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3 years ago

How do you know x when f(x)=2

Mathematics

2 answers:

AlexFokin [52]3 years ago

6 0

Its always 2 does not matter what is defined to x

frosja888 [35]3 years ago

3 0

Say your function is
f(x) = 2x + 3
If f(x) is 2 & you want to find x, replace f(x) with 2:
2=2x+3
And solve for x:
-1=2x
x=-1/2
Hope this helps!

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Help me plz this is my last question of the day but plz help me

gizmo_the_mogwai [7]

Answer:

Additive Inverse Property.

Step-by-step explanation:

the value that brings you back to the identity element under addition

So a+-a=0

Hope this helps!!!

~BBGLUVER

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3 years ago

Which is the equation of the given line in slope-intercept form

Tanya [424]

Answer: The Slope-Intercept Form equation is y=mx+b

i think the answer is y=3x+2 but i hope this helps

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3 years ago

If a cookie recipe that makes 2 dozen cookies calls for 2 1/4 cups of flour, how cups of flour do you need to make 5 dozen cooki

Schach [20]

Answer:

5. 5/8 flour needed to make 5 dozen cookies

Step-by-step explanation:

2 dozen= 24

5 dozen=60

It takes 1 1/8 flour for 1 dozen cookies so 1 1/8 x 5= 5.625 in fraction form that= 5 5/8

5 0

3 years ago

I need help with that

Digiron [165]

9514 1404 393

Answer:

AC ≈ 7.3

Step-by-step explanation:

Corresponding sides of similar triangles are proportional.

AC/BC = DF/EF

AC/18 = 15/37 . . . . fill in the given values

AC = 18(15/37) . . . multiply by 18

AC ≈ 7.3

4 0

3 years ago

"A cable TV company wants to estimate the percentage of cable boxes in use during an evening hour. An approximation is 20 percen

Marizza181 [45]

Answer:

The company should take a sample of 148 boxes.

Step-by-step explanation:

Hello!

The cable TV company whats to know what sample size to take to estimate the proportion/percentage of cable boxes in use during an evening hour.

They estimated a "pilot" proportion of p'=0.20

And using a 90% confidence level the CI should have a margin of error of 2% (0.02).

The CI for the population proportion is made using an approximation of the standard normal distribution, and its structure is "point estimation" ± "margin of error"

[p' ± $Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }$ ]

Where

p' is the sample proportion/point estimator of the population proportion

$Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }$ is the margin of error (d) of the confidence interval.

$Z_{1-\alpha /2} = Z_{1-0.05} = Z_{0.95}= 1.648$

$d= Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }$

$d *Z_{1-\alpha /2}= \sqrt{\frac{p'(1-p')}{n} }$

$(d*Z_{1-\alpha /2})^2= \frac{p'(1-p')}{n}$

$n*(d*Z_{1-\alpha /2})^2= p'(1-p')$

$n= \frac{p'(1-p')}{(d*Z_{1-\alpha /2})^2}$

$n= \frac{0.2(1-0.2)}{(0.02*1.648)^2}$

n= 147.28 ≅ 148 boxes.

I hope it helps!

3 0

4 years ago