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4 years ago

The diagram shows the cross-section ABCD of a sculpture in the shape of

Mathematics

1 answer:

zvonat [6]4 years ago

8 0

Answer:

891 cm³

Step-by-step explanation:

The volume (V) of the prism is calculated as

V = Ah ( A is the area of cross- section and h is the height (

The cross- section is a trapezium with area (A) calculated as

A = $\frac{1}{2}$ h ( a + b)

where h is the perpendicular height and a, b the parallel bases

Here h = 9, a = AB = 14 and b = CD = 8 , thus

A = $\frac{1}{2}$ × 9 × (14 + 8) = 4.5 × 22 = 99 cm² , thus

V = 99 × 9 = 891 cm³

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A math class consists of 25 students, 14 female and 11 male. Three students are selected at random to participate in a probabili

vovangra [49]

Answer:

(a) The probability that a male is selected, then two females is 0.4352.

(b) The probability that a female is selected, then two males is 0.3348.

(c) The probability that two females are selected, then one male is 0.4352.

(d) The probability that three males are selected is 0.0717.

(e) The probability that three females are selected is 0.1583.

Step-by-step explanation:

We are given that a math class consists of 25 students, 14 female and 11 male. Three students are selected at random to participate in a probability experiment.

(a) The probability that a male is selected, then two females is given by;

Number of ways of selecting a male from a total of 11 male = $^{11}C_1$

Number of ways of selecting two female from a total of 14 female = $^{14}C_2$

Total number of ways of selecting 3 students from a total of 25 = $^{25}C_3$

So, the required probability = $\frac{^{11}C_1 \times ^{14}C_2}{^{25}C_3}$

= $\frac{\frac{11!}{1! \times 10!} \times \frac{14!}{2! \times 12!} }{\frac{25!}{3! \times 22!} }$ { $\because ^{n}C_r = \frac{n!}{r! \times (n-r)!}$ }

= $\frac{1001}{2300}$ = 0.4352

(b) The probability that a female is selected, then two males is given by;

Number of ways of selecting a female from a total of 14 female = $^{14}C_1$

Number of ways of selecting two males from a total of 11 male = $^{11}C_2$

Total number of ways of selecting 3 students from a total of 25 = $^{25}C_3$

So, the required probability = $\frac{^{14}C_1 \times ^{11}C_2}{^{25}C_3}$

= $\frac{\frac{14!}{1! \times 13!} \times \frac{11!}{2! \times 9!} }{\frac{25!}{3! \times 22!} }$ { $\because ^{n}C_r = \frac{n!}{r! \times (n-r)!}$ }

= $\frac{770}{2300}$ = 0.3348

(c) The probability that two females is selected, then one male is given by;

Number of ways of selecting two females from a total of 14 female = $^{14}C_2$

Number of ways of selecting one male from a total of 11 male = $^{11}C_1$

Total number of ways of selecting 3 students from a total of 25 = $^{25}C_3$

So, the required probability = $\frac{^{14}C_2 \times ^{11}C_1}{^{25}C_3}$

= $\frac{\frac{14!}{2! \times 12!} \times \frac{11!}{1! \times 10!} }{\frac{25!}{3! \times 22!} }$ { $\because ^{n}C_r = \frac{n!}{r! \times (n-r)!}$ }