All categories

3 years ago

A stock market fell 75 points over a period of 5 days. What was the average change in the stock market each day?

Mathematics

2 answers:

nydimaria [60]3 years ago

6 0

Answer:

Step-by-step explanation:

In 5 days there was a 75 point loss

In 1 day there was a loss of x (which would be the average per day)

5/75 = 1/x Cross multiply

5x = 75 * 1

5x = 75 Divide by 5

5x/5 = 75/5 Do the division

x = 15

The average loss was 15 points a day.

seropon [69]3 years ago

5 0

Answer:

15 points

Step-by-step explanation:

Divide the 75 points by the five days to find the average points lost per day.

75 points / 5 days = 15 points per day

You might be interested in

Which distance measures 7 units?

Naddika [18.5K]

Answer:

first one

Step-by-step explanation:

4 0

3 years ago

Given the following sets:

hjlf

<h3>Answer: 10</h3>

===========================================================

Explanation:

Even though your teacher doesn't want you to list the items of the set, it helps to do so.

We'll be working with these two sets

A = {b, d, f, h, j, I, n, p, r, t}

C = {d, h, I, p, t}

When we union them together, we combine the two sets together. Think of it like throwing all the letters in one bin rather than two bins.

A u C = {b, d, f, h, j, I, n, p, r, t, d, h, I, p, t }

The stuff that isn't bolded is set A, while the stuff that is bolded is set C

After we toss out the duplicates, we end up with this

A u C = {b, d, f, h, j, I, n, p, r, t}

But wait, that's just set A. Notice how everything in set C can be found in set A. This indicates set C is a subset of set A.

That's why all of the stuff in bold was tossed out (because they were duplicates of stuff already mentioned).

Once we determine what set A u C looks like, we count out the number of items in that set to determine the final answer.

There are 10 items in {b, d, f, h, j, I, n, p, r, t} which means 10 is the final answer.

----------------------------

An alternative method is to use the formula below

n(A u C) = n(A) + n(C) - n(A and C)

n(A u C) = 10 + 5 - 5

n(A u C) = 10

The notation n(A and C) counts how many items are found in both sets A and C at the same time. But as mentioned earlier, this is identical to just counting how many items are in set C. So we'll have n(C) cancel out with itself.

In short, n(A u C) = n(A) = 10

8 0

3 years ago

Find the slope of the line passing through (9, -6) and (5, 10).

Marrrta [24]

Slope = (y2 - y1)/(x2 - x1)

Let's plug in our values: (10 - 9)/(5 - - 6)
This gives us 1/11 as our slope.

3 0

3 years ago

A survey said that 3 out of 5 students enrolled in higher education took at least one online course last fall. Explain your calc

marysya [2.9K]

Answer:

a) 60% probability that student took at least one online course

b) 40% probability that student did not take an online course

c) 12.96% probability that all 4 students selected took online courses.

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they took at least one online course last fall, or they did not. The probability of a student taking an online course is independent of other students. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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

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

And p is the probability of X happening.

3 out of 5 students enrolled in higher education took at least one online course last fall.

This means that $p = \frac{3}{5} = 0.6$

a) If you were to pick at random 1 student enrolled in higher education, what is the probability that student took at least one online course?

This is P(X = 1) when n = 1. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{1,1}.(0.6)^{1}.(0.4)^{0} = 0.6$

60% probability that student took at least one online course.

b) If you were to pick at random 1 student enrolled in higher education, what is the probability that student did not take an online course?

This is P(X = 0) when n = 1.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{1,1}.(0.6)^{0}.(0.4)^{1} = 0.4$

40% probability that student did not take an online course

c) Now, consider the scenario that you are going to select random select 4 students enrolled in higher education. Find the probability that all 4 students selected took online courses

This is P(X = 4) when n = 4.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 4) = C_{4,4}.(0.6)^{4}.(0.4)^{0} = 0.1296$

12.96% probability that all 4 students selected took online courses.

3 0

3 years ago

Jasmine spends $86.90 on blueberries if the blueberries cost $3.95 per pound how many pounds of blueberries did she buy

ivanzaharov [21]

Answer:

She bought 22 pounds of blueberries.

Step-by-step explanation:

$86.90 divided by $3.95 = 22 pounds

5 0

3 years ago

Read 2 more answers

A stock market fell 75 points over a period of 5 days. What was the average change in the stock market each day?​

A stock market fell 75 points over a period of 5 days. What was the average change in the stock market each day?