I can't remember for the life of me what these are

Solve 7d + d - 4d - 7 - 5d

Butoxors [25]

-1d-7 is the answer I'm pretty sure

6 0

4 years ago

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Write 16.3 as a mixed number

GarryVolchara [31]

Answer:

Step-by-step explanation:

The answer is 16 3/10

That's about as mixed as you can get it.

4 0

3 years ago

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6(8^2)* 45-98+456= Answers: 17890 17182 17638 18400

sukhopar [10]

Answer:

17638 its the correct one

8 0

3 years ago

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One number is 7 more than another. The difference between their squares is 161. What are the numbers?

olga nikolaevna [1]

Let's say the numbers are "a" and "b"

now, we know one of them is 7 more than the other, so hmmm let's say "b" is 7 more than "a", so b = a + 7

the difference of their squares is 161, alrite

$\bf a^2-b^2=161\qquad b=a+7\implies a^2-(a+7)^2=161
\\\\\\
a^2-(a^2+14a+49)=161\implies \underline{a^2-a^2}-14a-49=161
\\\\\\
-14a-49=161\implies -49-161=14a\implies -210=14a
\\\\\\
\cfrac{-210}{14}=a$

and surey you know how much that is

now, what's "b"? well, b = a + 7

3 0

4 years ago

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

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

So

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

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

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

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

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

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

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

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

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

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

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

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

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

So

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

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