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3 years ago

9

Three fourths of the 8th graders at Kingston middle school went on a class trip if 252 students went on the trip, how many 8th g

raders are there

1 answer:

Minchanka [31]3 years ago

3 0

Divide the three fourths of the eight graders by 252 and then simplify

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Which of the following options results in a graph that shows exponential growth?

poizon [28]

Answer:

$f(x)=0.7(5)^x$

Step-by-step explanation:

$\text{Exponential function: }y=a\cdot b^x$

It is two types of function.

Exponential Growth: If function increase as value of x increase.

In this case, a>0 and b>1

Exponential decay: If function decrease as value of x increase.

In this case, a>0 and 0<b<1

We need to choose exponential growth function.

$\text{Exponential Growth function: }f(x)=0.7(5)^x$

Here a=0.75>0 and b=5>1

Therefore, This function is exponential growth.

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3 years ago

Read 2 more answers

What is the constant of posprtainality in the equation y=0.41

KATRIN_1 [288]

Answer:

The constant of proportionality in the equation y=0.41 is 0.41

Step-by-step explanation:

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3 years ago

Julia is the president of the technology Club at school she has $568 to buy as many DVD players as she can for $75 each how many

goblinko [34]

You need to divide the total amount by $75, to see how many DVD players she could buy with the $568.
568÷75= 7 remaining 43
This means that Julia can buy 7 DVD players, and will have $43 extra dollars left over.

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3 years ago

it takes 29 minutes to ride your skateboard to school, give or take 4 minutes. what are the possible times (t) it takes you to g

lorasvet [3.4K]

29-4 = 25
29+4 = 33
So, from 25 to 33 minutes.

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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