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alex41 [277]
3 years ago
7

How to do this problem Please help

Mathematics
1 answer:
gregori [183]3 years ago
5 0
The answer is 64, because the amount is doubled each day.
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lines a and b are parallel. Line c is traversal. Find the measure of all angles formed by a,b, and c, given each senario. One of
madam [21]

Answer:

Because a straight line is 180°, the other angles to these 2 would each be 180°-150°=30°. The same angles would be between line b and line c because line b and line a are parallel and line c is the same. Therefore, the measures of all the angles would be 30° or 150°.

1 answer

Step-by-step explanation:

7 0
3 years ago
Does anyone know how to do this/have the answer? PLEASE
podryga [215]

15: You solve problems like this by finding the probability of each case, and then multiplying them all. For each of the last 4 question, she has probability 1/2 of guessing right. So, she guessed 4 consecutive questions with probability

\left(\dfrac{1}{2}\right)^4=\dfrac{1}{16}

16: Like before: you have pick a king with probability 4/52 = 1/13 (there are four kings - one for each suit, out of 52 cards in a standard deck), and you pick "I" from "INCREDIBLE" with probability 2/10 = 1/5 (there are two "I"s out of 10 letters). So, the probability of picking a king and then an "I" is

\dfrac{1}{13}\cdot\dfrac{1}{5}=\dfrac{1}{65}

17-20: The important bit of information here is that you replace the first ball. So, the first and second pick follow the exact same probability distribution, because they basically are two repetitions of the same experiment. So, for example, in ex. 17, the first ball is even with probability 15/30 = 1/2 (there are 15 even balls out of 30). Then, you have again probability 15/30 = 1/2 to pick an odd ball (there are also 15 odd balls out of 30). So, the probability of picking an even ball, replace it, and pick an odd ball is

\dfrac{1}{2}\cdot\dfrac{1}{2}=\dfrac{1}{4}

Exercises 18 to 20 follow the same scheme: find out the probability of the two events and multiply them.

21-26: Not we DON'T replace the balls, so the second pick will suffer the effects of the first one. Let's dive into ex. 21 for example. For the first pick, we want a 2-digits number. There are 21 of such balls (all balls except balls 1 to 9), so we pick a 2-digits ball with probability 21/30 = 7/10. For the second pick, we want the balls number 4. But we have to assume that we already picked the first ball, and we picked a 2-digits ball. So, for the second pick, we're choosing from a bag with 29 balls, and there is only one ball labeled 4. So, we pick the ball number 4 with probability 1/29. We deduce that the two events happen one after the other with probability

\dfrac{7}{10}\cdot\dfrac{1}{29}=\dfrac{7}{290}

Exercises 22 to 26 are similar: you find out the probability of the first event, and then you consider the new environment (i.e. you keep track of the first pick) when it comes to the probability distribution for the second pick).

5 0
3 years ago
Check all the factors of 2. * 0 1 2 3
Dovator [93]
1 and 2 are the only factors of 2
5 0
2 years ago
The histogram shows the ages of people in a community chorus. How many people in the chorus are 50 or older? Enter your answer i
Vilka [71]

Answer:

13

Step-by-step explanation:

The computation of the number of people in the chorus are 50 or older is as follows

According to the attached figure

In the age of 50 - 60 , there is 8 people

In the age of 60 - 70 , there is 4 people

In the age of 70 -80, there is 1 people

So, if we do the total that comes

= 8 people + 4 people + 1 people

= 13 people

Hence, the no of people would be 13

The same is relevant

3 0
3 years ago
I think it’s because the computer
zloy xaker [14]

Answer:

It's A and B

Step-by-step explanation:

Since she took 75% off of her commute time, she has 25% left. 25% as a decimal is 0.25 and 0.25 as a decimal is 1/4.

Also, since it says choose two answers, C and D are not equal to each other.

6 0
3 years ago
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