Help me please some help ................

A bag contains 15 marbles. The probability of randomly selecting a green marble is 1/5. The probability of randomly selecting a

slava [35]

Answer:

6 blue marbles

Step-by-step explanation:

Given;

Total number of marbles N = 15

probability of randomly selecting a green marble

P(G) = 1/5

The probability of randomly selecting a green marble, replacing it, and then randomly selecting a blue marble P(G∩B) = 2/25

P(G∩B) = P(G) × P(B)

P(B) = P(G∩B) ÷ P(G)

The probability of selecting a blue marble

P(B) = 2/25 ÷ 1/5 = 2/5

The number of blue marbles N(B) = P(B) × N

N(B) = 2/5 × 15 = 6

There are 6 blue marbles

7 0

3 years ago

Arrange these numbers in order from least to greatest. 0.125 2.5% 1/4

miv72 [106K]

Answer:

2.5%, 0.125, 1/4

Step-by-step explanation:

7 0

3 years ago

4-member curling team is randomly chosen from 6 grade 11 students and 8 grade 12 students. What is the probability that the team

Ivanshal [37]

Answer:

0.5944 = 59.44% probability that the team has at least 2 grade 11 students

Step-by-step explanation:

The students are chosen without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

We have that:

6 + 8 = 14 students, which means that $N = 14$

6 grade 11 students means that $k = 6$

Teams of 4 members means that $n = 4$

What is the probability that the team has at least 2 grade 11 students?

This is:

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

So

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 0) = h(0,14,4,6) = \frac{C_{6,0}*C_{8,4}}{C_{14,4}} = 0.0699$

$P(X = 1) = h(1,14,4,6) = \frac{C_{6,1}*C_{8,3}}{C_{14,4}} = 0.3357$

Then

$P(X < 2) = P(X = 0) + P(X = 1) = 0.0699 + 0.3357 = 0.4056$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.4056 = 0.5944$

0.5944 = 59.44% probability that the team has at least 2 grade 11 students

5 0

3 years ago

Need help, solve for x. sorry for bad pic.

Olin [163]

Answer:

x= 11.9 (3 s.f.)

Step-by-step explanation:

Please see attached picture for full solution.

7 0

3 years ago

I need this last answer please help

pychu [463]

I'm not 100% on this, but something like 90%
SSS stands for Side, Side, Side. This means if you have 2 triangles with equal sides, it would be this.
SAS stands for Side, Angle, Side. This means we know 2 sides are equal, as well as the provided angle. This wouldn't apply, because there IS no provided angle, you only have sides.
ASA stands for Angle, Side, Angle. This would mean that 2 angles are identical (of course, by 2, i mean 1 for each triangle, same applies for SAS) and one provided Side is identical. We, of course, don't know what the sides or the angles are in this case (this part always gets me)
We can eliminate A right off the bat, because it says QRP is equal to TRS, which isn't right. QRP is only equal to SRT, because the order actually DOES matter. We can go ahead and eliminate D as well. Since we have no provided sides or angles (of course we can eyeball it and see that are equal, but your not supposed to really do that) we can assume it is most likely B over C.
Your best answer provided would be QRP=SRT by SSS, or in other words, B
~Hope this helps, would appreciate brainliest!

7 0

3 years ago