Answer: P/2 - L = W
Step-by-step explanation:
P = 2L + 2W. We are isolating W. Subtract 2L from both sides to isolate the term first
P - 2L = 2W. Divide by 2 on both sides to isolate W.
P/2 - L = W
Answer:
Step-by-step explanation:
Total number of tickets sold = 3388
Total number of coach tickets = 3069
Total number of first-class tickets = Total number of tickets sold - Total number of coach tickets
= 3388 - 3069
= 319
Ratio of the number of first-class tickets to the total number of tickets = 319:3388
Answer:
Step-by-step explanation:
Given, Total no. of tickets sold = 3388
Total no. of coach tickets = 3069
Then, No. of first class ticket:
= 3388 - 3069
= 319
We need to find the ratio of first-class tickets to the total number of tickets: 319:3388
Answer:
Step-by-step explanation:
For this case we have the following expression:
(1)
And we want to find the value of ln K. If we apply natural log on both sides of the equation (1) we got:
Using the following property:
for x and y real numbers, x>0, y>0, then we have:
Now since the natural log and the exponentiation are inverse operations we have this:
And then the final expression for ln K is :
5*____= 1
Answer:
1
Step-by-step explanation:
just keep the first number and flip the sign and it would be 1 i think idrk
b. P(28c. P(X>35)
d. P(X>31)
e. the mileage rating that the upper 5% of cars achieve.
The upper 5% of cars have a mileage rating of 35.805 mpg
Z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:
z = (raw score - mean) / standard deviation
Given; mean of 33 mpg and a standard deviation of 1.7
a) For < 30:
z = (30 - 33)/1.7 = -1.76
P(x < 30) = P(z < -1.76) = 1 - 0.8413 = 0.0392
b) For < 28:
z = (28 - 33)/1.7 = -2.94
P(x < 28) = P(z < -2.94) = 0.0016
c) For > 35:
z = (35 - 33)/1.7 = 1.18
P(x > 35) = P(z > 1.18) = 1 - P(z < 1.18) = 1 - 0.8810 = 0.119
d) For > 31:
z = (31 - 33)/1.7 = -1.18
P(x > 31) = P(z > -1.18) = 1 - P(z < -1.18) = 0.8810
e) The upper 5% of cars achieve have a z score of 1.65, hence:
1.65 = (x - 33)/1.7
x = 35.805 mpg
The upper 5% of cars have a mileage rating of 35.805 mpg
Find out more on z score at: brainly.com/question/25638875
Answer:
a) P(X < 30) = 0.0392.
b) P(28 < X < 32) = 0.2760
c) P(X > 35) = 0.1190
d) P(X > 31) = 0.8810
e) At least 35.7965 mpg
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 question, we have that:
a. P(X<30)
This is the pvalue of Z when X = 30. So
has a pvalue of 0.0392.
Then
P(X < 30) = 0.0392.
b) P(28 < X < 32)
This is the pvalue of Z when X = 32 subtracted by the pvalue of Z when X = 28. So
X = 32
has a pvalue of 0.2776.
X = 28
has a pvalue of 0.0016.
0.2776 - 0.0016 = 0.2760.
So
P(28 < X < 32) = 0.2760
c) P(X>35)
This is 1 subtracted by the pvalue of Z when X = 35. So
has a pvalue of 0.8810.
1 - 0.8810 = 0.1190
So
P(X > 35) = 0.1190
d. P(X>31)
This is 1 subtracted by the pvalue of Z when X = 31. So
has a pvalue of 0.1190.
1 - 0.1190 = 0.8810
So
P(X > 31) = 0.8810
e. the mileage rating that the upper 5% of cars achieve.
At least the 95th percentile.
The 95th percentile is X when Z has a pvalue of 0.95. So it is X when Z = 1.645. Then
At least 35.7965 mpg
Answer:
19.06
Step-by-step explanation:
Take her wage and multiply by the percent increase
18*.06 =1.08
Add this to her original wage
18+1.06
19.06 is the new wage
4 m
5 m
6 m
Answer: answer is C
Step-by-step explanation: