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3 years ago

This cartoon expresses a viewpoint that led to...

Mathematics

1 answer:

Klio2033 [76]3 years ago

5 0

Answer:The answer is D

Step-by-step explanation:

the ratification of a constitutional amendment establishing term limits for presidents.

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I dont get this question.

SashulF [63]

It is 12 - 2/3 = 18, then solve :)
hope that helps :D

8 0

4 years ago

The price of gold has increased by 35% per year from 2000. In the year 2000, Harry bought a gold ring for $590. Which of the fol

Finger [1]

The function representing the price of the ring x years after 2000 is f(x) = (100 + 35)/100 *x + 590 = 135/100 *x + 590 = 1.35x + 590
i.e. f(x) = 1.35x + 590

3 0

3 years ago

Read 2 more answers

3) Washing your hands kills germs. If there are 275 germs chilling on your hands and

Romashka [77]

Answer:

So we can use geometric progression each time multiplying by 0.0475

so thats (275*0.0475)*10

So that means that we would get

130.625 so we subtract that from 275

275-130.625=144.375

That would be

Step-by-step explanation:

4 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

Quadrilateral ABCD is similar to quadrilateral EFGH. What is the length of<br> segment BC?

prohojiy [21]

So Basically, these are similar. So that means that (2x/5.5)=(3x-3/7.5). Then when you cross multiply, you find that x=11. So that means BA/FE=22/5.5=4. You can verify by checking (3*11-3)/7.4=4. So now that you know the scale factor is 4, just do 4*8=32.

6 0

3 years ago

Read 2 more answers