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ziro4ka [17]
3 years ago
15

Can someone help me 100 points have to turn them all into scientific notation

Mathematics
2 answers:
creativ13 [48]3 years ago
7 0

Answer:

5.7909000*10^7=Mercury

1.08208930*10^8=Venus

1.49600000*10^8=Earth

7.785472000*10^9=Jupiter

2.871000000*10^9=Uranus

4.495000000*10^9=Neptune

Step-by-step explanation:

All I was taught in math class was to count the places until it is between one and ten and how many places is your exponent after 10.

Natali5045456 [20]3 years ago
3 0

Answer:

See below in bold.

Step-by-step explanation:

Venus = 1.0820893 * 10^8 km.   The exponent 8 comes from counting the number of digits after the beginning 1 value.

Earth = 1. 496 * 10^8 km.

Mars = 2.2794 * 10^8 km.

Jupiter = 7.785472 * 10^8 km.

Saturn = 1.43344937 * 10^9 km.

Uranus = 2.871 * 10^9 km.

Neptune = 4.495 * 10^9 km.

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leaks can write a 500-work essay in a hour. if she writes a essay in 10 minutes, how many words will it contain?
gavmur [86]

Answer:

about 83 words

Step-by-step explanation:

60/6=10

500/6 about 83

7 0
3 years ago
A quality control expert at Glotech computers wants to test their new monitors. The production manager claims they have a mean l
tankabanditka [31]

Answer:

The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors

P(X⁻ ≥ 96.3) = 0.0087

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given that the mean of the Population = 95

Given that the standard deviation of the Population = 5

Let 'X' be the random variable in a normal distribution

Let X⁻ = 96.3

Given that the size 'n' = 84 monitors

<u><em>Step(ii):-</em></u>

<u><em>The Empirical rule</em></u>

         Z = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }

        Z = \frac{96.3 - 95}{\frac{5}{\sqrt{84} } }

       Z = 2.383

The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors

    P(X⁻ ≥ 96.3) = P(Z≥2.383)

                      =  1- P( Z<2.383)

                      =  1-( 0.5 -+A(2.38))

                      = 0.5 - A(2.38)

                     = 0.5 -0.4913

                    = 0.0087

<u><em>Final answer:-</em></u>

The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors

P(X⁻ ≥ 96.3) = 0.0087

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Answer:

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mina [271]
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8 0
3 years ago
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