The correct answer is C. F(x)=2 * (0.7)^x
Answer:
Part a) to find the maximum height of the snowball, you have to differentiate the function. Therefore you get ----> dh/dt= -32x-8 . Now equate this to zero and solve for x. x= (-1)/4 now sub this value in to find h(x) [note: i'm talking about the original function] . I got h max = h((-1)/4) = 9 which is the max height.
Step-by-step explanation:
I'm not too sure about the other questions. Sorry
Answer:
352 calls
Step-by-step explanation:
Each day he makes 7 calls and 36 days have gone by so you'll multiply 36 and 7.
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>
