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3 years ago

5

you deposit $5,000 in an account that pays 6% annual interest. find the balance after 5 years when the interest is compounded qu

arterly

1 answer:

NeTakaya3 years ago

8 0

Answer:

15,000

Step-by-step explanation:

5000-*0.6=3000 *5 =15,000

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If a candy machine contains 76 green candies, 42 orange candies, 58 red

sergiy2304 [10]

Answer:

P(candy chosen is red) = 58/200

Step-by-step explanation:

6 0

3 years ago

10. What is the distance between (8,7) and (3,-5)?

Anarel [89]

Answer:

distance between points=13 units

Step-by-step explanation:

distance between points= $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

put $(x_1,x_2)$ =(8,7) and $(y_1,y_2)$ =(3,-5)

distance between points= $\sqrt{(3-8)^2+(-5-7)^2}$

distance between points= $\sqrt{(-5)^2+(-12)^2}$

distance between points= $\sqrt{25+144}$

distance between points= $\sqrt{169}$

distance between points=13 units answer

8 0

3 years ago

Can you help I have tried 4 times

docker41 [41]

Answer:

5120oz/12=426.667 ounces.

Step-by-step explanation:

We know 1 pound = 16 oz. If we want to find out how many ounces are in 320 lbs, we must multiply 16 by 320 = 5120 ounces. Now we must divide by 12 to see how much she ate per month. 5120 divided by 12 is 426.667 or 426 and 2/3 ounces per month.

Alternatively, we can set up a proportion 1 lb / 16 ounces = 320 lb / x ounces. We cross multiply 1 times x = 320 times 15. On the left hand side we just have x and on the right hand side we get 5120. Hence, x=5120 ounces per year. To find how many she ate per month we must divide by 12 since there are 12 months in a year. So it comes out to 426 and 2/3 ounces per month.

Here are the equations: (1lb)(320lb)=(16oz)(320oz) implies 320lb=5120oz. Then divide 5120oz 12 to get how much she ate per month: 5120oz/12=426.667 ounces.

7 0

3 years ago

Solve by any method <br> {3x+2y=11 <br> {x-2=-4y<br> PLEEEAASSSEEEE HELP ME!

bixtya [17]

Answer:

Step-by-step explanation:

(x,y)=(4,-1/2)

3 0

3 years ago

Read 2 more answers

suppose a sample of 1305 tenth graders is drawn. Of the students sampled, 1044 read above the eighth grade level. Using the data

Inessa05 [86]

Answer: $(0.771,\ 0.829)$

Step-by-step explanation:

Given : Significance level : $\alpha: 1-0.99=0.01$

Critical value : $z_{\alpha/2}=2.576$

Sample size : n=1305

The proportion of tenth graders read above the eighth grade level. = $p=\dfrac{1044}{1305}=0.8$

The confidence interval for population proportion is given by :-

$p\pm z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}\\\\=0.8\pm(2.576)\sqrt{\dfrac{0.8(1-0.8)}{1305}}\\\\\approx0.8\pm0.029\\\\=(0.771,\ 0.829)$

Hence, the 99% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.771,0.829) .

4 0

3 years ago