Please answerrrrrrrrrrrr Pleaseeeee andddd thankkk youuuu. .

A box has three balls, one white and two red. We select one ball, put it back in the box, and select a second ball (sampling wit

Dennis_Churaev [7]

Answer:

See explaation

Step-by-step explanation:

Given

Represent the balls with the first letters

$W =1$

$R =2$

Solving (a): P(F) --- White balls twice

The event of F is:

$F = \{(W,W)\}$

So:

$P(F) = P(W) * P(W)$

$P(F) = \frac{n(W)}{n} * \frac{n(W)}{n}$

$P(F) = \frac{1}{3} * \frac{1}{3}$

$P(F) = \frac{1}{9}$

Solving (b): P(G) --- two different colors

The event of G is:

$G = \{(W,R),(R,W)\}$

So:

$P(G) = P(W) * P(R) + P(R) * P(W)$

$P(G) = \frac{n(W)}{n} * \frac{n(R)}{n} + \frac{n(R)}{n} * \frac{n(W)}{n}$

$P(G) = \frac{1}{3} * \frac{2}{3} + \frac{2}{3} * \frac{1}{3}$

$P(G) = \frac{2}{9} + \frac{2}{9}$

$P(G) = \frac{4}{9}$

Solving (c): P(H) --- White picked first

The event of H is:

$H = \{(W,R),(W,W)\}$

So:

$P(H) = P(W) * P(R) + P(W) * P(W)$

$P(H) = \frac{n(W)}{n} * \frac{n(R)}{n} + \frac{n(W)}{n} * \frac{n(W)}{n}$

$P(H) = \frac{1}{3} * \frac{2}{3} + \frac{1}{3} * \frac{1}{3}$

$P(H) = \frac{2}{9} + \frac{1}{9}$

$P(H) = \frac{3}{9}$

$P(H) = \frac{1}{3}$

Solving (d): F and G, mutually exclusive?

We have:

$F = \{(W,W)\}$

$G = \{(W,R),(R,W)\}$

Check for common elements

$n(F\ n\ G) = 0$

Hence, F and G are mutually exclusive

Solving (e): G and G, mutually exclusive?

We have:

$G = \{(W,R),(R,W)\}$

$H = \{(W,R),(W,W)\}$

Check for common elements

$n(G\ n\ H) = 1$

Hence, F and G are not mutually exclusive

7 0

3 years ago

What is the value of 5 in 756? write and draw to explain how you know

Dennis_Churaev [7]

The value of 5 is 10 because it is in the tens place. If you draw a place value chart it shows that 5 is in the tens place. You could also draw it with 5 blocks with ten cubes.

Happy studying ^_^

3 0

3 years ago

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1) Evaluate the expression 33(12 – 2) : 2. 33. (12 - 2) = 2 =

alekssr [168]

Answer:

Step-by-step explanation:

33(12-2)=33(10)=330

4 0

3 years ago

Ill give brainliest please help with this algebra!!

creativ13 [48]

Answer:

Each of these equations solves as 1, because each one of them is an instance of the same expression being divided by itself.

This will always give you a value of 1, as long as the denominator does not end up with a zero value.

Take for instance the third question:

$\frac{p^4}{p^4}\\= p^4 \times p^{-4}\\= p^{(4 - 4)}\\= p^0\\= 1$

This stands true with all three questions.

HOWEVER

I say this assuming that the 5 following the first brackets in the first question is meant to be an exponent, and not a multiple. Given that the norm is to make any numeric multiples precede the brackets, I assume it is an exponent. and we're good.

It's not using superscript though, which could mean that they want it multiplied by five instead of raised to the power of.

If that's case, we can solve it the same way we solved question 20. If the bases are the same, then when multiplying or dividing the terms, you can simply add or subtract the exponents respectively:

$\frac{(4x + 2y)\times5}{(4x + 2y)^5}\\= 5(4x + 2y) \times (4x + 2y)^{-5}\\= 5(4x + 2y)^1 \times (4x + 2y)^{-5}\\= 5(4x + 2y)^{1 - 5}\\= 5(4x + 2y)^{-4}\\= \frac{5}{(4x + 2y)^{4}}$

Again, this is probably not the correct answer for question 18, as that 5 is almost guaranteed to be meant as an exponent. If it is instead a coefficient though, then this would be the answer to it.

8 0

3 years ago

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In math class, the girl to boy ratio is 8 to 6. If there are 24 girls in the class, how many boys are there?

Studentka2010 [4]

Answer:

8 : 6 = 24 : 18

Step-by-step explanation:

If there were 24 girls in class, there would be 18 boys.

Girls: 8, 16, 24

Boys: 6, 12 , 18

4 0

3 years ago

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