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4 years ago

The greatest common factor of 24, 48m

Mathematics

1 answer:

Arturiano [62]4 years ago

3 0

Answer:

The greatest common factor of 24 and 48 is 24.

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What is the answer to i^{2n+1}

nekit [7.7K]

Answer:

(no solutions)

Step-by-step explanation:

Solve for n:

(0 + i)^(2 n + 1) = 0

Since i^z is never zero for any z element C, no solution exists for i^(2 n + 1) = 0:

Answer: (no solutions)

7 0

3 years ago

Jo's collection contains US, Indian and British stamps. If the ratio of US to Indian stamps is 5 to 2 and the ratio of Indian to

andreev551 [17]

Answer:(E). 25:2

Step-by-step explanation:

First ratio:

Us : India is 5 : 2

Second ratio

India : British is 5 : 1

multiply(5:2) by 5 and multiply (5:1) by 2 to balance the India stamps.

Therefore now

Us : India is 25 : 10

India : British is 10 : 2

The ratio of us : British is 25:2

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3 years ago

Can someone help me?

kirza4 [7]

Answer:

Second one: (1,2) (2,4) (3,6)

Step-by-step explanation:

y = 2x

2 = 1(2)

4 = 2(2)

6 = 2(3)

7 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

Urgent please help! will give brainliest

marishachu [46]

Answer:

The answer is A

Step-by-step explanation:

(-2,-3) (3,-4) (0,-1) (-7,3)

inverse means the opposite signs

(2,3) (-3,4) (0,1) (7,-3)

3 0

4 years ago