Answer:
Choice A: there is insufficient evidence to conclude that the mean for the fiber content is less than 2.5 gms.
Step-by-step explanation:
From the graphs given the more appropriate is one sample t- test
One sample t- test is used and the results are
Hypothesis Test Fiber
Sample Mean 2.62
Sample St. Dev 3.492
Degrees of Freedom 24
t- test statistic 0.1718
p- value 0.5675
the critical region for this test at ∝= 0.05 is t < -1.711
There is not enough evidence to reject the null hypothesis.
Since the calculated t= 0.1718 does not fall in the critical region the null hypothesis is accepted that the mean fiber content is not less than 2.5 grams.T
ΔACB and ΔMNB are similar. Therefore the corresponding sides are in proportion:

Substitute:

Answer: Hello!
Ok, because the bulbs are wired in series, then if only one fails, all the string fails.
Then we need to see the probability for the 20 bulbs to not fail.
If the probability for each bulb to fail is 0.02, then the complement (or the probability of working fine) is 1 - 0.02 = 0.98
then we have 20 bulbs, and each one has a probability of 0.98 of working alright, then the probability for all them to work alright is the multiplication of this probabilities, this is
= 0.6676
rounded up in the decimal, we have 0.668
then the correct answer is c.
Answer:
V = 23π/6
Step-by-step explanation:
V = 2π ∫ [a to b] (r * h) dx
y = −x² + 23x − 132
y = −(x² − 23x + 132)
y = −(x − 11) (x − 12)
Parabola intersects x-axis (line y = 0) at x = 11 and x = 12 ----> a = 11, b = 12
r = x
h = −x² + 23x − 132
V = 2π ∫ [11 to 12] x (−x² + 23x − 132) dx
V = 23π/6
Answer:
There is a 2.28% probability that it takes less than one minute to find a parking space. Since this probability is smaller than 5%, you would be surprised to find a parking space so fast.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X.
Also, a probability is unusual if it is lesser than 5%. If it is unusual, it is surprising.
In this problem:
The length of time it takes to find a parking space at 9 A.M. follows a normal distribution with a mean of 7 minutes and a standard deviation of 3 minutes, so
.
We need to find the probability that it takes less than one minute to find a parking space.
So we need to find the pvalue of Z when 



has a pvalue of 0.0228.
There is a 2.28% probability that it takes less than one minute to find a parking space. Since this probability is smaller than 5%, you would be surprised to find a parking space so fast.