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2 years ago

Hattie had $6600 to invest and wants to earn 4% interest per year. She will put some of the money into an

Mathematics

1 answer:

34kurt2 years ago

6 0

Answer:

The amount of money that should be invested at the rate of 12% is $900 and the amount of money that should be invested at the rate of 10% is $2,100

Step-by-step explanation:

Let

x ------> the amount of money that should be invested at the rate of 12%

3,000-x -----> the amount money that should be invested at the rate of 10%

we know that

The sum of the interest on each of the accounts must be equal to the total interest.

Solve for x

therefore

The amount of money that should be invested at the rate of 12% is $900 and the amount of money that should be invested at the rate of 10% is $2,100

Hope this helps you!!!! :D

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A random sample of n = 45 observations from a quantitative population produced a mean x = 2.5 and a standard deviation s = 0.26.

oee [108]

Answer:

P-value (t=2.58) = 0.0066.

Note: as we are using the sample standard deviation, a t-statistic is appropiate instead os a z-statistic.

As the P-value (0.0066) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the population mean μ exceeds 2.4.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the population mean μ exceeds 2.4.

Then, the null and alternative hypothesis are:

$H_0: \mu=2.4\\\\H_a:\mu> 2.4$

The significance level is 0.05.

The sample has a size n=45.

The sample mean is M=2.5.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=0.26.

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

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{0.26}{\sqrt{45}}=0.0388$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{2.5-2.4}{0.0388}=\dfrac{0.1}{0.0388}=2.58$

The degrees of freedom for this sample size are:

$df=n-1=45-1=44$

This test is a right-tailed test, with 44 degrees of freedom and t=2.58, so the P-value for this test is calculated as (using a t-table):

$P-value=P(t>2.5801)=0.0066$

As the P-value (0.0066) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the population mean μ exceeds 2.4.

7 0

3 years ago

Simplify the difference (-7×-5×^4+5)-(-7×^4-5-9×)

padilas [110]

-7x-5v^4+5+7x^4+5+9x-=
2x^4+2x+10

6 0

4 years ago

PLZZZ HELP ME IT WILL BE QUICK

sineoko [7]

Answer:

2i x 1.50b

Step-by-step explanation:

4 0

3 years ago

Please help in this question. I WILL mark brainliest. NO LINKS

deff fn [24]

Answer:

B, you are correct

Area of the bigger trapezoid = (6 + 2) x 1/2 x 6 = 8 x 1/2 x 6 = 4 x 6 = 24

Area of the smaller trapezoid = (4 + 2) x 1/2 x 3 = 6 x 1/2 x 3 = 3 x 3 = 9

<h2>Area of the shaded area = 24 - 9 = 15</h2>

<em>Hope that helps! :)</em>

<em>-Aphrodite</em>

Step-by-step explanation:

6 0

3 years ago

NEED HELP ASAP.....At the 2012 Summer Olympic Games in London, in the Men's Shot Put qualifying round, the distances ranged from

jeka57 [31]

Answer:

About 6 athletes.

Step-by-step explanation:

We are concerned only with the percent of athletes that are above one standard deviation.

Between one standard deviation and two standard deviations, there is 13.6% of the athletes.

From 2 to 3, there is 2.1%. Above 3, there is 0.2%.

The sum of those percentages is 15.9% of 40 = 0.159 · 40 = 6.36 or about 6 athletes.

Hope this helps.

6 0

4 years ago