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Anton [14]
3 years ago
8

Karen deposited $5,000 as a certificate of deposit (CD) in a bank for a period of 3 years. The CD pays simple interest of 15% pe

r year and pays interest every 6 months. How much interest does she get every 6 months?
$

(Hint: Use the formula i = prt, where i is the interest earned, p is the principal (starting amount), r is the interest rate expressed as a decimal, and t is the time in years.)
Mathematics
1 answer:
Serhud [2]3 years ago
3 0
375 every 6 months ..........
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Triangle ABC contains side lengths b=3 inches and c= 5 inches in two or more sentences describe whether or not it is possible fo
Scrat [10]

Answer:

Since sine of all angles are always less than one, this shows there is no possible way to have an angle C. Thus it is impossible to have a triangle ABC with the given properties of side lengths b=3 inches and c= 5 inches to have angle B =45 degrees.

Step-by-step explanation:

In the attached  drawing, each of the tic-marks are equal and

represent 1 inch each.  The angle B has measure 45.  We can

see by the arc that the line AC, which equals 3 inches, is

not long enough to reach the slanted side of the 45 angle.

Therefore triangle ABC is not possible.  We can also show

by the law of sines that no triangle ABC with the given

properties in possible.

5 0
3 years ago
What is the measures of the angles in a triangle?
Talja [164]

4 is 90

1 is 37

3

2 is 127

3 is 23

3 0
3 years ago
PLEASE ANSWER ASAP
Taya2010 [7]

I would say, well I THINK It's 2

4 0
3 years ago
Read 2 more answers
Consider the following hypothesis test:
postnew [5]

Answer:

a. P-value = 0.039.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that the population mean significantly differs from 100.

b. P-value = 0.013.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that the population mean significantly differs from 100.

c. P-value = 0.130.

The null hypothesis failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the population mean significantly differs from 100.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the population mean significantly differs from 100.

Then, the null and alternative hypothesis are:

H_0: \mu=100\\\\H_a:\mu\neq 100

The significance level is 0.05.

The sample has a size n=65.

The degrees of freedom for this sample size are:

df=n-1=65-1=64

a. The sample mean is M=103.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=11.5.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{11.5}{\sqrt{65}}=1.4264

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{103-100}{1.4264}=\dfrac{3}{1.4264}=2.103

 

This test is a two-tailed test, with 64 degrees of freedom and t=2.103, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=2\cdot P(t>2.103)=0.039

As the P-value (0.039) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that the population mean significantly differs from 100.

b. The sample mean is M=96.5.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=11.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{11}{\sqrt{65}}=1.3644

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{96.5-100}{1.3644}=\dfrac{-3.5}{1.3644}=-2.565

This test is a two-tailed test, with 64 degrees of freedom and t=-2.565, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=2\cdot P(t

As the P-value (0.013) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that the population mean significantly differs from 100.

c. The sample mean is M=102.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=10.5.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{10.5}{\sqrt{65}}=1.3024

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{102-100}{1.3024}=\dfrac{2}{1.3024}=1.536

This test is a two-tailed test, with 64 degrees of freedom and t=1.536, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=2\cdot P(t>1.536)=0.130

As the P-value (0.13) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the population mean significantly differs from 100.

8 0
3 years ago
Find the value of each variable. Round your answers to the nearest tenth. (Soh Coh Toa)
Butoxors [25]

Answer:

k = 8.4, j = 13.1

Step-by-step explanation:

Solve for k:

tan(40)=\frac{k}{10}\\10(tan(40))=k\\8.4=k

Solve for j:

8.4^2+10^2=j^2\\70.56+100=j^2\\170.56=j^2\\13.1=j

3 0
3 years ago
Read 2 more answers
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