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2 years ago

TRUE OR FALSE LOOK THE PHOTO RENAE HELP ME FIRST LOL

Mathematics

1 answer:

Lunna [17]2 years ago

7 0

Answer: theres no photo ?

Step-by-step explanation:

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A coin is tossed 5 times and a tail is the outcome each time. what is the probability of getting a tail on the sixth throw

Norma-Jean [14]

I believe it would be
1/5

5 0

2 years ago

alo had the third highest grade oint avwrage (GPA) of a group of five. His GPA was 79, amd ir was 12 points lower thank the high

soldier1979 [14.2K]

The fourth highest GPA is 6

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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

I'LL MARK YOU BRAINLIST !! HELPP PLS ...2nd time posting this sjsjsjka

maw [93]

Answer:

CORRECT ANSWER 5 AND 6

PLZ GIVE BRAINLIEST

3 0

2 years ago

Please answer ASAP! The Fergusons decided to book the room at the price that Mrs. Ferguson was given, $2075, because it included

Alex787 [66]

So what you should do is
15% off, then add tax or
2075 times 0.85 times 1.185=2090.04 about

you can't just add and subtract percents because they are percents of different things

what he did was
he found 15% of 2075 and 18.5% of 2075 then found the difference and added to 2075 to find his total

what he should have done is
found the discounted price (15% off) then found the tax on that

first way
2075 times 0.15=311.25
2075-311.25=1763.75
1763.75 times 0.185=326.29375
1763.75-326.29375=2090.04

2nd way (easy, if you understand commutative property and stuff)
100-15=85
1+0.185=1.185
2075*0.85*1.185=2090.04

answer should be $2090.04

7 0

3 years ago