Why is the answer C? Plz explain

Mathematics

1 answer:

Dafna11 [192]3 years ago

5 0

Set the equations each to each other and solve for x
f(x) = g(x)
x^2 - 3x = 2x+14
x^2-3x-2x-14 = 0
x^2-5x-14 = 0 .... see note below
(x-7)(x+2) = 0
x-7 = 0 or x+2 = 0
x = 7 or x = -2
The two solutions are x = 7 or x = -2

note: we technically could stop at this line and divert to another path. Instead of finding the actual solutions, we can use the discriminant formula
d = b^2 - 4ac
to find the value of d. It turns out that
d = (-5)^2 - 4(1)(-14)
d = 81
which is a positive number. So that shows we have two solutions.

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Why will the value of y for the function y equals 5x + 1 always be greater than that of y equals 4x + 1 with X greater than 1?

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