!

4 0

2 years ago

Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its average number of unoc

Bingel [31]

Answer:

We need a sample size of at least 719

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

How large a sample size is required to vary population mean within 0.30 seat of the sample mean with 95% confidence interval?

This is at least n, in which n is found when $M = 0.3, \sigma = 4.103$ . So

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.3 = 1.96*\frac{4.103}{\sqrt{n}}$

$0.3\sqrt{n} = 1.96*4.103$

$\sqrt{n} = \frac{1.96*4.103}{0.3}$

$(\sqrt{n})^{2} = (\frac{1.96*4.103}{0.3})^{2}$

$n = 718.57$

Rouding up

We need a sample size of at least 719

6 0

2 years ago

Suppose we have a group of 10 light users and 10 heavy users. what is the probability that exactly 3 of the 20 users are hiv pos

dmitriy555 [2]

Answer: The answer is 30%.

Step-by-step explanation: Given that there is a group of 10 light users and 10 heavy users. We are to find the probability that exactly 3 of the 20 users are HIV positive.

We have the following four possible cases -

(i) All 3 are light users.

(ii) 1 is a light user and 2 are heavy users.

(iii) 2 are light users and 1 is a heavy user.

(iv) All 3 are heavy user.

Since there is a 45% chance of a light user to be HIV positive and 55% chance of a heavy user to be HIV positive, so the required probability is given by

$p\\\\=\dfrac{3\times 0.45+1\times 0.45+2\times 0.55+2\times 0.45+1\times 0.55+3\times 0.55}{20}\\\\=\dfrac{6}{20}\\\\=0.3\\\\=30\%.$

Thus, the probability is 30.

4 0

2 years ago

the number of rainbow smelt in Lake Michigan had an average rate of change of −19.76 per year between 1990 and 2000. The bloater

Nataly_w [17]

Initial population of rainbow smelt: 227.
Initial population of bloater fish: 1,052.
t - number of the years;
The equation is:
227 - 19.76 t = 1,052 - 92.57 t
92.57 t - 19.76 t = 1,052 - 227
72.81 t = 825
t = 825 : 72.81
t = 11.33 years = 11 1/3 years = 11 years 4 months
Answer: After 11 years and 4 months.

5 0

2 years ago

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