All categories

3 years ago

Please help I need both company A and B!!!!

Mathematics

2 answers:

WARRIOR [948]3 years ago

7 0

Answer:

Company A: 40x + 50

Company B: 50x +25

Step-by-step explanation:

Company A's Fee is 50 dollars and there charge per hour (x) is 40, so since we don't know how many hours we substitute it for the variable x. Same explanation for Company B but with its data.

alexandr402 [8]3 years ago

5 0

This one was a little tricky when i did it but heres your answer

Answer: do 40x + 50

and 50x + 25

Step-by-step explanation:

You might be interested in

Alr, time to fight pico.

Troyanec [42]

Answer:

niceeeeeeeeeeeeeeeee!

Step-by-step explanation:

5 0

3 years ago

What is the remainder of 3,368,008 divided by 5574

zheka24 [161]

The answer is 604.235378543

4 0

3 years ago

Read 2 more answers

Round the product of 0.24 and 0.26 to the nearest hundredth. What is the answer?

Andreyy89

0.06 because the product is 0.0624

3 0

3 years ago

If a rectangular prism is sliced vertically by a plane, what is the shape of the resulting two dimensional figure?

Romashka [77]

I think 2 triangular prisms.

8 0

3 years ago

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