All categories

3 years ago

: $800 invested at a 9% interest rate for 8 years.

Mathematics

2 answers:

Korvikt [17]3 years ago

6 0

Answer:

576$

Step-by-step explanation:

$800×9/100×8=576$

Colt1911 [192]3 years ago

3 0

You would make $576. 800 at a 9% rate is $72 per year. 72x8 years is $576

You might be interested in

Simplify (2x+3y)+(7x-3y)

Radda [10]

(2x+3y)+(7x-3y) First, group like terms together, so (2x+7x) + (3y-3y). Then simplify within the parentheses. 2x+7x = 9x and 3y-3y = 0, so you have 9x, which cannot be simplified any further. The answer is 9x.

3 0

3 years ago

Read 2 more answers

Can someone help me my class is at 9

ololo11 [35]

Answer: 1/4

Step-by-step explanation: 20 is a fourth of 80

Please mark brainleist if I got it right. I really need it to level up

6 0

2 years ago

What is a number written to the thousandths place that would relate to 417.91 using the = symbol?

baherus [9]

Answer:

0.41791

Step-by-step explanation:

let x be the number written to the thousandth place then,

x times thousands = 417.91

or, x = 417.91/1000

therefore, x = 0.41791

and so

0.41791 thousand = 417.91

3 0

3 years ago

Read 2 more answers

A candidate for a US Representative seat from Indiana hires a polling firm to gauge her percentage of support among voters in he

mojhsa [17]

Answer:

(A) The minimum sample size required achieve the margin of error of 0.04 is 601.

(B) The minimum sample size required achieve a margin of error of 0.02 is 2401.

Step-by-step explanation:

Let us assume that the percentage of support for the candidate, among voters in her district, is 50%.

(A)

The margin of error, MOE = 0.04.

The formula for margin of error is:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}$

The critical value of z for 95% confidence interval is: $z_{\alpha/2}=1.96$

Compute the minimum sample size required as follows:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}\\0.04=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\(\frac{0.04}{1.96})^{2} =\frac{0.50(1-0.50)}{n}\\n=600.25\approx 601$

Thus, the minimum sample size required achieve the margin of error of 0.04 is 601.

(B)

The margin of error, MOE = 0.02.

The formula for margin of error is:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}$

The critical value of z for 95% confidence interval is: $z_{\alpha/2}=1.96$

Compute the minimum sample size required as follows:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}\\0.02=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\(\frac{0.02}{1.96})^{2} =\frac{0.50(1-0.50)}{n}\\n=2401.00\approx 2401$

Thus, the minimum sample size required achieve a margin of error of 0.02 is 2401.

8 0

3 years ago

PLS HELP 20 POINTS!! WILL GIVE BRAINIEST

umka2103 [35]

B- 18 units to the left of zero

7 0

2 years ago