Answer:
It would be B
Step-by-step explanation:
You can check each question by subsituting the x and y with points on the graph.
Example: (1,3)
y = 3x
3 = 1 x 3
3 = 3
Using the normal distribution, it is found that 2.64% of all the nails produced by this machine are unusable.
In a <em>normal distribution</em> with mean
and standard deviation
, the z-score of a measure X is given by:
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
In this problem:
- The mean is of 3 inches, hence
.
- The standard deviation is of 0.009 inches, hence
.
Nails that are <u>more than 0.02 inches</u> from the mean are unusable, hence:



The proportion is P(|Z| > 2.22), which is <u>2 multiplied by the p-value of Z = -2.22</u>.
Z = -2.22 has a p-value of 0.0132.
2 x 0.0132 = 0.0264
0.0264 x 100% = 2.64%
2.64% of all the nails produced by this machine are unusable.
You can learn more about the normal distribution at brainly.com/question/24663213
Answer:
2x³+17x²+25x-50
Step-by-step explanation:
Standard Form: All your numbers and variables in order of the highest exponent.
(2x²+7x-10)(x+5)
We want to distribute x and 5 to every single value.
2x³+7x²-10x+10x²+35x-50
Simplify.
2x³+17x²+25x-50
This is standard form because the highest values are on the left (x³ is larger than x²).
Umm I don’t know the answer to that yet because it looks complicated
Answer:
By using hypothesis test at α = 0.01, we cannot conclude that the proportion of high school teachers who were single greater than the proportion of elementary teachers who were single
Step-by-step explanation:
let p1 be the proportion of elementary teachers who were single
let p2 be the proportion of high school teachers who were single
Then, the null and alternative hypotheses are:
: p2=p1
: p2>p1
We need to calculate the test statistic of the sample proportion for elementary teachers who were single.
It can be calculated as follows:
where
- p(s) is the sample proportion of high school teachers who were single (
) - p is the proportion of elementary teachers who were single (
)
- N is the sample size (180)
Using the numbers, we get
≈ 1.88
Using z-table, corresponding P-Value is ≈0.03
Since 0.03>0.01 we fail to reject the null hypothesis. (The result is not significant at α = 0.01)