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zvonat [6]
3 years ago
13

Write the estimated weight of each dinosaur in scientific notation.

Mathematics
2 answers:
Lubov Fominskaja [6]3 years ago
6 0
Argentinosaurus- 2.2x10^5

Brachiosaurus- 1.0x10^5

Apatosaurus- 6.6x10^4

Diplodocus- 5.0x10^4

Camarasourus- 4.0x10^4

Cetioauriscus- 1.985X10^4

I hope this helps :3
Levart [38]3 years ago
6 0
<span>Argentinosaurus 220,000
</span><span>Apatosaurus 66,000
</span><span>Brachiosaurus 100,000
 </span><span>Cetioauriscus 19,850 
</span> <span>Camarasourus 40,000
 </span><span>Diplodocus 50,000</span> 
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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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1. bobby would be 21 and chris 18.


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