Ask question

Ask question

All categories

2 years ago

6

Three defective electric toothbrushes were accidentally shipped to a drugstore by the manufacturer along with 17 non-defective o

nes. What is the probability that the first two electric toothbrushes sold will be returned to the drugstore because they are defective

1 answer:

PSYCHO15rus [73]2 years ago

7 0

Sleep nxbd ns jus juz b hb s hs s shs sh

You might be interested in

Nadia selects three marbles from the jar to the right without replacing them. What is the probability that all three marbles are

Savatey [412]

Not very likely. Does a picture come with this question.

4 0

3 years ago

In what time will sum of money double itself at 5% per annum?

klemol [59]

Step-by-step explanation:

Mr.Josol : what will be the height of the ball after 2 seconds?

8 0

3 years ago

5m<6m+6 pls help me

torisob [31]

Let's solve your inequality step-by-step.

5m<6m+6

Step 1: Subtract 6m from both sides.

5m−6m<6m+6−6m

−m<6

Step 2: Divide both sides by -1.

−m / −1 < 6 / −1

m>−6

Answer:

m>−6

4 0

3 years ago

Suppose p(a) = 0.40 and p(a 

sveticcg [70]

Between the probability of union and intersection, it's not clear what you're supposed to compute. (I would guess it's the probability of union.) But we do know that

$P(A\cup B)+P(A\cap B)=P(A)+P(B)$

For parts (a) and (b), you're given everything you need to determine $P(B)$ .

For part (c), if $A$ and $B$ are mutually exclusive, then $P(A\cap B)=0$ , so $P(A\cup B)=P(A)+P(B)$ . If the given probability is $P(A\cup B)=0.55$ , then you can find $P(B)=0.15$ . But if this given probability is for the intersection, finding $P(B)$ is impossible.

For part (d), if $A$ and $B$ are independent, then $P(A\cap B)=P(A)\cdot P(B)$ .

8 0

3 years ago

2 divided by 1/3 in simpleast form

leva [86]

Answer:

6

Step-by-step explanation:

When dividing by a fraction flip it over and multiply

2 / (1/3) = 2 * 3/1

2 / (1/3) = 6/1

2 / (1/3) = 6

8 0

3 years ago