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Dafna11 [192]
2 years ago
9

Ms. Stapleton had 2,600 hens, ducks, and goats on her farm. 35% of them were hens, and 25% of them were goats. How many ducks di

d she have on her farm ?
Mathematics
1 answer:
DIA [1.3K]2 years ago
6 0
Hello!

ANSWER:

Ms. Stapleton had 1,040 ducks on her farm.

-------------------------------------------------------

EXPLANATION:

What we know:

We know that altogether Ms. Stapleton had 2,600 animals.

35% of which were hens, 25% were goats, and the rest were goats.

-------------------------------------------------------

How to solve:

To find out how many of the animals are ducks, we must first find out how many are hens and ducks and then subtract them from the total number of animals.

-------------------------------------------------------

Convert the percentages to decimals by dividing them by 100, then multiply.

35 ÷ 100 = 0.35

25 ÷ 100 = 0.25

2,600 × 0.35 = 910

2,600 × 0.25 = 650

910 + 650 = 1,560
-------------------------------------------------------

Ms. Stapleton had 910 hens and 650 goats on her farm, altogether 1,560. We must subtract:

2,600 - 1,560 = 1,040



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Answer:

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

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d. E=5,000

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

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E=-10,000+50,000(0.243)+100,000(0.027)+150,000(0.001)\\E=5,000

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3 0
2 years ago
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Answer:

A. $3992.50

Step-by-step explanation:

Credit line - $8,000.00

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8 0
2 years ago
the population of two towns were equal in a particular year. subsequently the population of one town increases by 8 % and the po
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8 0
2 years ago
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