The third quartile of this data set is B. 24.
1. Since it's m - 7 you would have 7 to both sides so you would in fact have m < 13. If you double check your answer, you see that if m is say 12 (because 12 is obviously less than 13), 12 - 7 < 6
2. Again, use the same process on this problem as the first one. Add 8 to each sides because it's you're subtracting 8 from n. So you end up with n > 13. Check your answer. Say n is 14. 14 - 8 > 5
3. This one is different because you are adding 5 to p. So in order to get p by itself, you need to subtract 5 from both sides. p < 5. Say p is 4, 4 + 5 < 10.
When working with problems like these, you need to isolate the variable on one side and get it by itself.
We will get the number of possible selections, and then subtract the number less than 25 cents.
We can choose the number of dimes 5 ways 0,1,2,3 or 4.
We can choose the number of nickels 4 ways 0,1,2 or 3.
We can choose the number of quarters 3 ways 0,1, or 2.
That's 5*4*3 = 60 selections
Now we must subtract from the 60 the number of selections of coins that are less than 25 cents. These will involve only dimes and nickels.
To get a selection of coin worth less than 25 cents:
If we use no dimes, we can use 0,1,2 on all 3 nickels.
That's 4 selections less than 25 cents. (that includes the choice of No coins at all in the 60, which we must subtract).
If we use exactly 1 dime , we can use 0,1,2, or all 3 nickels.
That's the 3 combinations less than 25 cents.
And there is 1 other selection less than 25 cents, 2 dimes and no nickels.
So that's 4+3+1 = 8 selections which we must subtract from the 60.
Answer 60-8 = 52 selections of coins worth 25 cents or more.
Answer:
Please see the answer below
Step-by-step explanation:
a. Since there’s no restrictions .Therefore , the number of ways = 7!*150 = 756000
b. The number of ways such that the 4 math books remain together
The pattern is as follows: MMMMEEE, EMMMMEE, EEMMMME, and EEEMMMM
Where M = Math’s Book and E= English Book.
Number of ways = 4!*8!*4*150= 86400 ways.
c. The number of ways such that math book is at the beginning of the shelf
The number of ways = 6!*4*150 = 432000
d. The number of ways such that math and English books alternate
The number of ways = 150*4!*3! =2160 ways
e. The number of ways such that math is at the beginning and an English book is in the middle of the shelf. The number of ways = 4*3*5!*150 =216000 ways.
