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3 years ago

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Christian is throwing a pizza party . he has 1 1/3pounds of cheese available and he needs 1/2 pounds of cheese to make each pizz

a . how many full for pizzas can he make with the amount of cheese he has?

1 answer:

Ivanshal [37]3 years ago

3 0

Answer:

2

Step-by-step explanation:

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Danielle stayed in three different cities (Washington, D.C., Atlanta, Georgia, and Dallas, Texas) for a total of 22 nights. She

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Step-by-step explanation:

you had putten 5 cities not 3

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I need help making a proof for geometry! Here is the question…

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2 years ago

The 8th and 20th terms of an arithmetic sequence are 20 and 56 respectively. What is the 1st term of the sequence.

Ostrovityanka [42]

Answer:

<h3> a₁ = -1</h3>

Step-by-step explanation:

$a_n=a_1+d(n-1)\\\\a_8=a_1+d\cdot7\\\\a_{20}=a_1+d\cdot19\\\\a_{20}-a_8=a_1+19d-(a_1+7d)=12d\\\\56-20=2d\\\\12d=36\\\\d=3\\\\\\ a_1+3\cdot7=20\\\\a_1=20-21=-1$

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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

What is the slope and y-intercept of 6x-5y=13

DaniilM [7]

Answer:

The slope is 6/5 and the y intercept is -13/5

Step-by-step explanation:

6x-5y=13

Solve for y

Subtract 6x from each side

6x-6x-5y=-6x+13

-5y = -6x+13

Divide by -5

-5y/-5 = -6x/-5 +13/-5

y = +6/5x -13/5

This is in slope intercept form y = mx+b where m is the slope and b is the y intercept

The slope is 6/5 and the y intercept is -13/5

5 0

3 years ago