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vodomira [7]
2 years ago
14

Of the labeled points, which represents the animal

Mathematics
1 answer:
Paha777 [63]2 years ago
8 0

The ratio of life expectancy to gestation period is greatest at point (A) A.

<h3>What is life expectancy?</h3>
  • Life expectancy is a statistical measure of how long an organism is expected to live based on its birth year, current age, and other demographic factors such as gender.
  • The most commonly used metric is life expectancy at birth (LEB), which has two definitions.

To find the labeled points, which represent the animal for which the ratio of life expectancy to gestation period is greatest:

  • The graph below shows life expectancy on the y-axis and gestation period on the x-axis.
  • The life expectancy to gestation period ratio for point A is 7/22.5 = 14/45.
  • For point B, the ratio is 8/45.
  • Because the y coordinate is greater at Y than at X, which has the same x coordinate, we only consider the ratio at D, which is 10/51.
  • Since 14/45 > 8/45, we only have to compare 14/45 and 10/51.
  • So, 14 × 51 = 714 and 45 × 10 = 450.
  • Then, 14/45 > 10/51.

Therefore, the ratio of life expectancy to gestation period is greatest at point (A) A.

Know more about life expectancy here:

brainly.com/question/4648168

#SPJ4

The correct question is given below:
Of the labeled points, which represent the animal for which the ratio of life expectancy to gestation period is greatest?

A) A

B) B

C) C

D) D

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The desired percentage of Silicon Dioxide (SiO2) in a certain type of aluminous cement is 5.5. To test whether the true average
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a)  Null hypothesis : H₀ : μ = 5.5

 Alternative Hypothesis : H₁ : μ < 5.5

b) The test statistic

        |t| = |-3.33| = 3.33

c) P - value lies between in these intervals

0.001 < P < 0.005

Step-by-step explanation:

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

Given data the Population mean 'μ' = 5.5

The small sample size 'n' = 16

The sample mean (x⁻) = 5.25

Given the  percentage of SiO2 in a sample is normally distributed with a sigma of 0.3.

<u><em> Null hypothesis : H₀ : μ = 5.5</em></u>

<u><em>  Alternative Hypothesis : H₁ : μ < 5.5</em></u>

 Level of significance ∝ = 0.01

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

 The test statistic

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

                             t = \frac{5.25 -5.5}{\frac{0.3}{\sqrt{16} } }

On calculation , we get

                            t = -3.33

                           |t| = |-3.33| = 3.33

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

<u><em>P - value</em></u>

<u><em>The degrees of freedom γ = n-1 = 16-1 =15</em></u>

The calculated value t = 3.33 (check t-table) lies between the 0.001 to 0.005

0.001 < P < 0.005

<u>Condition(i)</u>

P - value < ∝ then reject H₀

<u>Condition(ii)</u>

P - value > ∝ then Accept H₀

we observe that  0.001 < P < 0.005

P- value < 0.01

we rejected  H₀

<em>(or)</em>

The tabulated value  = 2.60 at 0.01 level of significance with '15' degrees of freedom

The calculated value t = 3.33 > 2.60 at 0.01 level of significance with '15' degrees of freedom

The null hypothesis is rejected

<u><em>Conclusion</em></u>:-

Accepted Alternative hypothesis H₁

The Claim that the true average is smaller than 5.5

<u><em></em></u>

             

4 0
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