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3 years ago

Decompose to find the product of 4 × 3.5.PLEASE HELP QUICK

Mathematics

2 answers:

Len [333]3 years ago

8 0

<u><em>Answer: =14</em></u>

Step-by-step explanation:

multiply by the numbers. product it's going to be multiply by the numbers, decimals, fractions, factors, and multiples.

you can also multiply without decimal point, and then put the decimal point as an answer.

4*35=140

=14.0

refine.

=14

Hope this helps!

Thanks!

Have a great day!

worty [1.4K]3 years ago

5 0

Answer:

14 is the answer to this question.

Step-by-step explanation:

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lozanna [386]

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(Q) Find the amount in the account at the end of 1 year.

(A) $8,190

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2 years ago

1 is 25% of what number?

Travka [436]

Answer:

Step-by-step explanation:

25% is the same as 1/4.

1 is 1/4 of what number?

Answer: 4

You can also answer it with an equation.

Let x = the number you are looking for.

1 is 25% of x can be written as the following equation.

25% of x = 1

0.25x = 1

Multiply both sides by 4.

x = 4

Answer: 4

8 0

2 years ago

A Math professor scores all tests with a maximum of 100 points. A student must have an average of 90 on six tests to receive an

nignag [31]

Answer:

No, he cannot

Step-by-step explanation:

Lets say the student takes the maximum socre on his last test, so he gets 100. Now, his average on the six texts will be:

(75+85+90+95+94+100)/6 = 89.83

So, if taking 100 is not enough to get to the average of 90, this means he would need a score greater than 100, which is impossible as 100 is the maximum. So, the student CAN'T get an A

3 0

3 years ago

What is the square root of 25?

jolli1 [7]

Answer:

It would be 5.

8 0

2 years ago

Read 2 more answers

23% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and as

zlopas [31]

Answer:

a) There is a 29.42% probability that the number of college students who say they use credit cards because of the rewards program is exactly two.

b) There is a 41.37% probability that the number of college students who say they use credit cards because of the rewards program is more than two.

c) There is a 69.49% probability that the number of college students who say they use credit cards because of the rewards program is between two and five, inclusive.

Step-by-step explanation:

There are only two possible outcomes. Either the student use credit cards because of the rewards program, or they use for other reason. So, we can solve this exercise using the binomial probability distribution.

Binomial probability distribution

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

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

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

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

And $\pi$ is the probability of X happening.

In this problem, we have that:

10 students are randomly selected, so $n = 10$ .

23% of college students say they use credit cards because of the rewards program. This means that $\pi = 0.23$

(a) exactly two

This is P(X = 2).

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

$P(X = 2) = C_{10,2}.(0.23)^{2}.(0.77)^{8} = 0.2942$

There is a 29.42% probability that the number of college students who say they use credit cards because of the rewards program is exactly two.

(b) more than two

This is $P(X > 2)$ .

Either a value is larger than two, or it is smaller of equal. The sum of the decimal probabilities of these events must be 1. So:

$P(X \leq 2) + P(X > 2) = 1$

$P(X > 2) = 1 - P(X \leq 2)$

In which

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

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

$P(X = 0) = C_{10,0}.(0.23)^{0}.(0.77)^{10} = 0.0733$

$P(X = 1) = C_{10,1}.(0.23)^{1}.(0.77)^{9} = 0.2188$

$P(X = 2) = C_{10,2}.(0.23)^{2}.(0.77)^{8} = 0.2942$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0733 + 0.2188 + 0.2942 = 0.5863$

$P(X > 2) = 1 - P(X \leq 2) = 1 - 0.5863 = 0.4137$

There is a 41.37% probability that the number of college students who say they use credit cards because of the rewards program is more than two.

(c) between two and five inclusive.

This is

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

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

$P(X = 2) = C_{10,2}.(0.23)^{2}.(0.77)^{8} = 0.2942$

$P(X = 3) = C_{10,3}.(0.23)^{3}.(0.77)^{7} = 0.2343$

$P(X = 4) = C_{10,4}.(0.23)^{4}.(0.77)^{6} = 0.1225$

$P(X = 5) = C_{10,3}.(0.23)^{5}.(0.77)^{5} = 0.0439$

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.2942 + 0.2343 + 0.1225 + 0.0439 = 0.6949$

There is a 69.49% probability that the number of college students who say they use credit cards because of the rewards program is between two and five, inclusive.

8 0

3 years ago