Answer:
99.1 is the 90th percentile of these scores.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 85
Standard Deviation, σ = 11
We are given that the distribution of score is a bell shaped distribution that is a normal distribution.
Formula:
We have to find the value of x such that the probability is 0.90
Calculation the value from standard normal z table, we have,
Putting values, we get,
Thus, 99.1 is the 90th percentile of these scores.
Answer:
18. Sqrt40 goes between 6 and 7.
19. Sqrt83 ~= 9
Step-by-step explanation:
For both of these problems, it's really useful to be very familiar with the perfect squares on the times table. 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225...
On #18, we're looking at the sqrt40. This will be between sqrt36 and sqrt49, because 40 is between 36 and 49. But sqrt36 is 6 and sqrt49 is 7. So sqrt40 is between 6 and 7.
For #19, the sqrt83 is far closer to sqrt81 than it is to sqrt100. Sqrt81 = 9. So sqrt83 is very close to 9, it would round to 9 and not 10.
39 squared = 39²
'Squared' means to put to the power of 2.
39*39 = 1521
Answer: 39² or 1521
Simplify first the equation,
R(x) = 25000 - 2500x + 6250x - 625x²
Then, we differentiate the equation and equate to zero.
dR(x) = -2500 + 6250 - 1250x = 0
The value of x from the equation is 3.
(1) The price is equal to 50 - 5(3) = $35.
(2) The possible maximum revenue:
R(x) = 2500 - 2500(3) + 6250(3) - 625(3²) = $8125
(3) number of yearbooks sold:
500 + 125(3) = 875 yearbooks