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3 years ago

What are the solution(s) of X2-4=0?

Mathematics

1 answer:

irinina [24]3 years ago

6 0

Answer:

-2 or 2

Step-by-step explanation:

So we have the equation:

$x^2-4=0$

Add 4 to both sides:

$x^2=4$

Take the square root of both sides:

$x=\pm 2$

Therefore, our answers are -2 or 2.

And we're done :)

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Plsss help it due today I don’t understand how to do this plssss don’t just take the points

tensa zangetsu [6.8K]

Answer:

d a m i dont even know this

Step-by-step explanation:

4 0

3 years ago

PLEASE HELP IT'S DUE IN LIKE 20 MINUTES

Alexeev081 [22]

9514 1404 393

Answer:

x +3x = 180
x = 45 . . . . . . . choice C
3x = 135 . . . . . choice F

Step-by-step explanation:

Let x represent the measure of the original angle in degrees. Three times its measure is 3x. That will be the supplement if the sum of the two angles is 180°.

x + 3x = 180 . . . . . . . . . equation for finding the angle

x = 180/4 = 45 . . . . . . divide both sides by 4

3x = 3(45) = 135

The first angle is 45°. The second angle is 135°.

4 0

3 years ago

Susie ordered her sister a get well card from a company that randomly selects a card from their inventory. The company has 50 to

Hunter-Best [27]

If you want it in fraction form it would be 4/5 but if you wanted it in the percentage it would be there is a 80% chance it will not be a get well card. Work: 4/5 x 2= 8/10 which converted into a percentage would be 80/100, or 80%. Hope this helped!

3 0

3 years ago

Read 2 more answers

You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confide

salantis [7]

Answer:

With 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is wider than the 90%.

Step-by-step explanation:

We are given that a random sample of 60 home theater systems has a mean price of$131.00. Assume the population standard deviation is$18.80.

Firstly, the pivotal quantity for 90% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean price = $131

$\sigma$ = population standard deviation = $18.80

n = sample of home theater = 60

$\mu$ = population mean

Here for constructing 90% confidence interval we have used One-sample z test statistics as we know about the population standard deviation.

So, 90% confidence interval for the population mean, $\mu$ is ;

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $131-1.645 \times {\frac{18.8}{\sqrt{60} } }$ , $131+1.645 \times {\frac{18.8}{\sqrt{60} } }$ ]

= [127.01 , 134.99]

Therefore, 90% confidence interval for the population mean is [127.01 , 134.99].

Now, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean price = $131

$\sigma$ = population standard deviation = $18.80

n = sample of home theater = 60

$\mu$ = population mean

Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about the population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $131-1.96 \times {\frac{18.8}{\sqrt{60} } }$ , $131+1.96 \times {\frac{18.8}{\sqrt{60} } }$ ]

= [126.24 , 135.76]

Therefore, 95% confidence interval for the population mean is [126.24 , 135.76].

Now, with 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is wider than the 90%.

7 0

3 years ago

Twelve pounds of beans are distributed equally into 8 bags to give out at the food bank.

siniylev [52]

Answer:

1 1/2 or 1.5

Step-by-step explanation:

12 divided by 8

brainliest?

4 0

3 years ago