Use ten to subtract

In 565656 years, daniel will be 555 times as old as he is right now. how old is he right now?

Taya2010 [7]

In 565656 years Daniel will be 555 times as old as he's right now how old is he right now

Let x be his current age.
"In 565656 years" can be represented by the expression x+565656.
"Will be 555 times as old as he's right now" can be represented bas the expression 555x.
The whole question means that the age of Daniel in 565656 years will be 555 times his age now so x+565656 = 555x.
Solving this equation will find x, so his current age.
x+565656 = 555x
You subtract an x of each side, to have variables on a side, and numbers on an other:
565656 = 555x - x
565656 = 554x
You divide each side by 554
x = 1021.

So Daniel is currently 1021 years old.

3 0

3 years ago

Preform the indicated operation. √75 + √12

Nezavi [6.7K]

Answer:

the exact form is $7 \sqrt{3}\\$ but the decimal form is 12.124 :)))

Step-by-step explanation:

6 0

3 years ago

A car travels 182 kilometers in 4 hours with the constant speed. how much time will it take traveling 42 kilometers

Schach [20]

Answer:

55.38 min

Step-by-step explanation:

A car travels,

182 km ⇒ 4 hours

And we have to find the time that it takes to travel 42 km .

Lets make a equation.

Let the unknown time be x

182 km ⇒ 4 hours

42 km ⇒ x

To make easy let's km in to meters and hours in to minutes

182000 m ⇒ 240 min

42000m ⇒ x

Now use cross multiplication.

$182000x=240*42000\\$

$182000x=10080000$

$\frac{182000x}{182000} =\frac{1008000}{182000}$

$x =55.38min$

Hope this helps you.

Let me know if you have any other questions :-)

6 0

3 years ago

I WILL GIVE BRAINLIEST TO CORRECT ANSWER<br> -x<-x+7(x-2)

BlackZzzverrR [31]

Answer:

x>2

Step-by-step explanation:

0<7x-14

-7x<-14

x>2

hope that's right

8 0

3 years ago

Read 2 more answers

A Common Measure of Student Achievement, The National Assessment of Educational Progress (NAEP) is the only assessment that meas

vlada-n [284]

Answer:

a) By the central limit theorem we know that if the random variable X="quantitative scores for young women", and we know that this distribution is normal.

$X \sim N(\mu, \sigma=60)$

And we are interested on the distribution for the sample mean $\bar X$ , we know that distribution for the sample mean is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And with that we have the assumptions required to apply the normal distribution to create the confidence interval since the distribution for $\bar X$ is normal.

b) $270.283\leq \mu \leq 279.717$

c) We are confident (99%) that the true mean for the quantitative scores for young women is between (270.283;3279.717)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=275$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma=60$ represent the population standard deviation

n=1077 represent the sample size

Part a

By the central limit theorem we know that if the random variable X="quantitative scores for young women", and we know that this distribution is normal.

$X \sim N(\mu, \sigma=60)$

And we are interested on the distribution for the sample mean $\bar X$ , we know that distribution for the sample mean is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And with that we have the assumptions required to apply the normal distribution to create the confidence interval since the distribution for $\bar X$ is normal.

Part b

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

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

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that $z_{\alpha/2}=2.58$

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

$275-2.58\frac{60}{\sqrt{1077}}=270.283$

$275+2.58\frac{60}{\sqrt{1077}}=279.717$

So on this case the 99% confidence interval would be given by (270.283;3279.717)

$270.283\leq \mu \leq 279.717$

Part c

We are confident (99%) that the true mean for the quantitative scores for young women is between (270.283;3279.717)

8 0

3 years ago