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olga nikolaevna [1]
3 years ago
7

The distance between the Sun and the Earth is 1 .496 x 1O^8km and distance between the Earth and lhe Moon is 3.84 x 10^8m. Durin

g solar eclipse. The Moon comes in between the Earth and the Sun. What is the distance between the Moon and the Sun at that particular time?
Mathematics
1 answer:
Dennis_Churaev [7]3 years ago
3 0

Answer:

1.49216*10^8 km

Step-by-step explanation:

The first thing is to pass all the distances to the same unit system, since the distance from the earth to the sun is in kilometers and from the earth to the moon is in meters, therefore we will pass the distance of the moon in kilometers:

3.84 x 10 ^ 8m * 1 km / 1000 m = 3.84 x 10 ^ 5 km

To calculate the distance between the moon and the sun, it would be the difference between the distance they give us in the statement, because they have a point in common which is the earth, it would be:

1.496 x 10 ^ 8 km - 3.84 x 10 ^ 5 km = 149216000 = 1.49216 * 10 ^ 8 km

Therefore the distance between the moon and the sun is 1,49216 * 10 ^ 8 km

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In tests of a computer component, it is found that the mean time between failures is 937 hours. A modification is made which is
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Answer:

Null hypothesis is \mathbf {H_o: \mu > 937}

Alternative hypothesis is \mathbf {H_a: \mu < 937}

Test Statistics z = 2.65

CONCLUSION:

Since test statistics is greater than  critical value; we reject the null hypothesis. Thus, there is sufficient evidence to support the claim that the modified components have a longer mean time between failures.

P- value = 0.004025

Step-by-step explanation:

Given that:

Mean \overline x = 960 hours

Sample size n = 36

Mean population \mu = 937

Standard deviation \sigma = 52

Given that the mean  time between failures is 937 hours. The objective is to determine if the mean time between failures is greater than 937 hours

Null hypothesis is \mathbf {H_o: \mu > 937}

Alternative hypothesis is \mathbf {H_a: \mu < 937}

Degree of freedom = n-1

Degree of freedom = 36-1

Degree of freedom = 35

The level of significance ∝ = 0.01

SInce the degree of freedom is 35 and the level of significance ∝ = 0.01;

from t-table t(0.99,35), the critical value = 2.438

The test statistics is :

Z = \dfrac{\overline x - \mu }{\dfrac{\sigma}{\sqrt{n}}}

Z = \dfrac{960-937 }{\dfrac{52}{\sqrt{36}}}

Z = \dfrac{23}{8.66}

Z = 2.65

The decision rule is to reject null hypothesis   if  test statistics is greater than  critical value.

CONCLUSION:

Since test statistics is greater than  critical value; we reject the null hypothesis. Thus, there is sufficient evidence to support the claim that the modified components have a longer mean time between failures.

The P-value can be calculated as follows:

find P(z < - 2.65) from normal distribution tables

= 1 - P (z ≤ 2.65)

= 1 - 0.995975     (using the Excel Function: =NORMDIST(z))

= 0.004025

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Step-by-step explanation:

8(1/2x + 5) = 200

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x = 40

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