B.30
cause 60 is 30 halted so yeah
Remark
First of all you have to declare the meaning of g(f(x)) After you have done that, you have to make the correct substitution.
Givens
f(x) = 4x^2 + x + 1
g(x) = x^2 - 2
Discussion
What the given condition g(f(x)) means is that you begin with g(x). Write down g(x) = x^2 - 2
Wherever you see an x on either the left or right side of the equation, you put fix)
Then wherever you see f(x) on the right you put in what f(x) is equal to.
Solution
g(x) = x^2 - 2
g(f(x)) = (f(x))^2 - 2
g(f(x)) = [4x^2 + x + 1]^2 - 2
f(x)^2 =
4x^2 + x + 1
<u>4x^2 + x + 1</u>
16x^4 + 4x^3 + 4x^2
4x^3 + x^2 + x
<u> 4x^2 + x + 1</u>
16x^4 + 8x^3 + 9x^2 + 2x + 1
Answer
g(f(x)) = 16x^4 + 8x^3 + 9x^2 + 2x + 1 - 2
g(f(x)) = 16x^4 + 8x^3 + 9x^2 + 2x - 1
From the box plot, it can be seen that for grade 7 students,
The least value is 72 and the highest value is 91. The lower and the upper quartiles are 78 and 88 respectively while the median is 84.
Thus, interquatile range of <span>the resting pulse rate of grade 7 students is upper quatile - lower quartle = 88 - 78 = 10
</span>Similarly, from the box plot, it can be seen that for grade 8 students,
The
least value is 76 and the highest value is 97. The lower and the upper
quartiles are 85 and 94 respectively while the median is 89.
Thus, interquatile range of the resting pulse rate of grade 8 students is upper quatile - lower quartle = 94 - 85 = 9
The difference of the medians <span>of the resting pulse rate of grade 7 students and grade 8 students is 89 - 84 = 5
Therefore, t</span><span>he difference of the medians is about half of the interquartile range of either data set.</span>
I'd say the ball player would get 275 hits.
