All categories

3 years ago

Can someone help me please

Mathematics

1 answer:

DochEvi [55]3 years ago

8 0

Answer:

Step-by-step explanation:

You might be interested in

Plz help with this one I have ten minutes to turn it in! Will give u points! And mark u BRAINLY

Neporo4naja [7]

I think it’s -3,0,0 but i’m not 100% sure

8 0

3 years ago

Read 2 more answers

An automated egg carton loader has a 1% probability of cracking an egg, and a customer will complain if more than one egg per do

vaieri [72.5K]

Answer:

a) Binomial distribution B(n=12,p=0.01)

b) P=0.007

c) P=0.999924

d) P=0.366

Step-by-step explanation:

a) The distribution of cracked eggs per dozen should be a binomial distribution B(12,0.01), as it can model 12 independent events.

b) To calculate the probability of having a carton of dozen eggs with more than one cracked egg, we will first calculate the probabilities of having zero or one cracked egg.

$P(k=0)=\binom{12}{0}p^0(1-p)^{12}=1*1*0.99^{12}=1*0.886=0.886\\\\P(k=1)=\binom{12}{1}p^1(1-p)^{11}=12*0.01*0.99^{11}=12*0.01*0.895=0.107$

Then,

$P(k>1)=1-(P(k=0)+P(k=1))=1-(0.886+0.107)=1-0.993=0.007$

c) In this case, the distribution is B(1200,0.01)

$P(k=0)=\binom{1200}{0}p^0(1-p)^{12}=1*1*0.99^{1200}=1* 0.000006 = 0.000006 \\\\ P(k=1)=\binom{1200}{1}p^1(1-p)^{1199}=1200*0.01*0.99^{1199}=1200*0.01* 0.000006 \\\\P(k=1)= 0.00007\\\\\\P(k\leq1)=0.000006+0.000070=0.000076\\\\\\P(k>1)=1-P(k\leq 1)=1-0.000076=0.999924$

d) In this case, the distribution is B(100,0.01)

We can calculate this probability as the probability of having 0 cracked eggs in a batch of 100 eggs.

$P(k=0)=\binom{100}{0}p^0(1-p)^{100}=0.99^{100}=0.366$

5 0

3 years ago

A circle has a circumference of 42 in.what is the area of the circle

Solnce55 [7]

Answer:

140.37 in²

Step-by-step explanation:

C = 2πr

42 = 2πr

21 = πr

r = 21/π

A = πr²

A = π × (21/π)²

≈ 140.37 in²

3 0

3 years ago

Read 2 more answers

taurus [48]

search for the formula for net income

Step-by-step explanation:

that would help u more.

6 0

3 years ago

What is the greatest common factor of 32 and 48

Bond [772]

Turn each number into the product of it's prime factors.

32=16x2=2x2x2x2x2=2^5

48=24*2=6x4x2=2x3x2x2x2

Pick the highest number that occurs. In this case it is 2. Now we have to see how many times it appears in both. It appears 5 times in 32 and 4 times in 48. 4 is the highest number of times it appears in the numbers so:

2^4=2x2x2x2=16

The Greatest Common Factor (GCF) of 32 and 48 is 16.

4 0

3 years ago

Read 2 more answers