Answer:
57.93% probability that a trip will take at least 35 minutes.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean 
and standard deviation 
, the zscore of a measure X is given by:
The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
What is the probability that a trip will take at least 35 minutes
This probability is 1 subtracted by the pvalue of Z when X = 35. So
has a pvalue of 0.4207
1 - 0.4207 = 0.5793
57.93% probability that a trip will take at least 35 minutes.
Answer: Yes my bestfriend
Step-by-step explanation:
A tilted or slanted cylinder has the same volume as its equivalent right cylinder, provided that
(a) the two bases of the tilted cylinder are parallel,
(b) the base areas of the tilted and right cylinders are equal,
(c) the vertical height of the right and tilted cylinders are equal.
This fact can be verified by slicing the tilted cylinder into thin, flat washers, as in integral calculus.
Because all 3 conditions are satisfied, the volume of the tilted cylinder is 450 cm³.
Answer: 430 cm³
#4 Find angles 1, 2, 3 A SO 13 2 A) 90, 40, 50 B) 90, 50, 40 C) 90, 50, 90 D) 90, 40, 90
maksim [4K]
Answer:
HI UR CUTE
Step-by-step explanation:
I JUST FARTED
Answer:
The probability will be 0.3085 or 0
Step-by-step explanation:
Given:
True mean=12.5
Sample mean =12.6
Standard deviation=0.2
Samples=100
To Find:
Probability that exceeds 12.6 ounces.
Solution:
Calculate the Z-score for given means and standard deviation.
So
Z-score= (true mean -sample mean)/standard deviation.
Z-score=(12.5 -12.6)/0.2
=-0.1/0.2
=-0.5
Now Using Z-table
P(X≥-0.5)=p(Z≥-0.5)=0.3085
Hence Probability that sample mean weight exceeds will be 0.3085
OR
By using Normal distribution with sampling ,it will be as follows
Z=(X-u)/[Standard deviation/Sqrt(No of samples)]
Z=(12.6-12.5)/(0.2/Sqrt(100)
Z=0.1/0.2/10
Z=5
So P(X≥12.6 )=P(Z≥5)=1
Pr(Z≥5)=1-1=0.
(Refer the attachment )
Hence Probability of getting ounces greater than 12.6 is '0'.
The sampling is of 0.02 size hence graphically it looks likely.
as shown in attachment.
