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Lady_Fox [76]
2 years ago
13

PLEASE HELP DUE IN 15 MNUTES MARKING BRAINLIEST AND 15 POINTS YOU DONT HAVE TO EXPLAIN MUCH

Mathematics
1 answer:
hram777 [196]2 years ago
5 0

Answer:

see below

Step-by-step explanation:

11a   -2(x^2 -3)

Distribute

-2*x^2 -2(-3)

-2x^2 +6

11b  2x(x+4)

 Distribute

2x*x +2x*4

2x^2 +8x

12 The perimeter is the sum of all sides

The top is equal to the bottom and the left side is equal to the right

x-10 + x+20+x-10+x+20

Combine like terms

4x +20

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Sum of $12000 from a financing company for 3 years at 15% per annum. What is the total interest he paid in 3 years
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So there are many factors to this, but lets assume the basics:

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An automobile manufacturer who wishes to advertise that one of its models achieves 30 mpg (miles per gallon) decides to carry ou
Katyanochek1 [597]

Answer:

t=\frac{29.35-30}{\frac{1.365}{\sqrt{6}}}=-1.17  

p_v =P(t_{(5)}  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.  

We can say that at 5% of significance the true mean is not significantly less than 30 so then the claim that true average fuel efficiency is (at least) 30 mpg makes sense

Step-by-step explanation:

Data given and notation  

The mean and sample deviation can be calculated from the following formulas:

\bar X =\frac{\sum_{i=1}^n x_i}{n}

s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)}{n-1}}

\bar X=29.35 represent the sample mean  

s=1.365 represent the sample standard deviation  

n=6 sample size  

\mu_o =30 represent the value that we want to test  

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is lower than 30 or no, the system of hypothesis are :  

Null hypothesis:\mu \geq 30  

Alternative hypothesis:\mu < 30  

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{29.35-30}{\frac{1.365}{\sqrt{6}}}=-1.17  

P-value  

We need to calculate the degrees of freedom first given by:  

df=n-1=6-1=5  

Since is a one-side left tailed test the p value would given by:  

p_v =P(t_{(5)}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.  

We can say that at 5% of significance the true mean is not significantly less than 30 so then the claim that true average fuel efficiency is (at least) 30 mpg makes sense

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