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AVprozaik [17]
3 years ago
6

A CHICKEN COOP HOLDS A TOTAL OF 10 HENS OR 20 CHICKS. IF 2/5THS OF THE HENS IN A FULL COOP ARE REMOVED, HOW MANY NEW CHICKS COUL

D FIT IN THE CHICKEN COOP?
Mathematics
1 answer:
LiRa [457]3 years ago
5 0

Answer:

8 new chicks can be fitted in the coop.

Step-by-step explanation:

A chicken coop holds 10 hens or 20 chicks.

That means space in the coop for 10 hens = space for 20 chicks

Or space for 1 hen = space for 2 chicks

Now \frac{2}{5}th of the hens were removed from the coop.

So, number of hens removed from the coop = \frac{2}{5}\times 10

= 4 hens

And space for 4 hens in the coop = space for the 8 chicks

Therefore, 8 new chicks can be fitted in the coop.

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In a continuous series, mean(x) = 80, assumed mean(A) = 60 and<br>EF=40 then find the value of EFd.​
Aneli [31]

Using the given information, the value of Σfd is 800

<h3>Calculating mean using Assumed mean </h3>

From the question, we are to determine the value of Σfd

The formula for mean, using the assumed mean method is given by

\bar x = A + \frac{\sum fd}{\sum f}

Where \bar x is the mean

A is the assumed mean

From the given information,

\bar x = 80

A = 60

\sum f = 40

Putting the parameters into the equation, we get

80 = 60 + \frac{\sum fd}{40}

80-60 = \frac{\sum fd}{40}

20 = \frac{\sum fd}{40}

\sum fd = 20 \times 40

\sum fd = 800

Hence, the value of Σfd is 800

Learn more on Mean here: brainly.com/question/20118982

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2 years ago
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(6,2)

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Test a pair from each table by substituting their values into the given equation and solving.

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