Solve 3x – 7y = 25 5x – 8y = 27 using matrices. (25, 27) (27, 25) (-1, -4) (4, 1)

1 answer:

8 0

<span>This is somewhat similar to something called Cramer's Rule http://www.1728.org/cramer.htm

A) 3x – 7y = 25
a =3 b=-7 e=25
B) 5x – 8y = 27</span>
c=5 d=-8 f=27

dn = |a b |
      |c d |

dn = |3 -7|
   |5 -8

dn = (3*-8) -(5*-7)

dn = -24 +35

dn = 11

x = |e b|
      |f d|
divided by dn

x = |25 -7|
      |27 -8|
divided by dn

x = -200 -(-189) / 11
x = -11 / 11
x = -1

y = |a e |
   |c f |
divided by dn

y = |3 25|
   |5 27|
divided by dn

y = 81 -125 / dn = -44 / 11 =-4

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The Wisconsin Dairy Association is interested in estimating the mean weekly consumption of milk for adults over the age of 18 in

Stolb23 [73]

Answer:

$ME = 1.653 *\frac{7.9}{\sqrt{300}}= \pm 0.75$

±0.75 ounce

Step-by-step explanation:

Assuming this complete question: The Wisconsin Dairy Association is interested in estimating the mean weekly consumption of milk for adults over the age of 18 in that state. To do this, they have selected a random sample of 300 people from the designated population. The following results were recorded: xbar=34.5 ounces, s=7.9 ounces Given this information, if the leaders wish to estimate the mean milk consumption with 90 percent confidence, what is the approximate margin of error in the estimate?

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=34.5$ represent the sample mean

$\mu$ population mean (variable of interest)

s=7.9 represent the sample standard deviation

n=300 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=300-1=299$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,199)".And we see that $t_{\alpha/2}=1.653$