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2 years ago

Light travels at 3.0 x 108 km/s. How far does light travel in 5 minutes?

Mathematics

1 answer:

Sergio039 [100]2 years ago

4 0

It’s distance times speed over time. So 108km/s x 3.0 = 324 divided by 5 which equals 64.8 km/s

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Which expression is equivalent to −1/ 12x−1 /3?

alexandr402 [8]

Answer:

$\frac{1}{12} (-x-4)$

Step-by-step explanation:

the expression we have is:

$-\frac{1}{12} x-\frac{1}{3}$

To find an equivalent expression, it would be helpful to express 1/3 so that it has a 12 in the denominator:

$\frac{1}{3}=\frac{4}{12}$

so we substitute this:

$-\frac{1}{12} x-\frac{4}{12}$

and now we have 1/12 in both terms :

$-\frac{1}{12}x-\frac{1*4}{12}$

so we can factor it (we take it out of each term: in the first we are left with -x and in the second with -4):

$\frac{1}{12} (-x-4)$

the equivalent expression is: $\frac{1}{12} (-x-4)$

5 0

3 years ago

During the basketball season,Josh pitched to 270 batters. he struck out 70% how many batters did he strike out

never [62]

Josh struck out 189 batters.

3 0

3 years ago

Which classification describes the system of linear equations?

katen-ka-za [31]

Both equations are the same
y=−4x+4 ----> y+4x=4,
so consistent dependent

5 0

3 years ago

Determine the amount of money you would have if you invested $15,000 dollars for 11 years at 3% annual interest compounded conti

tamaranim1 [39]

Answer:

The amount after 3 years of investment is $20763 .

Step-by-step explanation:

Given as :

The principal invested = p =$15,000

The rate of interest = r = 3% compounded annually

The time period = t = 11 years

Let The Amount after 3 years = $ A

From Compounded method

Amount = Principal × $(1+\dfrac{\textrm rate}{100})^{\textrm time}$

Or, A = p × $(1+\dfrac{\textrm r}{100})^{\textrm t}$

Or, A = $15,000 × $(1+\dfrac{\textrm 3}{100})^{\textrm 11}$

Or, A = $15,000 × $(1.03)^{11}$

Or, A = $15,000 × 1.3842

Or, A = $20763

So, Amount = A = $20763

Hence The amount after 3 years of investment is $20763 . Answer

8 0

2 years ago

A local car dealer claims that 25% of all cars in San Francisco are blue. You take a random sample of 600 cars in San Francisco

sammy [17]

Answer:

No, we can't reject the dealer's claim with a significance level of 0.05.

Step-by-step explanation:

We are given that a local car dealer claims that 25% of all cars in San Francisco are blue.

You take a random sample of 600 cars in San Francisco and find that 141 are blue.

Let p = proportion of all cars in San Francisco who are blue

SO, Null Hypothesis, $H_0$ : p = 25% {means that 25% of all cars in San Francisco are blue}

Alternate Hypothesis, $H_A$ : p $\neq$ 25% {means that % of all cars in San Francisco who are blue is different from 25%}

The test statistics that will be used here is One-sample z proportion statistics;

T.S. = $\frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of 600 cars in San Francisco who are blue = $\frac{141}{600}$ = 0.235

n = sample of cars = 600

So, test statistics = $\frac{0.235-0.25}{{\sqrt{\frac{0.235(1-0.235)}{600} } } } }$

= -0.866

Now at 0.05 significance level, the z table gives critical values of -1.96 and 1.96 for two-tailed test. Since our test statistics lies within the range of critical values of z so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that 25% of all cars in San Francisco are blue which means the dealer's claim was correct.

4 0

3 years ago