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3 years ago

What is the least common multiple of 4 and 8? Enter your answer in the box.

Mathematics

1 answer:

jeyben [28]3 years ago

7 0

Answer:

The least common would be 8

Step-by-step explanation:

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What are the possible values of x for the following functions? f(x)=2-x/x(x-1)

nevsk [136]

The possible values of x for the following functions are values on real number except 0 and 1

<h3>Domain of a function</h3>

The domain of a function are the values of the independent variable for which it exists.

Given the function below

f(x)=2-x/x(x-1)

The function does not exist at the. point where the denominator is zero. From the function given, the function does not exist when;

x(x -1) = 0

x = 0 and x = 1

Hence the possible values of x for the following functions are values on real number except 0 and 1

Learn more on domain of a function here; brainly.com/question/1770447

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1 year ago

I need the answerrrr asappppp

Tresset [83]

Answer:

48 inch

Step-by-step explanation:

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2 years ago

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What is -4/5x-2=18 what is x help quick

mart [117]

Answer:

If I'm right x= -25

Step-by-step explanation:

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2 years ago

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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

Add three data values to the following data set so the mean increases by 10 but the median does not change. The numbers are 42 3

777dan777 [17]

In order the numbers are 20, 29, 32, 37, 42.
So that means the median is 32. In order for the median to stay the same one of the three data values must be 32 as well as adding a number to each side of the ends. One 32 or less another 32or more.
The mean right now is: 20+29+32+37+42=160 and there are 5 values.
Mean: 160/5=32
For it to increase by 10 it has to be 42.
extra 3 numbers I used were 31, 32, & 113 (those are your answer)

3 0

3 years ago