a bag contains 5 blue marbles,3 red marbles, and 2 yellow marbles. You select a marble at random. What is p (yellow)

Aiden refills three token machines in an arcade. He puts twice the number in machine A as in machine B, and in machine C , he pu

Sonja [21]

It is a possibility of being 16 or 5/6

7 0

3 years ago

Read 2 more answers

In the standard (x, y) coordinate plane which equation represents a line through the point (6, 1) and perpendicular to the line

Nata [24]

Answer:

d. -2/3x+5

Step-by-step explanation:

Because the line is perpendicular, the slope must be the inverse of the original slope. The inverse of 3/2 is -2/3. To find the b value, you plug (6,1) into the equation y=-2/3x+b

1=-2/3(6)+b

1=-4+b

5=b

The final equation is y=-2/3x+5

6 0

3 years ago

A map of a highway has a scale of 2 inches =45 miles. The length of the highway on the map is 6 inches. There are 11 rest stops

Elenna [48]

Answer:

each rest stop is 0.45 inches apart on the new map.

Step-by-step explanation:

First, we need to figure out how long is the highway. since 6/2 = 3, we know that the length of the highway is equal to 45*3. 45*3 = 135.

Now since the beginning stop doesn't count(it's at the 0 miles point and normal dividing would not count 0), so we know that there is 10 stops.

Since on the new map there was 30 miles in a inch, 135 miles on the new map would equal to 4.5 inches(135 = 120+15 = 4*30+0.5*30). 4.5/10 = 0.45 inches. So basically the rest stops are 0.45 inches apart.

4 0

3 years ago

Nathan usually drinks 33 ounces of water per day. He read that he should drink 56 ounces of water per

neonofarm [45]

Answer:

Step-by-step explanation:

i don't know just want points

7 0

3 years ago

For married couples living in a certain suburb, the probability that the husband will vote on a bond referendum is 0.21, the pro

leva [86]

Answer:

a) 0.34 = 35% probability that at least one member of a married couple will vote.

b) 0.7143 = 71.43% probability that a wife will vote

c) 0.0968 = 9.68% probability that a husband will vote

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

This is used for itens B and C. For item a, we treat the probabilities as Venn sets.

A) What is the probability that at least one member of a married couple will vote?

I am going to say that:

Event A: Husband votes.

Event B: Wife votes.

The probability that the husband will vote on a bond referendum is 0.21

This means that $P(A) = 0.21$

The probability that the wife will vote on the referendum is 0.28

This means that $P(B) = 0.28$

The probability that both the husband and the wife will vote is 0.15.

This means that $P(A \cap B) = 0.15$

At least one votes:

This is $P(A \cup B)$ , which is given by:

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

So

$P(A \cup B) = 0.21 + 0.28 - 0.15$

$P(A \cup B) = 0.34$

0.34 = 35% probability that at least one member of a married couple will vote.

B) What is the probability that a wife will vote, given that her husband will vote?

Here, we use conditional probability:

Event A: Husband votes:

Event B: Wife votes

The probability that the husband will vote on a bond referendum is 0.21

This means that $P(A) = 0.21$

Intersection of events A and B:

Intersection between husband voting and wife voting is both voting, which means that $P(A \cap B) = 0.15$

The desired probability is:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.15}{0.21} = 0.7143$

0.7143 = 71.43% probability that a wife will vote.

C) What is the probability that a husband will vote, given that his wife does not vote?

Event A: Wife does not vote.

Event B: Husband votes.

The probability that the wife will vote on the referendum is 0.28

So 1 - 0.28 = 0.62 probability that she does not vote, which means that $P(A) = 0.62$

Probability of husband voting and wife not voting:

0.21 probability husband votes, 0.15 probability wife votes, so 0.21 - 0.15 = 0.06 probability husband votes and wife does not, so $P(A \cap B) = 0.06$

The desired probability is:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.06}{0.62} = 0.0968$

0.0968 = 9.68% probability that a husband will vote

8 0

3 years ago