68 divided by 4 is 17 and 17 times 7 (the amount of people watching) is 119
The Equation of the Line that Passes through (4, 2) and is Parallel to 3x – 2y = –6 is 3x – 2y = 8
95% of red lights last between 2.5 and 3.5 minutes.
<u>Step-by-step explanation:</u>
In this case,
- The mean M is 3 and
- The standard deviation SD is given as 0.25
Assume the bell shaped graph of normal distribution,
The center of the graph is mean which is 3 minutes.
We move one space to the right side of mean ⇒ M + SD
⇒ 3+0.25 = 3.25 minutes.
Again we move one more space to the right of mean ⇒ M + 2SD
⇒ 3 + (0.25×2) = 3.5 minutes.
Similarly,
Move one space to the left side of mean ⇒ M - SD
⇒ 3-0.25 = 2.75 minutes.
Again we move one more space to the left of mean ⇒ M - 2SD
⇒ 3 - (0.25×2) =2.5 minutes.
The questions asks to approximately what percent of red lights last between 2.5 and 3.5 minutes.
Notice 2.5 and 3.5 fall within 2 standard deviations, and that 95% of the data is within 2 standard deviations. (Refer to bell-shaped graph)
Therefore, the percent of red lights that last between 2.5 and 3.5 minutes is 95%
Answer:
Size of each group = 
Step-by-step explanation:
Total number of paint brushes = x
Since each the three groups are equally sized, to get the number of paintbrushes in a group, we will have to divide the total number of paint brushes by 3
∴ We have, number of paint brushes in a group = 
= 
this is also the same thing as saying
×
This shows that the size of the new group is the same as
the size of the original group
Answer:
Step-by-step explanation
Hello!
Be X: SAT scores of students attending college.
The population mean is μ= 1150 and the standard deviation σ= 150
The teacher takes a sample of 25 students of his class, the resulting sample mean is 1200.
If the professor wants to test if the average SAT score is, as reported, 1150, the statistic hypotheses are:
H₀: μ = 1150
H₁: μ ≠ 1150
α: 0.05
![Z= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } } ~~N(0;1)](https://tex.z-dn.net/?f=Z%3D%20%5Cfrac%7BX%5Bbar%5D-Mu%7D%7B%5Cfrac%7BSigma%7D%7B%5Csqrt%7Bn%7D%20%7D%20%7D%20~~N%280%3B1%29)

The p-value for this test is 0.0949
Since the p-value is greater than the level of significance, the decision is to reject the null hypothesis. Then using a significance level of 5%, there is enough evidence to reject the null hypothesis, then the average SAT score of the college students is not 1150.
I hope it helps!