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2 years ago

How do I calculate these division facts and what’s the answer 63 ÷ 7 56 ÷ 7

Mathematics

2 answers:

bixtya [17]2 years ago

3 0

Answer:

63 divided by 7 is 9

56 divided by 7 is 8

Step-by-step explanation:

I don't really know how to explain it, because it's kind of complicated to do it how I answered it. I will try, but what you do is write out the equation using the little line division thing with the divisor and the dividend. First for 63 divided by 7, how many times can 7 go into 6? 0, so put a 0 on top of the 6. Then, minus 6 by 0, which is 6. Then, drop down the 3 in the ones place to where the difference of the subtraction problem is, which of 6. Then, we try to find a product that is 63 or close to 63 in a multiplication problem, and one of those factors has to be 7. 7 x 9 = 63, so put a 9 on top of the 3 at the beginning and minus 63 - 63, which is 0, meaning it has no remainder. Very hard to explain digitally, look up a video for help.

gizmo_the_mogwai [7]2 years ago

3 0

Yeah so I don't know the answer but I think the answer is bla bla bla

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julsineya [31]

Answer:

a. $P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}\\$

b. $E(x) = 0.3$

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Step-by-step explanation:

The probability that the company won x bids follows a binomial distribution because we have n identical and independent experiments with a probability p of success and (1-p) of fail.

So, the PMF of X is equal to:

$P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}\\$

Where p is 0.1 and it is the chance of winning. Additionally, n is 3 and it is the number of bids. So the PMF of X is:

$P(x)=\frac{3!}{x!(3-x)!}*0.1^{x}*(1-0.1)^{n-x}\\$

For binomial distribution:

$E(x)=np\\S(x)=\sqrt{np(1-p)}$

Therefore, the company can expect to win 0.3 bids and it is calculated as:

$E(x) = np = 3*0.1 = 0.3$

Additionally, the standard deviation of the number of bids won is:

$S(x)=\sqrt{np(1-p)}=\sqrt{3(0.1)(1-0.1)}=0.5196$

Finally, the probability to won 1, 2 or 3 bids is equal to:

$P(1)=\frac{3!}{1!(3-1)!}*0.1^{1}*(1-0.1)^{3-1}=0.243\\P(2)=\frac{3!}{2!(3-2)!}*0.1^{2}*(1-0.1)^{3-2}=0.027\\P(3)=\frac{3!}{3!(3-3)!}*0.1^{3}*(1-0.1)^{3-3}=0.001$

So, the expected profit for the company is equal to:

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Because there is a probability of 0.243 to win one bid and it will produce 50,000 of income, there is a probability of 0.027 to win 2 bids and it will produce 100,000 of income and there is a probability of 0.001 to win 3 bids and it will produce 150,000 of income.

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