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erica [24]
2 years ago
8

Alona has some wood cutouts of two different shapes - square and triangle. Each square cutout is 6 mm thick and each triangular

cutout is 8 mm thick. Both cutouts have the same area of cross-section. Alona makes an upright stack using 24 square cutouts. She is making an oblique stack using the triangular cutouts. How many triangular cutouts should she put in the oblique stack so that both stacks have the same volume?
Mathematics
1 answer:
kobusy [5.1K]2 years ago
7 0

Answer:

18

Step-by-step explanation:

Volume is a product of cross-sectional area and height and relresented as V=Ah where V is volume, A is cross-sectional area and h is height.

For the square, since each has a thickness of 6mm then for 24 of them the height will be 24*6=144 mm

Since the volume and cross sectional area of the triangle should be the same, then the height should also be the same.

Since for the triangle each has a thickness of 8mm then the number requred to achieve a height of 144 mm will be 144/8=18 stacks

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tatiyna

Correct question is ;

The grade appeal process at a university requires that a jury be structured by selecting eight individuals randomly from a pool of nine students and eleven faculty.​ (a) What is the probability of selecting a jury of all​ students? (b) What is the probability of selecting a jury of all​ faculty? (c) What is the probability of selecting a jury of six students and two ​faculty?

Answer:

A) 7.144 × 10^(-5)

B) 0.00131

C) 0.0367

Step-by-step explanation:

We are given;

Number of students = 9

Number of faculty members = 11

A) Now, the number of ways we can select eight students from 9 =

C(9, 8) = 9!/(8! × 1!) = 9

Also, number of ways of selecting 8 individuals out of the total of 20 = C(20,8) = 20!/(8! × 12!) = 125970

Thus, probability of selecting a jury of all​ students = 9/125970 = 7.144 × 10^(-5)

B) P(selecting a jury of all faculty) = (number of ways to choose 8 faculty out of 11 faculty)/(Total number of ways to choose 8 individuals out of 20 individuals) = [C(11,8)]/[C(20,8)] = (11!/(8! × 3!))/125970 = 0.00131

C) P(selecting a jury of six students and two faculty) = ((number of ways to choose 6 students out of 9 students) × (number of ways to choose 2 faculty out of 11 faculty))/(Total number of ways to choose 8 individuals out of 20 individuals) = [(C(9,6) × C(11,2)]/125970

This gives;

(84 × 55)/125970 = 0.0367

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