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QuestionFind the batting average greater than 2×0.1+6×0.01+9×0.0012 \times 0.1 + 6 \times 0.01 + 9 \times 0.001. Options: A 0.27, B 0.2, C 0.264, D 0.25.

Studdy Solution

STEP 1

Assumptions1. The batting average of the player is given by the expression ×0.1+6×0.01+9×0.001 \times0.1+6 \times0.01+9 \times0.001 . We need to compare this average with the given options0.27,0.,0.264,0.25

STEP 2

First, we need to calculate the player's batting average. We can do this by adding the products of the given expression.
Batting average=2×0.1+6×0.01+9×0.001Batting\, average =2 \times0.1+6 \times0.01+9 \times0.001

STEP 3

Now, calculate the player's batting average.
Batting average=2×0.1+6×0.01+9×0.001=0.269Batting\, average =2 \times0.1+6 \times0.01+9 \times0.001 =0.269

STEP 4

Now that we have the player's batting average, we can compare it with the given options. The options greater than the player's batting average are those which are greater than0.269.

STEP 5

Comparing the player's batting average with option A0.27>0.2690.27 >0.269So, option A is greater than the player's batting average.

STEP 6

Comparing the player's batting average with option B0.2<0.2690.2 <0.269So, option B is not greater than the player's batting average.

STEP 7

Comparing the player's batting average with option C0.264<0.2690.264 <0.269So, option C is not greater than the player's batting average.

STEP 8

Comparing the player's batting average with option D0.25<0.2690.25 <0.269So, option D is not greater than the player's batting average.
So, the batting average which is greater than the player's batting average is0.27 (option A).

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