I don't know if this is the equation you need. But here it is.
Y - 4 = -2 (x + 1)
Answer: It is B and C.
Step-by-step explanation:
If you divde the -4 and -5 for choice B, it will become 4/5 since it's an absolute value.
For Choice C, it's also the same but before the fractions were negative and there is a negative sign which means it's also the aboslute value which is 4/5.
Hope this helps! <3
Answer:
Paired samples t- test is appropriate for Dean's hypothesis
Step-by-step explanation:
Given Data :
Student 1 2 3 4 5 6 7 8
Marketing 82 86 74 93 90 76 87 100
Finance 76 91 70 79 96 70 85 81
We are supposed to find the test which is appropriate for Dean's hypothesis
We are given the grades of two different subjects for same student
We perform the paired t test because the final grades for the Marketing and Finance is taken for the same student.
So, Option B is correct
Hence Paired samples t- test is appropriate for Dean's hypothesis
Answer:
P(10 ≤ x ≤ 12) = 0.4274
Step-by-step explanation:
Population mean = u = 10
Population Standard Deviation =
= 9
Sample size = n = 43
Sample mean(
) is equal to the population mean. So,
Sample mean =
= 10
Sample standard deviation(
) is equal to population standard deviation divided by square root of sample size. So,
Sample standard deviation =
= 
We have to find the probability that for a random sample of n = 43, the value lies between 10 and 12 i.e. P(10 ≤ x ≤ 12)
P(10 ≤ x ≤ 12) = P(x ≤ 12) - P( x ≤ 10)
We can find P(x ≤ 12 ) and P(x ≤ 10) by converting these values to z scores.
The formula for z score is:

For x =12, we get:

For x =10, we get:

So,
P(x ≤ 12) - P( x ≤ 10) = P(z ≤ 1.457) - P(z ≤ 0)
From the z table,
P(z ≤ 1.457) = 0.9274
P(z ≤ 0) = 0.5
So,
P(x ≤ 12) - P( x ≤ 10) = P(z ≤ 1.458) - P(z ≤ 0) = 0.9274 - 0.5 = 0.4274
So,
P(10 ≤ x ≤ 12) = P(x ≤ 12) - P( x ≤ 10) = 0.4274
Therefore,
The probability that for a random sample of size 43, the mean lies between 10 and 12 is 0.4274.