F(x)=x^2 what is g(x)

Korolek [52]

The answer to this question is true. If i'am right this from k12 and edg-enuity. have a good day bro. ;)

4 0

2 years ago

Can someone help me solve for X :( I’m stressing out please HELP !

Leni [432]

Answer:

7

Step-by-step explanation:

28 ÷ 16 = 1.75

1.75 × 4 = 7

The way I did it is by finding out how much bigger 28 is than 16 by dividing and then I multiplied the answer (1.75) by 4 to get the bigger answer. The problem is showing you that the triangles are the same just different sizes.

6 0

2 years ago

1) Lithium isotope rations are important to medicine, the 6Li/7Li ratio in a standard reference material was measured several ti

uysha [10]

Answer:

1) $0.0826052-2.776\frac{0.000013424}{\sqrt{5}}=0.082588$

$0.0826052+2.776\frac{0.000013424}{\sqrt{5}}=0.0826219$

b) $ME= 2.776\frac{0.000013424}{\sqrt{5}}=0.0000166653$

And we want 2/3 of the margin of error so then would be: $2/3 ME = 0.00001111$

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

And on this case we have that ME =0.00001111016 and we are interested in order to find the value of n, if we solve n from equation (1) we got:

$n=(\frac{z_{\alpha/2} s}{ME})^2$ (2)

Replacing we got:

$n=(\frac{2.776(0.000013424)}{0.00001111})^2 =11.25 \approx 12$

So the answer for this case would be n=12 rounded up to the nearest integer

Step-by-step explanation:

Information given

0.082601, 0.082621, 0.082589, 0.082617, 0.082598

We can calculate the sample mean and deviation with the following formulas:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

$s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X=0.0826052$ represent the sample mean

$\mu$ population mean

s=0.000013424 represent the sample standard deviation

n=5 represent the sample size

Part 1

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

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

The degrees of freedom, given by:

$df=n-1=5-1=4$

The Confidence level is 0.95 or 95%, and the significance would be $\alpha=0.05$ and $\alpha/2 =0.025$ , the critical value would be using the t distribution with 4 degrees of freedom: $t_{\alpha/2}=2.776$

Now we have everything in order to replace into formula (1):

$0.0826052-2.776\frac{0.000013424}{\sqrt{5}}=0.082588$

$0.0826052+2.776\frac{0.000013424}{\sqrt{5}}=0.0826219$

Part 2

The original margin of error is given by:

$ME= 2.776\frac{0.000013424}{\sqrt{5}}=0.0000166653$

And we want 2/3 of the margin of error so then would be: $2/3 ME = 0.00001111$

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

And on this case we have that ME =0.00001111016 and we are interested in order to find the value of n, if we solve n from equation (1) we got:

$n=(\frac{z_{\alpha/2} s}{ME})^2$ (2)

Replacing we got:

$n=(\frac{2.776(0.000013424)}{0.00001111})^2 =11.25 \approx 12$

So the answer for this case would be n=12 rounded up to the nearest integer

3 0

3 years ago

How do you find the holes, vertical, and horizontal asymptote of an equation?

AfilCa [17]

This is a great question, but it's also a very broad one. Please find and post one or two actual rational expressions, so we can get started on specifics of how to find vertical and horiz. asymptotes.

In the case of vert. asy.: Set the denom. = to 0 and solve for x. Any real x values that result indicate the location(s) of vertical asymptotes.

4 0

2 years ago

What does JK stand for when doing angles??? Offering 29 pts and brainliest answer

MaRussiya [10]

Hello.

JK (or any other combination of letters) doesn't stand for anything in geometry. The purpose of these letters is to reference certain angles and sides of a figure without having to describe the angles/sides in relation to other angles/sides to know exactly which angle or side is being referenced.

Hope this helps!

3 0

3 years ago

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