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3 years ago

Rename 3/4 and 2/5 using 20 as the common denominator

Mathematics

1 answer:

Korvikt [17]3 years ago

5 0

Answer:

Let’s use 2 methods to do this:

First, let’s try cross multiplying:

3/4=x/20

2/5=x/20

20 times 3 = 60

60 divided by 4 is 15, so 3/4 = 15/20

20 times 2 = 40

40 divided by 5 is 8, so 2/5 = 8/20

The next method:

What time 4 equals to 20? What times 5 equals to 20?

4 times FIVE equals to 20

For the fraction of 3/4, multiply the numerator and the denominator by five to get your answer of 15/20.

5 times FOUR equals to 20

For the fraction of 2/5, multiply the numerator and the denominator by four to get your answer of 8/20.

Hope I helped!

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Location is known to affect the number, of a particular item, sold by an auto parts facility. Two different locations, A and B,

Mama L [17]

We have two samples, A and B, so we need to construct a 2 Samp T Int using this formula:

$\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$

In order to use t*, we need to check conditions for using a t-distribution first.

Random for both samples -- NOT STATED in the problem ∴ proceed with caution!
Independence for both samples: 130 < all items sold at Location A; 180 < all items sold at Location B -- we can reasonably assume this is true
Normality: CLT is not met; n < 30 for both locations A and B ∴ proceed with caution!

Since 2/3 conditions aren't met, we can still proceed with the problem but keep in mind that the results will not be as accurate until more data is collected or more information is given in the problem.

Solve for t*:

We need the tail area first.

$\displaystyle \frac{1-.9}{2}= .05$

Next we need the degree of freedom.

The degree of freedom can be found by subtracting the degree of freedom for A and B.

The general formula is df = n - 1.

df for A: 13 - 1 = 12
df for B: 18 - 1 = 17
df for A - B: |12 - 17| = 5

Use a calculator or a t-table to find the corresponding t-score for df = 5 and tail area = .05.

t* = -2.015

Now we can use the formula at the very top to construct a confidence interval for two sample means.

$\overline {x}_A=39$
$s_A=8$
$n_A=13$
$\overline {x}_B = 55$
$s_B=2$
$n_B=18$
$t^{*}=-2.015$

Substitute the variables into the formula: $\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$ .