I need help with this question.This is due in like 7 mins :(

A bag contains 40 marbles. The probability of selecting a red marble at random is 1/5. How many red marbles are in the bag?

goblinko [34]

Answer:

8 red marbles.

Step-by-step explanation:

All together we have 40 marbles. The fraction of marbles that are red is 1/5 which means that we should divide the total number of marbles, which is 40 by the denominator ( Bottom number of fraction ) 5 so that we get the number of groups that have 1 red marble in them. In this case we have 8 groups.

40/5=8

You always divide by the denominator ( Bottom number of fraction ) first and then you multiply your answer by the numerator ( Top number of fraction ) .

If each group has 1 red marble in it and we have 8 groups, it means that we have 8 red marbles in the bag.

1 * 8 = 8 red marbles in the bag

If the numerator (number above the fraction line) was more than one, then this would mean that we have more than 1 marble in each group of marbles.

Example:

If we had 3/5 we would have 3 marbles in each group of marbles. This means that instead of doing 1 * 8 we would have to do 3 * 8 = 24. We would have 24 red marbles in the bag.

Easy and quick way if you do not need the explanation:

Another way we could do this is by simply multiplying the fraction ( 1/5 ) by the total number of marbles ( 40 ). This would still give us an answer of 8 red marbles in the bag. 40/5=8, 8*1=8

7 0

3 years ago

We have n = 100 many random variables xi 's, where the xi 's are independent and identically distributed bernoulli random variab

Alex777 [14]

Recall that for a random variable $X$ following a Bernoulli distribution $\mathrm{Ber}(p)$ , we have the moment-generating function (MGF)

$M_X(t)=(1-p+pe^t)$

and also recall that the MGF of a sum of i.i.d. random variables is the product of the MGFs of each distribution:

$M_{X_1+\cdots+X_n}(t)=M_{X_1}(t)\times\cdots\times M_{X_n}(t)$

So for a sum of Bernoulli-distributed i.i.d. random variables $X_i$ , we have

$M_{\sum\limits_{i=1}^nX_i}(t)=\displaystyle\prod_{i=1}^n(1-p+pe^t)=(1-p+pe^t)^n$

which is the MGF of the binomial distribution $\mathcal B(n,p)$ . (Indeed, the Bernoulli distribution is identical to the binomial distribution when $n=1$ .)

8 0

3 years ago

Jason has five zucchinis he grew in his garden. He wants to share them equally among three of his neighbors. How many zucchinis

Irina18 [472]

Answer:

$1 \frac{2}{3}$

Step-by-step explanation:

Given: Jason has 5 Zucchinis.

Jason want to share Zucchinis equally among three of his neighbour.

Now, find the share of each neighbour out of 5 zucchinis

As given zucchini is divided equally.

Share of each neibour= $\frac{Total\ zucchinis}{number\ of\ neighbour}$

∴ Share of each neibour= $\frac{5}{3}$

next converting the improper fraction into mixed fraction.

If we divide 5 by 3 we get quotient as 1 and remainder as 2

∴ share for each neighbour will be $1\frac{2}{3}$

Hence, each neighbour of Jason will be $1\frac{2}{3}$

4 0

4 years ago

What is the quotient of (x3 3x2 5x 3) Ă· (x 1)? x2 4x 9 x2 2x x2 2x 3 x2 3x 8.

Rina8888 [55]

The quotient of expression is $\rm x^2+2x+3$ .

Given that,

Expression; $\rm\dfrac{ (x^3 + 3x^2 + 5x + 3) }{ (x + 1)}$

We have to determine,

The quotient of expression?

According to the question,

To determine the quotient of expression following all the steps given below.

Simplify the expression,

$\rm =\dfrac{ (x^3 + 3x^2 + 5x + 3) }{ (x + 1)}\\\\= \dfrac{ (x^3 + 2x^2 + 3x + x^2+2x+ 3) }{ (x + 1)}\\\\ = \dfrac{ x(x^2+ 2x+ 3) + 1(x^2+2x+ 3) }{ (x + 1)}\\= \dfrac{ (x^2+ 2x+ 3) (x+1) }{ (x + 1)}\\\\= x^2+2x+3$

Hence, The required quotient of expression is $\rm x^2+2x+3$ .

For more details refer to the link given below.

brainly.com/question/12895249

4 0

2 years ago

Hey guys can I get a little help on this problem please, thank you.

DedPeter [7]

Point g because it’s in the middle.

midpoint= the middle.

5 0

3 years ago

I need help with this question.This is due in like 7 mins :(​

I need help with this question.This is due in like 7 mins :(