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son4ous [18]
3 years ago
6

last month , he sold 50 chickens and 30 ducks for $550 . This month , he sold 44 chickens and 36 ducks for $532. How much does a

chicken cost , and how much does a duck cost ?
Mathematics
2 answers:
Gelneren [198K]3 years ago
6 0

Answer:

chicken cost $8 and ducks cost $5

Step-by-step explanation:

Kazeer [188]3 years ago
4 0
Hello there, my fellow human being!
So, the chicken costs $8 and the duck costs $5, here's why.

Let's say x is the cost of a chicken while y is the cost of a duck.
We can make two linear equations using the information above.
Last month, he sold 50 chickens and 30 ducks for $550: 50x + 30y= 550
This month, he sold 44 chicken and 36 ducks for $532: 44x + 36y = 532

50x/10 + 30y/10 = 440/10
44x/4 + 36y/4 = 532/4

5x + 3y = 44
11x + 9y = 133

So, now that we have our answer simplified, we have to use elimination to solve this system of equations. But, first we need to make sure that at least one of our variables is able to be canceled out.
Let's multiply this equation by -3.
(5x + 3y = 44) * -3.

-15x-9y=-165.

11x + 9y = 133
-15x - 9y = -165
______________
-4x/4 = -32/4
x = 8
11x + 9y = 133
11(8) + 9y = 133
88 + 9y = 133
-88 -88
________________
9y/9 = 45/9
y = 5.
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A new car is purchased for 19700 dollars. The value of the car depreciates at
Usimov [2.4K]

Answer:

Step-by-step explanation:

We can plug this into a formula and get

5,500 = 19,700(.9075)^{?}

the .9075 comes from 1 - the depreciation value of 9.25% which converts to .0925 in decimal form.

? is what we are trying to solve for.

We can divide both sides by 19,700 and we get

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3 years ago
a random sample of 4 claims are selected from a lot of 12 that has 3 nonconforming units. using the hypergeometric distribution
Sloan [31]

Answer:

The probability that the sample will contain exactly 0 nonconforming units is P=0.25.

The probability that the sample will contain exactly 1 nonconforming units is P=0.51.

.

Step-by-step explanation:

We have a sample of size n=4, taken out of a lot of N=12 units, where K=3 are non-conforming units.

We can write the probability mass function as:

P(x=k)=\frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}

where k is the number of non-conforming units on the sample of n=4.

We can calculate the probability of getting no non-conforming units (k=0) as:

P(x=0)=\frac{\binom{3}{0}\binom{9}{4}}{\binom{12}{4}}=\frac{1*126}{495}=\frac{126}{495} = 0.25

We can calculate the probability of getting one non-conforming units (k=1) as:

P(x=1)=\frac{\binom{3}{1}\binom{9}{3}}{\binom{12}{4}}=\frac{3*84}{495}=\frac{252}{495} = 0.51

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UNO [17]
12/21 in its simplest form is 4/7.

To find this out we need to divide the numerator and denominator by the GCF of 12 and 21 which is 3.

12/21

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21 ÷ 3 = 7
4/7

12/21 in its simplest form is 4/7.

12/21 = 4/7
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Mode : The value which occures most frequently in a data set.

12,9,12,11,10,18,7,19,13,19

Mean = 13 (Average)

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Step-by-step explanation:

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