Answer:
- <u><em>About 0.22</em></u>
Explanation:
There are two sets:
- Set W of incoming seniors who took AP World History, and
- Set E of incoming seniors who took AP European History
And there is a subset, which is the intersection of those two sets:
- Subset W ∩ E of senior students who took both.
The incoming seniors who are allowed to enroll in AP U.S. History, call them the subset S, is the set of those students that belong to W or E or both W ∩E.
By property of sets:
- S = W + E - W∩E = 175 + 36 - 33 = 178
Then, 178 out of 825 incoming seniors took one or both courses, and the desired probability of a randomly selected incoming senior is allowed to enroll in AP U.S. History is:
In order to solve that, let's call 845 100%, as we're solving it terms of 845. 845 is 100%, then it follows that 8.45 is 1%.
Now we can see how many 8.45's go into 608, which comes out to 71.95266...%, which can be rounded nicely to 72%
The million doesn't matter.
Let x be a random variable representing the number of skateboards produced
a.) P(x ≤ 20,555) = P(z ≤ (20,555 - 20,500)/55) = P(z ≤ 1) = 0.84134 = 84.1%
b.) P(x ≥ 20,610) = P(z ≥ (20,610 - 20,500)/55) = P(z ≥ 2) = 1 - P(z < 2) = 1 - 0.97725 = 0.02275 = 2.3%
c.) P(x ≤ 20,445) = P(z ≤ (20,445 - 20,500)/55) = P(z ≤ -1) = 1 - P(z ≤ 1) = 1 - 0.84134 = 0.15866 = 15.9%
Answer:
104
Step-by-step explanation:
100%= 1.00
so
420%= 4.20
Multiply 4.20×20
84
now add the original 20
104. She has 104 boxes in total.
Hope this helps! If you have any questions on how I got my answer feel free to ask. Stay safe!
