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2 years ago

All parts please, i need this by the end of today and pls write answer in comments

Mathematics

1 answer:

ale4655 [162]2 years ago

7 0

Answer:

$= \frac{78}{360} \\ = \frac{13}{60}$

$let \: total \: number \: be \: x \\ \frac{13}{60} x = 1164 \\ x = \frac{1164 \times 60}{13} \\ x = 5372 \: movies$

A year has (4 × 12) weeks

= 48 weeks.

In 1 week = 9,568,000 movies

48 weeks = (48 × 9,568,000) movies

Average movies = 459.264 movies

Since 9.568 million movies are rented.

How about in one year ( 48 weeks )

Convert 1 year to weeks and compare with 1 week

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A pen company averages 1.2 defective pens per carton produced (200 pens). The number of defects per carton is Poisson distribute

nlexa [21]

Answer:

a. P(x = 0 | λ = 1.2) = 0.301

b. P(x ≥ 8 | λ = 1.2) = 0.000

c. P(x > 5 | λ = 1.2) = 0.002

Step-by-step explanation:

If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:

$P(k)=\frac{\lambda^{k}e^{-\lambda}}{k!}= \frac{1.2^{k}\cdot e^{-1.2}}{k!}$

a. What is the probability of selecting a carton and finding no defective pens?

This happens for k=0, so the probability is:

$P(0)=\frac{1.2^{0}\cdot e^{-1.2}}{0!}=e^{-1.2}=0.301$

b. What is the probability of finding eight or more defective pens in a carton?

This can be calculated as one minus the probablity of having 7 or less defective pens.

$P(k\geq8)=1-P(k$

$P(0)=1.2^{0} \cdot e^{-1.2}/0!=1*0.3012/1=0.301\\\\P(1)=1.2^{1} \cdot e^{-1.2}/1!=1*0.3012/1=0.361\\\\P(2)=1.2^{2} \cdot e^{-1.2}/2!=1*0.3012/2=0.217\\\\P(3)=1.2^{3} \cdot e^{-1.2}/3!=2*0.3012/6=0.087\\\\P(4)=1.2^{4} \cdot e^{-1.2}/4!=2*0.3012/24=0.026\\\\P(5)=1.2^{5} \cdot e^{-1.2}/5!=2*0.3012/120=0.006\\\\P(6)=1.2^{6} \cdot e^{-1.2}/6!=3*0.3012/720=0.001\\\\P(7)=1.2^{7} \cdot e^{-1.2}/7!=4*0.3012/5040=0\\\\$

$P(k$

c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?

We can calculate this as we did the previous question, but for k=5.

$P(k>5)=1-P(k\leq5)=1-\sum_{k=0}^5P(k)\\\\P(k>5)=1-(0.301+0.361+0.217+0.087+0.026+0.006)\\\\P(k>5)=1-0.998=0.002$

5 0

3 years ago

Select the graph of the solution. Click until the correct graph appears.

faust18 [17]

Answer:

C. is your answer

Step-by-step explanation:

4 0

2 years ago

1,200 laptop ; 7.25% tax can someone help me!!

ziro4ka [17]

Answer:

$2,070

Step-by-step explanation:

1200*0.0725=870

the amount of tax is $870

1200+870=2070

7 0

2 years ago

An airplane is preparing to land an airport. It is 34,800 feet above the ground and descending at the rate of 3,000 feet per min

sveta [45]

Answer:

6 mins, and the altitude will be 16,800 ft

Step-by-step explanation:

1st: Subtract 3000 from 34800

2nd: Multiply 2 to 2800

Keep doing this until you get the same altitude for each one.

3 0

3 years ago

What is the relationship among the numbers and symbols in the expression (6 – 4) • 2? Choose all answers that are correct. A. Th

mash [69]

The answer is option B "The difference between 6 and 4 is doubled in 2." Since how 6-4=2 then 2×2=4 which means it is split into two different parts.

Would be solved this way:

 (6 – 4) • 2 (Using PEMDAS)
Use P:
6-4=2
Skip E:
Use M:
2×2=4
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Hope this helps!

4 0

3 years ago