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Montano1993 [528]
2 years ago
6

A-line passes through the points (2, -2) and (-6,2). the point (a,-4) is also in the line. what is the value of A

Mathematics
2 answers:
Tomtit [17]2 years ago
8 0
I believe the answer would be d.6
zepelin [54]2 years ago
6 0

Answer:

d 6  

it is not A B or c its

Step-by-step explanation:

edge 2020

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The department of transportation is repaving 3 1/3 miles of the interstate. If the workers can repave 2/3 of a mile per day, how
Crazy boy [7]

Answer:

5 days

Step-by-step explanation:

(3 1/3 miles)/(2/3 miles/day) = (10/3)/(2/3) days = 10/2 days = 5 days

_____

time = distance / speed

8 0
3 years ago
The product of 5 and y added to 3
olganol [36]
Subtract 5 and 3 then add 3 to the number you get? Idk I think thats what you do
8 0
3 years ago
Read 2 more answers
Each answer choice below shows an equation and a value for x. Which of the following are true? Select three that apply.
Nataliya [291]

Replace x in each equation and see which ones are true.

The answers are:

C. If 5x + 10 =20 then x =2

D. If 6x-12=48, then x=10

F. If 8x(x-4)=32, then x=8

4 0
3 years ago
For a triangle with the length of its sides of 5/36, 1/3 and 13/36, determine if it is a right triangle?
Anuta_ua [19.1K]

Yes it is. Input the numbers into the pythagorean theorem,

{a}^{2}  +  {b}^{2}  =  {c}^{2}

and if they equal each other it is a right triangle.

8 0
3 years ago
The local oil changing business is very busy on Saturday mornings and is considering expanding. A national study of similar busi
eimsori [14]

Answer:

t=\frac{4.2-3.6}{\frac{1.4}{\sqrt{16}}}=1.714

Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131  

Rejection Zone: t_{calculated} or t_{calculated}>2.131

In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.

Step-by-step explanation:

Previous concepts  and data given  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

\bar X=4.2 represent the sample mean  

s=1.4 represent the sample standard deviation  

n=16 represent the sample selected  

\alpha=0.05 significance level  

State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to check if the mean is different from 3.6, the system of hypothesis would be:    

Null hypothesis:\mu = 3.6    

Alternative hypothesis:\mu \neq 3.6    

If we analyze the size for the sample is < 30 and we know the population deviation so 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{4.2-3.6}{\frac{1.4}{\sqrt{16}}}=1.714

Critical values

On this case since we have a bilateral test we need to critical values. We need to use the t distribution with df=n-1=16-1=15 degrees of freedom. The value for \alpha=0.05 and \alpha/2=0.025 so we need to find on the t distribution with 15 degrees of freedom two values that accumulates 0.025 of the ara on each tail. We can use the following excel codes:

"=T.INV(0.025,15)" "=T.INV(1-0.025,15)"

And we got t_{crit}=\pm 2.131    

So the decision on this case would be:

Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131  

Rejection Zone: t_{calculated} or t_{calculated}>2.131

Conclusion    

In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.

3 0
3 years ago
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