All categories

2 years ago

What is the monthly payment rate for a $20,000 loan for 4 years with an annual interest rate of 4.8%

Mathematics

1 answer:

yarga [219]2 years ago

5 0

Answer:

96048

Step-by-step explanation:

times 20000 by 4.8 then times 4 by 12 and add your answers

You might be interested in

Victoria has $200 of her birthday gift money saved at home, and the amount is modeled by the function h(x) = 200. She reads abou

olchik [2.2K]

If Victoria puts 200 dollars a savings account she will earn 1.05 for x amount of time.<span>
</span>

7 0

2 years ago

Read 2 more answers

When the net is folded into the rectangular prism shown beside it, which letters will be on the front and back of the rectangula

earnstyle [38]

<h2>Answer:</h2>

<u>The correct option is </u><u>The letter on the front will be N. The letter on the back will be L. </u>

<h2>Step-by-step explanation:</h2>

When we fold the given net, we will get Q,P,M and N on sides. Side M will come to the top, side Q on the right side, side P on the left and side O on the bottom. The side which comes to the front will be N of the observer and similarly the side L will come to the back of the rectangular prism.

4 0

2 years ago

Read 2 more answers

A daffodil grows 0.05m every day. Plot the growth of the flower if the initial length of the daffodil is 0.8m and hence give the

Vanyuwa [196]

Answer:

1.2m

Step-by-step explanation:

You must first find out how much the daffodil grew over the 8 days:

0.05 x 8 = 0.4

Then you must add how much it grew to the original height:

0.4 + 0.8 = 1.2

Hope this helps you out! : )

8 0

2 years ago

Which statement is true regarding the functions on the graph?

Mrrafil [7]

Answer: f(3)=g(3)
I believe that is correct.
Tell me if I’m wrong

3 0

2 years ago

Two machines are used for filling glass bottles with a soft-drink beverage. The filling process have known standard deviations s

stellarik [79]

Answer:

a. We reject the null hypothesis at the significance level of 0.05

b. The p-value is zero for practical applications

c. (-0.0225, -0.0375)

Step-by-step explanation:

Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.

Then we have $n_{1} = 25$ , $\bar{x}_{1} = 2.04$ , $\sigma_{1} = 0.010$ and $n_{2} = 20$ , $\bar{x}_{2} = 2.07$ , $\sigma_{2} = 0.015$ . The pooled estimate is given by

$\sigma_{p}^{2} = \frac{(n_{1}-1)\sigma_{1}^{2}+(n_{2}-1)\sigma_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(25-1)(0.010)^{2}+(20-1)(0.015)^{2}}{25+20-2} = 0.0001552$

a. We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} \neq 0$ (two-tailed alternative).

The test statistic is $T = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}}$ and the observed value is $t_{0} = \frac{2.04 - 2.07}{(0.01246)(0.3)} = -8.0257$ . T has a Student's t distribution with 20 + 25 - 2 = 43 df.

The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value $t_{0}$ falls inside RR, we reject the null hypothesis at the significance level of 0.05

b. The p-value for this test is given by $2P(T$ 0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.

c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)

$(\bar{x}_{1}-\bar{x}_{2})\pm t_{0.05/2}s_{p}\sqrt{\frac{1}{25}+\frac{1}{20}}$ , i.e.,

$-0.03\pm t_{0.025}0.012459\sqrt{\frac{1}{25}+\frac{1}{20}}$

where $t_{0.025}$ is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So

$-0.03\pm(2.0167)(0.012459)(0.3)$ , i.e.,

(-0.0225, -0.0375)

8 0

2 years ago