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3 years ago

Solve for x: |3x|=12

2 answers:

6 0

Clear the absolute-value bars by splitting the equation into its two cases, one for the Positive case and the other for the Negative case.

The Absolute Value term is |x-3|

For the Negative case we'll use -(x-3)

For the Positive case we'll use (x-3)

Step 3 :

Solve the Negative Case

-(x-3) = 12

Multiply

-x+3 = 12

Rearrange and Add up

-x = 9

Multiply both sides by (-1)

x = -9

Which is the solution for the Negative Case

Step 4 :

Solve the Positive Case

(x-3) = 12

Rearrange and Add up

x = 15

Which is the solution for the Positive Case

Step 5 :

Wrap up the solution

x=-9

x=15

Dennis_Churaev [7]3 years ago

3 0

Answer:

x = 4; x = -4

Step-by-step explanation:

|3x| = 12

3x = 12

3x/3 = 12/3

x = 4

3x = -12

3x/3 = -12/3

x = -4

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The radii of two circles are in the ratio of 3 to 1. Find the area of the smaller circle if the area of the larger circle is 27

Naddik [55]

ANSWER
$3\pi sq.\: in.$

EXPLANATION

Let R be the radius of the bigger circle and r, be the radius of the smaller circle.

Then their ratio is given as,

$R:r=3:1$

We can rewrite it as fractions to get,

$\frac{R}{r} = \frac{3}{1}$

We make R the subject to get,

$R = 3r$

The area of the bigger circle can be found using the formula,

$Area=\pi {r}^{2}$

This implies that,

$Area=\pi ({3r})^{2}$

$Area=9\pi {r}^{2}$

But it was given in the question that, the area of the bigger circle is 27π.

$27\pi=9\pi {r}^{2}$

We divide through by 9π to get,

$3 = {r}^{2}$

This means that,
$r = \sqrt{3}$

The area of the smaller circle is therefore

$= \pi {( \sqrt{3}) }^{2}$

$= 3\pi$

8 0

4 years ago

Read 2 more answers

In a certain country, the number of charter schools is 4 less than twice the number of alternative schools. We know that there a

aleksandr82 [10.1K]

C = 48
c = 2a - 4

48 = 2a - 4
48 + 4 = 2a
52 = 2a
52/2 = a
26 = a <==

4 0

3 years ago

Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its average number of unoc

Bingel [31]

Answer:

We need a sample size of at least 719

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

How large a sample size is required to vary population mean within 0.30 seat of the sample mean with 95% confidence interval?

This is at least n, in which n is found when $M = 0.3, \sigma = 4.103$ . So

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.3 = 1.96*\frac{4.103}{\sqrt{n}}$

$0.3\sqrt{n} = 1.96*4.103$

$\sqrt{n} = \frac{1.96*4.103}{0.3}$

$(\sqrt{n})^{2} = (\frac{1.96*4.103}{0.3})^{2}$

$n = 718.57$

Rouding up

We need a sample size of at least 719

6 0

3 years ago

Simplify <img src="https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B405%7D" id="TexFormula1" title="\sqrt[3]{405}" alt="\sqrt[3]{405}" a

Natali5045456 [20]

3 outside 3/15 inside

5 0

3 years ago

10.4.1 .WP The manager of a fleet of automobiles is testing two brands of radial tires and assigns one tire of each brand at ran

Sunny_sXe [5.5K]

Answer:

[-0.65;2.54]km

Step-by-step explanation:

Hello!

So, you need to find a 99%CI for the difference in mean life of two brands of radial tires. Since he assigned one tire of each brand at random to the two rear wheels of each car, in other words, every car tested had one rear tire of each brand at the same time, this test can be considered to be of paired samples.

Brand 1 Brand 2 X₁-X₂

car 1: 36,925 ; 34,318 ; 2.607

car 2: 45,300 ; 42,280 ; 3.020

car 3: 36,240 ; 35,500 ; 0.740

car 4: 32,100 ; 31,950 ; 0.150

car 5: 37,210 ; 38,015 ; -.0805

car 6: 48,360 ; 47,800 ; 1.160

car 7: 38,200 ; 37,810 ; 0.390

car 8: 33,500 ; 33,215 ; 0.285

n= 8

With this in mind, we define the study variable as Xd= X₁-X₂

Where X₁ corresponds to the lifespan, in km, of a tire from Brand 1

and X₂ corresponds to the lifespan, in km, of a tire from Brand 2

so Xd will be the difference between the lifespan of the tires from Brand 1 and Brand 2.

This variable Xd~N(μd;δd²) (p-value for normality test is 0.4640)

To calculate the CI the best statistic is the Student's t with the following formula:

t= (xd[bar] - μd)/(Sd/√n) ~t₍ₙ₋₁₎

sample mean: xd[bar]= 0.94

standard deviation: Sd= 1.29

$t_{8;0.995}$ = 3.355

xd[bar] ± $t_{8;0.995}$ *(Sd/√n)

⇒ 0.94 ± 3.355*(1.29/√8)

[-0.65;2.54]km

The Confidence Interval can be compared to a pair of bilateral hypothesis. If we were to determine the following hypothesis

H₀:μd=0

H₁:μd≠0

Using the level of significance of 0.01 (complementary to the confidence level)

As the calculated confidence interval contains zero, we do not reject the null hypothesis, that is, there is no significant difference between the two tire brands.

I hope you have a SUPER day!

5 0

3 years ago