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photoshop1234 [79]

3 years ago

15

Kiersten runs a website that helps people learn programming. Every month, Kiersten receives a subscription fee of $10 from each

of the subscribers to her website. Last month, the website had 100 subscribers. This month, 30 new members joined the website, and 10 members cancelled their subscription.
How much money will Kiersten’s online business earn this month?

1 answer:

just olya [345]3 years ago

3 0

So what you do is add 30 to 100 and then take away 10 leaving you with 120. You then multiply by 10 giving you 1,200 dollars a month. Wish I was making that much money a month :)

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At Ersatz University, the graduating class of 480 includes 96 guest students from Latvia. A sample of 10 students is selected at

grandymaker [24]

Answer:

A).3222

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they are from Latvia, or they are not. We are choosing students without replacement. However, since the sample size is large, we can use the binomial distribution to approximate the hypergeometric 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.

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

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.

480 students, of which 96 are from Latvia:

This means that $p = \frac{96}{480} = 0.2$

Sample of 10 students:

This means that $n = 10$

Probability that the sample contains at least three Latvian students.

Either there are less than three students from Latvia, or there are at least three. The sum of the probabilities of these events is 1. So

$P(X < 3) + P(X \geq 3) = 1$

We want $P(X \geq 3) = 1 - P(X < 3)$ , in which:

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

So

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

$P(X = 0) = C_{10,0}.(0.2)^{0}.(0.8)^{10} = 0.1074$

$P(X = 1) = C_{10,1}.(0.2)^{1}.(0.8)^{9} = 0.2684$

$P(X = 2) = C_{10,2}.(0.2)^{2}.(0.8)^{8} = 0.3020$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.1074 + 0.2648 + 0.3020 = 0.6742$

So

$P(X \geq 3) = 1 - P(X < 3) = 1 - 0.6742 = 0.3258$

The closest option is A.

4 0

3 years ago

Can anyone help me with this question ?

Assoli18 [71]

I’m pretty sure the answer is k= 10

8 0

3 years ago

What is 90 divided by 214

babunello [35]

4205 is the answer I think

3 0

2 years ago

Read 2 more answers

Rationalize the numerator or denominator, and simplify:

allochka39001 [22]

$\dfrac{\sqrt[3]{x+h}-\sqrt[3]x}h\times\dfrac{\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}}{\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}}=\dfrac{(\sqrt[3]{x+h})^3-(\sqrt[3]x)^3}\cdots$
$=\dfrac{x+h-x}\cdots=\dfrac h\cdots$

The $h$ s then cancel, leaving you with the $\cdots=\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}$ term.

If it's not clear what I did above, consider the substitution $a=\sqrt[3]{x+h}$ and $b=\sqrt[3]x$ . Then

$a^3-b^3=(a-b)(a^2+ab+b^2)$
$\implies\dfrac{a-b}h=\dfrac{a-b}h\times\dfrac{a^2+ab+b^2}{a^2+ab+b^2}=\dfrac{a^3-b^3}{h(a^2+ab+b^2)}$

4 0

3 years ago

In a triangle angle a=40 angle b= 45 angle c= 90 find <br> angle of x

cluponka [151]

Answer:angle x=5

Step-by-step explanation:

Sum of triangles =180 ,x+90+45+40=180, x+175=180 ,x =5

6 0

3 years ago