HELP PLEASE look at the picture

Among the contestants in a competition are 46 women and 23 men. If 5 winners are randomly selected, what is the probability that

eimsori [14]

Answer:

When 5 winners are randomly selected, the probability that they are all men is 0.003.

Step-by-step explanation:

Among the contestants in a competition are 46 women and 23 men.

Total persons are = $46+23=69$

When one winner is chosen, there are 68 people left, when second winner is chosen there are 67 people left and so on.

So, the probability that they are all men is given by:

$\frac{23}{69} \times\frac{22}{68} \times\frac{21}{67} \times\frac{20}{66} \times\frac{19}{65}$

= $\frac{4037880}{1348621560}$ = 0.002999 ≈ 0.003

Therefore, when 5 winners are randomly selected, the probability that they are all men is 0.003.

7 0

3 years ago

If events X and Y are independent, what must be true? Check all that apply. P(Y | X) = 0 P(X | Y) = 0 P(Y | X) = P(Y) P(Y | X) =

lana [24]

if events X and Y are independent, then for intersection we multiply the probability

P(Y∩X) = P(Y) * P(X)

We know that

$P(Y|X) =\frac{P(YintersectionX)}{P(X)}$

Now we replace P(Y) * P(X) for P(Y∩X)

$P(Y|X) =\frac{P(Y)*P(X)}{P(X)}$

Cancel out P(X)

So $P(Y|X) = P(Y)$

Like that

$P(X|Y) =\frac{P(XintersectionY)}{P(Y)}$

Now we replace P(X) * P(Y) for P(X∩Y)

$P(X|Y) =\frac{P(X)*P(Y)}{P(Y)}$

Cancel out P(Y)

So $P(X|Y) = P(X)$

P(Y | X) = P(Y) and P(X | Y) = P(X) are true

3 0

3 years ago

Read 2 more answers

Use induction to show that 12 + 22 + 32 + ... + n2 = n(n+1)(2n+1)/6, for all n > 1.

dlinn [17]

Answer with Step-by-step explanation:

Let $P(n)=1^2+2^2+3^2+.....+n^2=\frac{n(n+1)(2n+1)}{6}$

Substitute n=2

Then $P(2)=1+2^2=5$

$P(2)=\frac{2(2+1)(4+1)}{6}=5$

Hence, P(n) is true for n=2

Suppose that P(n) is true for n=k >1

$P(k)=1^2+2^2+3^2+...+k^2=\frac{k(k+1)(2k+1)}{6}$

Now, we shall prove that p(n) is true for n=k+1

$P(k+1)=1^2+2^2+3^2+...+k^2+(k+1)^2=\frac{(k+1)(k+2)(2k+3)}{6}$

LHS

$P(k+1)=1^2+2^2+3^2+.....+k^2+(k+1)^2$

Substitute the value of P(k)

$P(k+)=\frac{k(k+1)(2k+1)}{6}+(k+1)^2$

$P(k+1)=(k+1)(\frac{k(2k+1}{6})+k+1)$

$P(k+1)=(k+1)(\frac{2k^2+k+6k+6}{6})$

$P(k+1)=(k+1)(\frac{2k^2+7k+6}{6})$

$P(k+1)=(k+1)(\frac{2k^2+4k+3k+6}{6})$

$P(k+1)=\frac{(k+1)(k+2)(2k+3)}{6}$

LHS=RHS

Hence, P(n) is true for all n >1.

Hence, proved

4 0

3 years ago

An equation 1/6 x equals 2/3

Kobotan [32]

1/6 x = 2/3 |multiply both sides by 6

x = 12/3

x = 4

5 0

3 years ago

A survey in September 2020 of 1000 gas stations found that the price charged for a gallon of regular gas could be closely approx

sleet_krkn [62]

Basically given normal distribution we have to draw the bell curve with standard deviations which should look like(picture below) so draw something like that with center number being mean (3.53) and each positive standard deviation(SD) being + 0.2:
Ex 1 SD is 3.73, 2SD is 3.93, -1SD is 3.33

Now to solve the questions
A. If we know there’s 1000 gas stations and 3.93 is just 2 SD we can just refer to the picture below.
Basically because we’re finding the amount of values that are greater than that it should be percentage to the right of 2 SD. Which is 2.2% or 0.022 Then multiply that by the number of gas station.
So 0.022 * 1000 = 22.
B. This is basically the values between -2SD and 2 SD which is the percentage between them. Add 13.6 *2 + 34.1*2 = 95.2% or 0.952
Now multiply by 1000 * 0.952=952 gas stations.
C. For this one we just find percentage to the left of -1 SD or simpler just subtract half(the right half) and the percentage in 1 SD from 1 to find the remainder which is the value left of -1SD.
1-0.5 - 0.341 = 0.159
Now multiply by 1000 * 0.159 = 159 stations.

Hopefully this helps.

5 0

2 years ago