Please help ASAP!! Will give brainliest to correct answer!

There are 15 technicians and 11 chemists working in a research lab. They need to form a 5 person safety committee. In how many w

Burka [1]

<h2>There are a total 65,780 ways in which the committee can be made of 5 members.</h2>

Step-by-step explanation:

Here, the total number of technicians in the lab = 15

The total number of chemists in the lab = 11

So, the total members in the lab = 15 + 11 = 26

So, there are total 26 eligible candidates in the lab from which committee of 5 people has to be made.

Now, since there is <u>no restrictions</u> on the choice, so the committee can be formed in nay way, such that:

It may or may not have any technician.

It may or may not have may chemists.

So, the number of ways 5 member can randomly be chosen from a committee of 26 people = $^{26} C_5$

$^{26} C_5$ = $^{26} C_5 = \frac{26!}{5! \times 21!} = 65,780$

Hence, there are a total 65,780 ways in which the committee can be made of 5 members.

5 0

3 years ago

Heather works in an appliance store. She earns $1,500 per month plus a commission of 4% on her monthly sales. Her weekly sales a

Nataly_w [17]

Answer: $1980

Step-by-step explanation:

First, let’s find out her monthly sales by adding up the 4 weekly sales.

2400+ 3200 + 5200 + 1200 = 12000

Next, find the amount of commission on 12000

4%/100=0.04

12000*0.04=480

Next add her commission to her monthly wage of 1500

1500+480=1980

3 0

3 years ago

Suppose that 3 balls are chosen without replacement from an urn consisting of 3 white and 7 red balls. Assume moreover that the

zimovet [89]

Answer:

P(X1=1, X2=1) = 1/15

P(X1=1, X2=0) = 7/30

P(X1=0, X2=1) = 7/30

P(X1=0, X2=0) = 7/15

Step-by-step explanation:

Let Xi = 1 if the i-th white ball is selected. In this question the 3 white balls are marked 1,2 and 3.

We need to know the possible combination between X1 and X2 i.e. for the white ball 1 and 2 being chosen in the event.

We also need to note that the event is dependent which that after a ball is being chosen, it will not be put back hence affecting the probability of picking the next ball.

Consider all the possible combination between X1 and X2

a) both being chosen P(X1=1, X2=1)

= (3/10) x (2/9) = 1/15

Note that the first probability is the probability before any ball is being picked. The chances for ball white to be pick is 3/10 (3 white ball from the total 10 balls).

After 1 white ball being selected, that ball is not again out back into the urn making white ball 2 and total ball 9. Hence probability of picking another white ball is 2/9

b) only X1 chosen P(X1=1, X2=0)

= (3/10) x (7/9) = 7/30

After the white ball was picked, the probability of white not being pick again is the same as red being picked. Since there is still 7red balls and a total of 9 balls, the probability is 7/9

c) only X2 chosen P(X1=0, X2=1)

= (7/10) x (3/9) = 7/30

The white is not being picked first, making the probability of picking red is 7/10. Then the probability of white being picked is 3/9

d) both not chosen

P(X1=0, X2=0)

= (7/10) x (6/9) = 7/15

In other word only red being chosen. So the first probability is 7 red out of 10 balls (7/10), and the next red ball being picked next is 6/9

7 0

3 years ago

What is 4 1/5 x 5/14

Bogdan [553]

4 1/5*5/14
4 1/5=21/5 since 4*5+1=21,
So 21/5*5/14=21/14 since the 5's cancel out,
simplify, you get 3/2.
Answer: 3/2.