Convert <img src="https://tex.z-dn.net/?f=17%5E%5Cfrac%7B3%7D%7Bx%7D" id="TexFormula1" title="17^\frac{3}{x}" alt="17^\frac{3}{x

The equations of the line is y=17.63x+27.52. What is the best interpretation of the slope of the line?

nevsk [136]

Answer:

slope=17.63

Step-by-step explanation:

the eq. of line in slope intercept form is y=mx+c

where m is the slope and c is y intercept.

comparing

m=17.63

8 0

4 years ago

Read 2 more answers

Select the correct answer. Given: `Delta`ABC, where AB = BC Prove: `"m"/_BAC = "m"/_BCA` Statement Reason 1. Let `Delta`ABC be a

Leto [7]

Since BD is the angle bisector of ∠ ABC,

∠ ABD = ∠ CBD --- (1)

Now, in triangles Δ ABD and Δ CBD,

BD = BD (common side)

∠ ABD = ∠ CBD from (1)

AB = BC (Given)

Hence, Δ ABD ≅ Δ CBD (SAS Rule).

5 0

3 years ago

Mooooooooo 4/9-4/10 =

Anon25 [30]

Answer:

0.04444444444

Step-by-step explanation:

7 0

3 years ago

What is the equation of the line showed in this graph?

Irina-Kira [14]

Answer:

y=1

Step-by-step explanation

3 0

3 years ago

A major credit card company has determined that customers charge between $100 and $1100 per month. If the monthly amount charged

Alexxx [7]

Answer:

a) $E(X) = \frac{a+b}{2} = \frac{100+1100}{2}=600$

b) First we need to calculate the variance given by this formula:

$Var(X) = \frac{(b-a)^2}{12}= \frac{(1100-100)^2}{12} = 83333.33$

And the deviation would be:

$Sd(X) = \sqrt{83333.33}= 288.675$

c) $P(600 < X< 889) = P(X$

And using the cdf we got:

$P(600 < X< 889)= F(889) -F(600) = \frac{889-100}{1000} -\frac{600-100}{1000}= 0.789- 0.5= 0.289$

d) $P(311 < X< 889)= F(889) -F(311) = \frac{889-100}{1000} -\frac{311-100}{1000}= 0.789- 0.211= 0.578$

Step-by-step explanation:

For this case we define the random variable X who represent the customers charge, and the distribution for X on this case is:

$X \sim Unif (a= 100, b=1100)$

Part a

For this case the average is given by the expected value and we can use the following formula:

$E(X) = \frac{a+b}{2} = \frac{100+1100}{2}=600$

Part b

First we need to calculate the variance given by this formula:

$Var(X) = \frac{(b-a)^2}{12}= \frac{(1100-100)^2}{12} = 83333.33$

And the deviation would be:

$Sd(X) = \sqrt{83333.33}= 288.675$

Part c

For this case we want to find the percent between 600 and 889, so we can use the cumulative distribution function given by:

$F(x) = \frac{X -100}{1100-100}= \frac{x-100}{1000}, 100 \leq X \leq 1100$

And we can find this probability:

$P(600 < X< 889) = P(X$

And using the cdf we got:

$P(600 < X< 889)= F(889) -F(600) = \frac{889-100}{1000} -\frac{600-100}{1000}= 0.789- 0.5= 0.289$

Part d

$P(311 < X< 889) = P(X$

And using the cdf we got:

$P(311 < X< 889)= F(889) -F(311) = \frac{889-100}{1000} -\frac{311-100}{1000}= 0.789- 0.211= 0.578$

6 0

3 years ago