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2 years ago

13

Enter the values for the highlighted variables that show how to subtract the rational expressions correctly

2 answers:

I am Lyosha [343]2 years ago

8 0

Answer:

a = 6

b = 2

c = 6

d = 2

e = 6

f = 6

g=1

Step-by-step explanation:

a = 6

x^2 + 6x is equal to x(x+6)

b=2

Denominator and numerator of the first term are multiplied by x.

c=6

Second term is multiplied by (x-6)/(x-6)

d=2

Now that they have the same denominator, the two terms are combined. 2 is the coefficient of the first term

e=6

In the same way as d is carried over from b, e is carried over from c.

f = 6

2x - x + 6 = x + 6

g = 1

We factor out the (x+6) from the numerator and denominator

Lina20 [59]2 years ago

4 0

A = 6
x^2 + 6x is equal to x(x+6)
b=2
Denominator and numerator of the first term are multiplied by x.
c=6
Second term is multiplied by (x-6)/(x-6)
d=2
Now that they have the same denominator, the two terms are combined. 2 is the coefficient of the first term
e=6
In the same way as d is carried over from b, e is carried over from c.
f = 6
2x - x + 6 = x + 6
g = 1
We factor out the (x+6) from the numerator and denominator

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then,

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Step-by-step explanation:

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AleksandrR [38]

Answer:

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$(240-210) +1.984 \sqrt{\frac{8.6^2}{100}+\frac{5.7^2}{100}}=32.047$

And the confidence interval for the difference is between:

$27.953 \leq \mu_1 -\mu_2 \leq 32.047$

Step-by-step explanation:

We have the following info given:

$\bar X_1 = 240$ sample mean for medium Pizzas from Prim's

$\bar X_2 = 210$ sample mean for medium Pizzas from Pizza Place

$s_1 =8.6$ sample deviation for Prim's

$s_2 =5.7$ sample deviation for Pizza Palca

$n_1 =n_2 = 100$ sample size selected for each case

The confidence interval for the difference of means is given by:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}$

And for the 95% confidence we need a significance level of $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ , the degrees of freedom are given by:

$df= n_1 +n_2 -2= 100+100-2 =98$

And the critical value would be

$t_{\alpha/2}= 1.984$

And replacing we got:

$(240-210) -1.984 \sqrt{\frac{8.6^2}{100}+\frac{5.7^2}{100}}=27.953$

$(240-210) +1.984 \sqrt{\frac{8.6^2}{100}+\frac{5.7^2}{100}}=32.047$

And the confidence interval for the difference is between:

$27.953 \leq \mu_1 -\mu_2 \leq 32.047$

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alexgriva [62]

Answer:

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