All categories

2 years ago

The speed of light is 3 x 105 kilometers per second. What is that in standard form?

Mathematics

2 answers:

monitta2 years ago

8 0

$v=3*10^{5} \frac{km}{s} =300 \ 000 \frac{km}{s}$

spayn [35]2 years ago

4 0

Answer:

speed of light is 3 x $10^{8}$ meters per second.

Step-by-step explanation:

Given : The speed of light is 3 x $10^{5}$ kilometers per second.

To find : What is that in standard form.

Solution : We have given 3 x $10^{5}$ kilometers per second.

First we convert it in to meter per sec.

We can write

3 x $10^{5}$ = 300000 kilometer per second .

We know that 1 km = 1000 meter.

Then 300000000 meter per second

Now, 3 x $10^{8}$ meter per second.

Therefore, speed of light is 3 x $10^{8}$ meters per second.

You might be interested in

Which value of x makes this equation true? –5(– 20 ) = 35 A. -11 B. -3 C. 13 D. 27

kirza4 [7]

Answer:

Step-by-step explanation:

its right

3 0

2 years ago

Read 2 more answers

Two ways technology can affect the environment

Lelu [443]

Two ways technology can affect the environment is by air pollution and toixc water

6 0

3 years ago

What is that value of (4-2)3-3x4

Over [174]

Answer:

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

One month, Jon worked 3 hours less than Chaya, and Angelica worked 4 hours more than Chaya. Together they worked 196 hours. Find

OleMash [197]

джон работал на 1 час меньше

3 0

3 years ago

Read 2 more answers

The daily revenues of a cafe near the university are approximately normally distributed. The owner recently collected a random s

lbvjy [14]

Answer:

The sample size to obtain the desired margin of error is 160.

Step-by-step explanation:

The Margin of Error is given as

$MOE=z_{crit}\times\dfrac{\sigma}{\sqrt{n}}$

Rearranging this equation in terms of n gives

$n=\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2$

Now the Margin of Error is reduced by 2 so the new M_2 is given as M/2 so the value of n_2 is calculated as

$n_2=\left[z_{crit}\times \dfrac{\sigma}{M_2}\right]^2\\n_2=\left[z_{crit}\times \dfrac{\sigma}{M/2}\right]^2\\n_2=\left[z_{crit}\times \dfrac{2\sigma}{M}\right]^2\\n_2=2^2\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2\\n_2=4\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2\\n_2=4n$

As n is given as 40 so the new sample size is given as

$n_2=4n\\n_2=4*40\\n_2=160$

So the sample size to obtain the desired margin of error is 160.

4 0

3 years ago