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By trial and error, x = 3 and y = 1 

Since the ratio of girls to teachers==7 :  3

But the ratio of boys to girls==2 :  3

Then the ratio of boys + girls to teachers= [7  +  (2/3 * 7)] :  3 ==[7  + 4  + 2/3] :  3 ==[11 2/3 * 3] :  [3 * 3]==35 : 9

The area of a trapezoid is represented by the formula:   \(\frac{a+b}{2}*h\)

Where a and b represent the two bases, and h represents the height.

All we need to do here is plug in the values:

\(\frac{12+15}2{}*6\)

And then solve the expression:

\(\frac{27}{2}*6\)

= 81 cm^2

The square root of 27 is \(3\sqrt3\).

The square root of 12 is \(2\sqrt3\).

The squre root of 300 is \(10\sqrt3\).

Multiplying all these together gives \(180\sqrt3\).

This can be reduced into the following system, where s is the amount of strong solution and w is the amount of weak solution:

s + w = 6.25

25%w + 75%s = 35% x 6.25 = 2.1875

Multiply the bottom by four to get rid of the percentages:

w + 3s = 8.75

Subtract the first equation from the second:

2s = 2.50

s = 1.25

w = 5.00

Answer: \(2\)

Solution:

First of all, I assume that this is the question that you are asking. \(\frac14<\frac{x}{5}<\frac23\)

The first step would be to multiply the middle fraction by five to isolate the variable:

\(\frac54

Now all that needs to be done is to count every integer in between 5/4 and 10/3, which is literally just 2 and 3. This makes the desired answer \(2\) integers.

m - n = 4*sqrt(7).

S==Students

B/G ==2/3,

G/T==7/3, solve for B,G

B = (14 T)/9 and G = (7 T)/3

S/T ==(14 T)/9 + (7 T)/3==35T / 9

Ratio of students to teachers==35 : 9

None, (x+2)^2 is an upward pointing parabola with a vertex at (-2,0)

Here is a visualization:

Combine like terms: \({1 - i \over 8i -2}\)

Multiply by \({8i + 2 \over 8i + 2}\): \({(1 -i)(8i + 2) \over {(8i - 2)(8i + 2)}}\)

Expand: \({-8i^2 + 6i + 2 \over 64i^2 - 4}\)

Substitute \(i^2 = -1\): \({8 + 6i + 2 \over -64 - 4}\)

Combine like terms: \({10 + 6i \over -68}\)

Simplify: \(\color{brown}\boxed{- 5 - 3i \over 34}\)

The only way Meena loses is if Bob wins the next 2 points. The probability of Bob winning the next 2 points is \(({1 \over 3})^2 = {1 \over 9}\).

So, the probability Meena wins is \(1 - {1 \over 9} = \color{brown}\boxed{8 \over 9}\)

No, I agree with your answer...

There are \(10^4 = 10000\) ways to choose the 4 letters, and there are only 2 ways to choose the word "MATH" (there are 2 T's), so the probability is \(\color{brown}\boxed{1 \over 5000}\)

The first letter we have to choose is M which is our only option. This gives us a probabililty of 1/10.

Next, the only letter where we can get our A is from the second letter. Again, this gives a probability of 1/10. 

When choosing the letter T, however, we have to consider the fact that there are two T's in MATHCOUNTS. This gives a probability of 1/5 of choosing it next.

Again, there is only one H so there is 1/10 probability

Multiplying all of these gives us: 1/5000.

Which I think is somehow wrong,  someone please correct me. 

The probability that the first letter is M is 1/9.  The probability that hte second letter is A is 1/8.  The probability that the third letter is T is 2/7.  The probability that the fourth letter is H is 1/6.  Multiply these products, and you get 1/1512.

If the base is odd, the number will never be even, because there are 3 odd numbers. The sum of these numbers will always be odd, because\(\text{odd} \times \text{odd} = \text{odd}\), and \(\text{odd} + \text{odd} + \text{odd} = \text{odd}\) 

However, if the base is even, all the individual digits will yield an even number because \(\text{even} \times \text{odd} = \text{even}\), and \(\text{even} + \text{even} = \text{even}\)

So, how many even digits are there between 7 and 100?

There are \(6 \times 20 = 120 \) total outcomes. 

Of these, \(5 \times 16 = 80\) outcomes have neither Alice nor Bob rolling a multiple of 5

So, there are \(120 - 80 = 40\) successes, so the probability is \({40 \over 120} = \color{brown}\boxed{1 \over 3}\)

If I understand your question:
This is the "closed form" of each term:
a_n = 1/2 (6 n + (-1)^(n + 1) + 13). 
And the number 235236(n) is a multiple of 7 in the following bases:
n = 10, 12, 16, 18, 22, 24, 28, 30, 34, 36, 40, 42, 46, 48, 52, 54, 58, 60, 64, 66, 70, 72, 76, 78, 82, 84, 88, 90, 94, 96, 100.

guest.