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S_A_V [24]
2 years ago
9

Mark wants to buy 5 new video games from

Mathematics
1 answer:
vlada-n [284]2 years ago
4 0
He will have to pay about $82.34 at the checkout.
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1 p

2 np

3 np

4 p

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A living room measures 24 feet by 15 feet. An adjacent square dining room measures 13 feet on each side. If carpet costs $6.98 p
blsea [12.9K]

Answer:

total cost of putting carpet in both rooms is $3692.42

Step-by-step explanation:

given data

living room length = 24 feet

living room width = 15 feet

dining room is square so length and width will be same

dining room length = 13 feet

dining room width = 13 feet

carpet costs =  $6.98 per square foot

to find out

what is the total cost of putting carpet in both rooms

solution

first we find the total area of both room that is

area of living room = length × width      .................1

area of living room = 24 × 15

area of living room = 360 feet²      ...........................2

and

areas of dining room = length × width

area of dining room = 13 × 13

area of dining room = 169 feet²    ...............................3

so total area = add equation 1 and 2

total area = 360 feet² + 169 feet²  

total area = 529 feet²  

and

total cost of putting carpet in both rooms are = total area × carpet costs    .................4

put here value

total cost of putting carpet in both rooms  = 529  ×  $6.98

total cost of putting carpet in both rooms  =$3692.42

6 0
3 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
2 years ago
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