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Sat Math Problems

Glenn Said:

Can anyone help me on these SAT I Math Problems?

We Answered:

Uhh, the first one is most definitely not C. It is D. 87 is divisible by 3, leaving 12/87 not a reduced fraction.

Douglas Said:

3 SAT Math Problems, Please Help!?

We Answered:

1. You know 9/15 are seniors. 1/3 of the remaining 6/15 is 2/15
so 11/15 is the sum of seniors and juniors. The remaining 4/15 is 80 so divide that by 4 to get the equivalent of 1/15 which is 20. 20*9=180. So that is your answer.

2. It is C. You can figure out his rate per day by dividing 3/4 by two since he worked two days and finished 3/4. His rate per day is 3/8 and since he finished 6/8 already he does not need a full day so you can eliminate D and E. You can also get rid of 1/4 a day because 3/24 is not enough to finish the rest. same can be said for 1/2 a day because 3/16 is not enough so the only possible answer is 2/3.
3.Same principle as the first problem , 10/15 for susan and a 1/5 of the remaining is 1/15 together that is 11/15 Bill has 4/15 with his 48 divide by 4 you have 12. 12*10 =120 there you go .

Priscilla Said:

Can someone tell me formulas for figuring out SAT I math problems?

We Answered:

Well, in the example you provided with the phone number, that would be solved fairly easily. Since the first, third, and fourth numbers are always going to be the same, you just have to figure out how many different combinations there can be for the remaining numbers.

Visual (lets say constant is 1):
1X1-1XXX

theres 4 numbers left, and each can be anything from a 1 to a 9, or nine different possibilities for each number. The solution would be 9^4, or 6561 different combinations.

Irma Said:

How do I solve this sat math problem and problems like it. Please show work?

We Answered:

The basic formula for arranging the 5 items is ?P? or permutations of 5 things taken 5 at a time. This is 5! or 5 x 4 x 3 x 2 x 1 = 120 Now just throw away the ones that start with C, which is 1/5 of them or 24. And throw away the ones that end in C, again 1/5 of them or 24. You have then 120 - 24 - 24 = 72.

Or, you can think of filling the 5 slots. Fill the first and last slots first, 4 choices for the first slot (A, B, D, E, no C) and 3 choices for the last (after using 1 for the first). Now fill the middle slots, 3 choices (used 2 already for the 1st and last, but now you have C available), then 2 choice, then 1 choice. Or,
4 x 3 x 2 x 1 x 3 for the choices in each slot. Product is 72.

Rene Said:

I need help with two SAT Math problems?

We Answered:

1. well the only possibilities for n are 5, 7, 9, and 11.
5+3=8
7+3=10
9+3=12
11+3=14
since none of them add up to 23, the probability is 0

however, if you meant n+3p=23 (which i assume you did, but i could be wrong) plug in each value for n, and solve for p. if the value you arrive at belongs to the set of p, it is a possibility.
5+3p=23
p=6 yes
7+3p=23
p=5.3333333 no
9+3p=23
p=4.66666667 no
11+3p=23
p=4 no

so there is only one possibility (5,6) and are 20 total combinations. thus the probability is 1/20 or .05 or 5%.

2. A(picture)=A(total)-A(margin)
=8.5*10-(8.5-3)(10-3)
=85-(5.5)(7)
=85-38.5
=46.5 square inches.

Clayton Said:

I need help on a couple of SAT math problems?!?

We Answered:

10 = 4 + 6 ratio 4:6::2:3
15= 6 + 9 ratio 6:9::2:3
Therefore for 30, just double 6 and 9 and then double again for 60

All that's left is 12. Look at 2 and 3. No multiples add up to 12.

18?18 = 18?9•2 = 18•3?2 = 54?2 and 54•2 = 108. Can't use 18 and 18 because r>T

Halfway is the arithmetic mean (average) (53 + 62)/2 = 115/2 = 57.5

Isosceles triangle, so sides are either 50-30-30 or 50-50-30. First is smaller (obviously) and 50+30+30 = 110

I think you mistyped. >20 but <20?? Impossible. In any event you'll be dividing somethings into 360.

h(2m) = 14 + (2m)²/4 = 9m

4m²/4 – 9m + 14 = 0

m² – 9m + 14 = 0

(m–2)(m–7)=0

m = 2 or m = 7

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