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3 years ago

A farmer saw some chickens and pigs in a field. He counted 51 heads and 172 legs. Determine

Mathematics

1 answer:

Sliva [168]3 years ago

3 0

Answers:

16 chickens

35 pigs

========================================================

Explanation:

x = number of chickens

y = number of pigs

There are 51 heads, so x+y = 51 which solves to y = 51-x

2x = number of legs just from the chickens (2 legs per chicken)

4y = number of legs just from the pigs (4 legs per pig)

2x+4y = 172 legs total

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Plug in y = 51-x and solve for x.

2x+4y = 172

2x+4(51-x) = 172

2x+204-4x = 172

-2x+204 = 172

-2x = 172-204

-2x = -32

x = -32/(-2)

x = 16

Now use this to find y

y = 51-x

y = 51-16

y = 35

Since x = 16 and y = 35, this means there are 16 chickens and 35 pigs.

----------

Check:

x+y = 16+35 = 51 which works

2x+4y = 2*16+4*35 = 172 and that also works.

Both equations are confirmed.

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Read 2 more answers

An experiment to investigate the survival time in hours of an electronic component consists of placing the parts in a test cell

mixer [17]

Answer:

a) The sample mean is of 49 and the sample standard deviation is of 11.7.

b) The range of the true mean at 90% confidence level is of 9.62 hours.

c) The prediction interval, at a 90% confidence level, of it's failure time is between 39.38 hours and 58.62 hours.

Step-by-step explanation:

Question a:

Sample mean:

$\overline{x} = \frac{34+40+46+49+61+64}{6} = 49$

Sample standard deviation:

$s = sqrt{\frac{(34-49)^2+(40-49)^2+(46-49)^2+(49-49)^2+(61-49)^2+(64-49)^2}{5}} = 11.7$

The sample mean is of 49 and the sample standard deviation is of 11.7.

b)Determine the range of the true mean at 90% confidence level.

We have to find the margin of error of the confidence interval. Since we have the standard deviation for the sample, the t-distribution is used.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 6 - 1 = 5

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 5 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.9}{2} = 0.95$ . So we have T = 2.0.150

The margin of error is:

$M = T\frac{s}{\sqrt{n}}$

In which s is the standard deviation of the sample and n is the size of the sample. So

$M = 2.0150\frac{11.7}{\sqrt{6}} = 9.62$

The range of the true mean at 90% confidence level is of 9.62 hours.

(c)If a seventh sample is tested, what is the prediction interval (90% confidence level) of its failure time.

This is the confidence interval, so:

The lower end of the interval is the sample mean subtracted by M. So it is 49 - 9.62 = 39.38 hours.

The upper end of the interval is the sample mean added to M. So it is 49 + 9.62 = 58.62 hours.

The prediction interval, at a 90% confidence level, of it's failure time is between 39.38 hours and 58.62 hours.

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