Answer:
The mean and standard deviation of the corrected pH measurements are 6.63 and 3.8025 respectively.
Step-by-step explanation:
We can correct the values of the mean and standard deviation using the properties of the mean and the variance.
To the original value X we have to add 0.2 and multiply then by 1.3 to calculate the new and corrected value Y:

The mean and standard deviation of the original value X are 4.9 and 1.5 respectively.
Then, we can apply the properties of the mean as:

For the standard deviation, we apply the properties of variance:

The properties that have been applied are:

Let's set up an inequality. t will equal the amount of hours.
123.80 [greater than or equal to] 47 + 9.60t
After isolating t, your answer is t less than or equal to 8. So they can rent the meeting room for at most 8 hours.
Answer: D
Step-by-step explanation:
8x^3+1 is the answer because we can rewrite it as this : (2x)^3 + 1^3
Answer: No, the normal curve cannot be used.
Step-by-step explanation:
The theorem of the Normal approximation states that if X is B(n,p) then for large n X is N(np, np(1-p)).
The accuracy of this approximation is good
i. for n > [10/p(1-p)]
ii. p is close to 1/2
Hence given p= 4% = 0.04,
q = 1 - 0.04 = 0.96
Let N = [10/p(1-p)]
We find N = 10/p(1-p) = 10/(0.04× 0.96)
N ~= 260
Since n < 260 and p < 0.5
The approximation is not a good one