Answer:
The probability that in a given day the store receives four or less bad checks is 0.70.
The probability that in a given day the store receives more than 3 bad checks is 0.50.
Step-by-step explanation:
The data provided shows the number of bad checks received by the management of a grocery store for a period of 200 days.
The probability distribution and the cumulative probability distribution are shown in the table attached below.
Let the number of bad checks received in a day be represented by <em>X</em>.
Compute the probability that in a given day the store receives four or less bad checks as follows:
P (X ≤ 4) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)

Thus, the probability that in a given day the store receives four or less bad checks is 0.70.
Compute the probability that in a given day the store receives more than 3 bad checks as follows:
P (X > 3) = 1 - P (X ≤ 3)
= 1 - [P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3)]
![=1-[0.04+0.06+0.10+0.30]\\=1-0.50\\=0.50](https://tex.z-dn.net/?f=%3D1-%5B0.04%2B0.06%2B0.10%2B0.30%5D%5C%5C%3D1-0.50%5C%5C%3D0.50)
Thus, the probability that in a given day the store receives more than 3 bad checks is 0.50.
This is asking you to interpret the slope. In this case, the +20 is just a baseline number and irrelevant to the question. Your slope is 10 which means for every car washed, they gain $10.
Answer:
(f^−1)′(6)=1/(f'(f^-1(6)))
(f^−1)′(6)=1/(f'(f^-6)))
I hope this helps.
Answer:
The minimum score required for an interview is 77.252
Step-by-step explanation:
We solve this using z score formula
z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.
Top 15% of the candidates is a ranking that is equivalent to = 100 - 15% = 85th percentile.
The z score of 85th percentile = 1.036
Mean = 70
Standard Deviation = 7
Minimum score = raw score = ???
Hence:
1.036 = x - 70/7
Cross Multiply
1.036 × 7 = x - 70
7.252 = x - 70
x = 70 + 7.252
x = 77.252
The minimum score required for an interview is 77.252