All categories

3 years ago

The nearest Black hole is 7800 light years from Earth. Light travels at 186,000 per sec how many miles away is it?

1 answer:

5 0

Distance=speed*time
speed of light in miles=3.6 km/hr
7800 light years=7800(9.4607×10^12) km
thus the distance in miles will be as follows:
1 mile=1.60934
thus the distance from each to black hole is:
[7800(9.4607×10^12)]/1.60934
=4.5853244×10^16 miles

You might be interested in

A triangle with two vertices located at (5, −8) and (5, 4) has an area of 48 square units. Determine one possible location of th

RoseWind [281]

Answer:x=1.5

Step-by-step explanation:

Given

Two points co-ordinates are given

A=(5,-8)

B=(5,4)

let other point be C =(x,y)

Area of triangle is given by

$A=\begin{vmatrix}5& -8&1\\5& 4&1\\x& y&1\end{vmatrix}$

$A=5\left ( 4-y\right )+8\left ( 5-x\right )+1\left ( 5y-4x\right )$

A=20-5y+40-8x+5y-4x

A=60-8x

and A=48 square unit

48=60-8x

x=1.5

As area is independent of y therefore at x=1.5 any value of y will give area of 48 square unit

7 0

3 years ago

The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and th

const2013 [10]

Step by step:

Perimeter = Summation of all sides = AB+BC+CD+DA = 2x+2+x+2x+2+x=(6x+4)

⇒6x+4=154

⇒6x=154−4

⇒ $\frac{150}{6x}$

⇒25=x

The breadth is x=25

The length of the rectangular pool = (2×25+2)=50+2=52m

<em>breadth </em>is 25 and the<em> length </em>is 52

Hope it helps!

4 0

3 years ago

Read 2 more answers

Suppose that the population of the scores of all high school seniors who took the SAT Math (SAT-M) test this year follows a Norm

Vlad1618 [11]

Answer:

Confidence = $1-\alpha=1-0.01=0.99$

And then the confidence level would be given by 99%

Step-by-step explanation:

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".

Assuming the X follows a normal distribution

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

The distribution for the sample mean is given by:

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

$\bar x =512$ represent the sample mean

$\sigma=100$ represent the population standard deviation

n= 100 sample size selected.

The confidence interval is given by this formula:

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

The marginof error for this case is given by Me=25.76. And we know that the formula for the margin of error is given by:

$Me=z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$

$25.76=z_{\alpha/2} \frac{100}{\sqrt{100}}$

And we can find the critical value $z_{\alpha/2}$ like this:

$z_{\alpha/2}=\frac{25.76(\sqrt{100})}{100}=2.576$

And we know that on the right tail of the z score =2.576 we have $\alpha/2$ of the total area. We can find the area on the right of the z score using this excel code:

"=1-NORM.DIST(2.576,0,1,TRUE)" or using a table of the normal standard distribution, and we got 0.004998= $\alpha/2$ , so then $\alpha=0.00498*2=0.009995 \approx 0.01$ , and then we can find the confidence like this: