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4vir4ik [10]
3 years ago
13

You play video games online. When you sign up with the game site, you get 200 points. You earn 20 more points for each hour that

you play. How many hours do you have to play to get to 500 points?
10 hours
25 hours
15 hours
35 hours
Mathematics
2 answers:
Savatey [412]3 years ago
8 0
500 = 200 + 20h is our equation.
Subtract 200 from both sides
300 = 20h
Divide by 20
300/20 = 15.
It will take 15 hours to get 500 points.
Hope this helps!
mr Goodwill [35]3 years ago
3 0
Well we are starting off with 200 just for signing up. So now we subtract the 200 from 500 because we want the total amount of hours. Now 500 divided by 20 will get you 15 hours of time. So your answer is 15 hours. Hope this helped
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Simplify 5m-2(6m-5)+2<br><br>a. -7m-8<br>b. -7m+12<br>c. 17m+12<br>d. 17m-8
fgiga [73]

The answer is B. -7m+12.


First distribute the -2 to 6m and -5.

5m-2(6m-5)+2

5m+(-2*6m)+(-2*-5)+2

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After that, combine the like terms.

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7 0
3 years ago
I could use some help!!
12345 [234]

Answer:

84

Step-by-step explanation:

The interquartile range is obtained using the relation :

Third quartile (Q3). - First quartile (Q1)

From. The boxplot :

Q3 = 96

Q1 = 12

Interquartile range (IQR) = Q3 - Q1 = 96 - 12 = 84

6 0
2 years ago
Samples of 20 parts from a metal punching process are selected every hour. Typically, 1% of the parts require rework. Let X deno
Roman55 [17]

Answer:

a) P(X>np+3\sqrt{np(1-p)}=0.017

b) P(x>1)=0.190

c) P(Y>1)=0.651

Step-by-step explanation:

This a binomial experiment where success is denoted by parts that need rework.

X ∼ B(n, p); n = 20; p = 0.01

The expected value of X is: E(X) = np =20×0.01= 0.2

The variance is: Var(X) = np(1 − p) = 0.2 × 0.99 = 0.198,

The standard deviation SD(X)= \sqrt{0.198} ≈ 0.445

a) P(X>np+3\sqrt{np(1-p)}=P(X>0.2+3×0.445)=P(X>1.535)=P(X≥2)

Probability function is given by:

\frac{n!}{x!(n-x)!} *p^x*(1-p)^{(n-x)}

P(X≥2)=1-P(X<2)=1-P(X=1)-P(X=0)= 1 - \frac{20!}{1!(20-1)!} *(0.01)^{1}*(1-0.01)^{(20-1)}-\frac{20!}{0!(20-0)!} *(0.01)^{0}*(1-0.01)^{(20-0)}

P(X≥2)=1-0.165-0.818=0.017

b) p=0.04

P(x>1)=P(x≥2)= 1 - P(x=1) - P(x=0)= 1 - \frac{20!}{0!(20-1)!} *(0.04)^{1}*(1-0.04)^{(20-1)} - \frac{20!}{0!(20-0)!} *(0.04)^{0}*(1-0.04)^{(20-0)}

P(x>1)= 1 - 0.368 - 0.442=0.190

c) In this case we consider p=0.19 (Probability that X exceeds 1)

In this experiment Y is the number of hours and n= 5 hours.

Then, we check the probability in each hour:

P(Y>1)=1- P(Y=0)

P(Y=0)=\frac{5!}{0!(5-0)!} *(0.19)^{0}*(1-0.19)^{(5-0)}=0.349

P(Y>1)=1-0.349=0.651

3 0
3 years ago
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