Hello! Your answer was 19! There is no possible way for me to show you how to work this out using base ten blocks! Hopefully you can work it out in your head next time! :)
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>
Answer:
1/9, decimal form = 0.1 repeating
Answer:
Sure
Step-by-step explanation:
X
1. 2x^2 - 8 = 2(x^2 - 4) = 2(x+2)(x-2)
2. 2x^2 + 8x + 6 = 2(x^2 + 4x + 3) = 2(x+3)(x+1)
3. 3n^2 + 9n -30 = 3(n^2 + 3n - 10) = 3(n+5)(n-2)
XII
1. x^2 + 2x + xy + 2y = x(x+2) + y(x+2) = (x+2)(x+y)
2. 3a^2 - 2b - 6a + ab = 3a^2 - 6a + ab - 2b = 3a(a - 2) + b(a - 2) = (a-2)(3a+b)
3. t^3 - t^2 + t - 1 = t^2(t - 1) + (t - 1) = (t-1)(t^2+1)
4. 10 + 2t - 5s - st = 10 - 5s + 2t - st = 5(2 - s) + t(2-s) = (5+t)(2-s)
