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2 years ago

Don’t steal points Pls I’m doing a final need it quickly

Mathematics

2 answers:

Keith_Richards [23]2 years ago

7 0

It’s the third one.

Tcecarenko [31]2 years ago

4 0

Answer:

the 3rd one

Step-by-step explanation:

A proportional relationship has to have a Stright line. Hope this helps :)

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Kate can run 5 blocks per minute she wants to run 80 blocks how long will it take her to finish if she has already run 20 blocks

ki77a [65]

12 minutes because 80-20 is 60 and 60 divides by 5 is 12

3 0

3 years ago

Read 2 more answers

Help please with #3

Alenkasestr [34]

Answer: x = 2a over b + c

Step-by-step explanation:

1. Multiply both sides by two. This cancels out 2 and turns a into 2a. It should look like 2a = bx + cx

2) Since both b and c have x’s, we turn it into (b + c)x

3. Then divide both sides by b+c to get 2a over b + c = x

6 0

2 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

Help ASAP:Solve : 0.84+91.279

jeka94

Answer:

92.119

Step-by-step explanation:

0.84+91.279=92.119

Hope this helps!

6 0

3 years ago

Read 2 more answers

The following options are available for a car wash

Margaret [11]

Answer:

£576

(would really, reallly appreciate the brainliest)

Step-by-step explanation:

8 * 60 * 3/5 + 12 * 60 * 2/5

3 0

2 years ago