Average rate of change = [H(100) - H(80)] / (100 - 80)
H(100) = 0.003(100)^2 + 0.07(100) - 0.027 = 0.003(10000) + 0.07(100) - 0.027 = 30 + 7 - 0.027 = 36.973
H(80) = 0.003(80)^2 + 0.07(80) - 0.027 = 0.003(6400) + 0.07(80) - 0.027 = 19.2 + 5.6 - 0.027 = 24.773
Average rate of change = (36.973 - 24.773)/(100 - 80) = 12.2/20 = 0.61
Answer: B
Answer:
Step-by-step explanation:
We have been given that there are 4 red marbles and 7 blue marbles and 5 yellow marbles in a bag. Monica will randomly pick two marbles out of the bag replacing the first marble before picking the second marble.
Since Monica will replace the first marble before picking the second marble, therefore, probability of both events will be independent and probability of occurring one event will not affect the probability of second event's occurring.
Since the probability of two independent compound events is always the product of probabilities of both events.
Now let us find probability of picking a red marble out of 16 (4+7+5) marbles.
Probability of picking blue ball out of 16 (4+7+5) marbles:
Now let us find probability of Monica picking a red and then a blue marble.
Therefore, the probability of picking a red and then blue marble is .
Here, regrouping is basically carrying.
64+43 shown vertically would be:
64
+43
-------
107
4+3 is 7, so seven is in the ones place, but that's not the point.
60+40 is 100, so you regroup by carrying the one to the hundreds place.
Hope I helped!
Their lines intersect at this point, which mean that they did the same thing, so it can't be A or D, and we dont know the total number of pages so we can't for sure say it's B.
So it's C, because we can see that the point is on 10 for the x-axis, which represents # of nights, and 150 for the y-axis, which represents # of pages read.
Answer:
a)0.099834
b) 0
Step-by-step explanation:
To solve for this question we would be using , z.score formula.
The formula for calculating a z-score is is z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.
A candy maker produces mints that have a label weight of 20.4 grams. Assume that the distribution of the weights of these mints is normal with mean 21.37 and variance 0.16.
a) Find the probability that the weight of a single mint selected at random from the production line is less than 20.857 grams.
Standard Deviation = √variance
= √0.16 = 0.4
Standard deviation = 0.4
Mean = 21.37
x = 20.857
z = (x-μ)/σ
z = 20.857 - 21.37/0.4
z = -1.2825
P-value from Z-Table:
P(x<20.857) = 0.099834
b) During a shift, a sample of 100 mints is selected at random and weighed. Approximate the probability that in the selected sample there are at most 5 mints that weigh less than 20.857 grams.
z score formula used = (x-μ)/σ/√n
x = 20.857
Standard deviation = 0.4
Mean = 21.37
n = 100
z = 20.857 - 21.37/0.4/√100
= 20.857 - 21.37/ 0.4/10
= 20.857 - 21.37/ 0.04
= -12.825
P-value from Z-Table:
P(x<20.857) = 0
c) Find the approximate probability that the sample mean of the 100 mints selected is greater than 21.31 and less than 21.39.
