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.

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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

8 0

2 years ago

likoan [24]

Answer:

1/4

Step-by-step explanation:

add the ratios

3+5=8

240 Divided by 8=30

3x30=90

5x30=150

Ann(90)

Bob150)

90/2=45 (remeber Ann gave half)

150/10=15 (remeber Bob gave a tenth)

45+15=60 (Colin received this amount)

60/240=1/4

8 0

4 years ago

An auto manufacturing company wanted to investigate how the price of one of its car models depreciates with age. The research de

sertanlavr [38]

Answer:

Price=150.414-16.626 Age

$34.032

We can't estimate the price of 18 year old car using the above regression equation because the price of 18 year old car falls outside the scope of the model

Step-by-step explanation:

Age(X) Price(Y)

8 18

3 94

6 50

9 21

2 145

5 42

6 36

3 99

The regression equation can be written as

y=a+bx

$b=\frac{sum(x-xbar)(y-ybar)}{sum(x-xbar)^{2} }$

a=ybar-bxbar

xbar=sumx/n

xbar=(8+3+6+9+2+5+6+3)/8=5.25

ybar=sumy/n

ybar=(18+94+50+21+145+42+36+99)/8=63.125

x y x-xbar y-ybar (x-xbar)^2 (x-xbar)(y-ybar)

8 18 2.75 -45.125 7.5625 -124.09375

3 94 -2.25 30.875 5.0625 -69.46875

6 50 0.75 -13.125 0.5625 -9.84375

9 21 3.75 -42.125 14.0625 -157.96875

2 145 -3.25 81.875 10.5625 -266.09375

5 42 -0.25 -21.125 0.0625 5.28125

6 36 0.75 -27.125 0.5625 -20.34375

3 99 -2.25 35.875 5.0625 -80.71875

Total 43.5 -723.25

$b=\frac{sum(x-xbar)(y-ybar)}{sum(x-xbar)^{2} }$

b=-723.25/43.5

b=-16.626

a=ybar-bxbar

a=63.125-(16.626)*5.25

a=150.414

Thus, the required regression equation is

y=150.414-16.626 x

Price=150.414-16.626 Age

For predicting price of 7 years old car we put x=7 in the estimated regression equation

y=150.414-16.626 x

y=150.414-16.626(7)

y=150.414-116.382

y=34.032

The predicted price of a 7 year old car is $34.032.

We can't estimate the price of 18 year old car using the above regression equation because the price of 18 year old car falls outside the scope of the model. The estimate of price is only valid for 2 to 9 year old car.

8 0

3 years ago

Your class earned $120.00 Saturday afternoon by washing cars to raise money for the class trip. This is 1/4 of the money needed

Ilya [14]

Answer:

$360.00

(I hope this was helpful)

5 0

3 years ago