What is the solution to 5-2x^2=-15?

Terrence gets a loan with a $50 processing fee. The loan is for $700 with an interest rate of 8% for one

agasfer [191]

Answer:

15.14%

Step-by-step explanation:

The formula for APR is stated thus:

APR=fees+interest/principal/n*365*100

principal is the loan amount of $700

fees is the processing fees on the loan which is $50

interest amount=principal*interest %=$700*8%=$56

n is the number of days of the loan which is a year i.e 365 days

APR=($50+$56)/$700/365*365*100

APR=$106/$700/365*365*100

APR=0.151428571 /365*365*100

APR=0.151428571 *100=15.14%

The annual percentage rate on the loan is 15.14% which represents the actual cost on the loan not just the interest cost of 8% annually

4 0

3 years ago

My son is having such a hard time with 6th grade math. and I'm no help at all. someone help please

lisabon 2012 [21]

I'd be more than happy to help. Im a sophmore and Love math

7 0

3 years ago

Read 2 more answers

90 3/7 was a demical

LenKa [72]

90.42857143 or
90.4 to nearest decimal

6 0

3 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

3 years ago

Help please and thank you so much! <br> i inserted the picture of the questions

Bond [772]

1. Points F and C
2. 2
3. (3,4)
4. (1,-2)
5. Quadrant 4
6. (-3,-3)
(I also added a photo of a graph with the labeled quadrants to help you)

6 0

2 years ago