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timofeeve [1]
3 years ago
5

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

Mathematics
1 answer:
Phantasy [73]3 years ago
6 0
Isolate x.

-2x^{2}=-20
x^{2}=10
x=\sqrt{10}


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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

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90 3/7 was a demical
LenKa [72]
90.42857143 or
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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(T0 (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
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1. Points F and C
2. 2
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