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3 years ago

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Suppose an interdisciplinary committee of 3 faculty members is to be selected from a group consisting of 4 men and 5 women. the

probability that all three faculty selected are men is approximately _______

1 answer:

yaroslaw [1]3 years ago

6 0

you answer to the problem is 3/9

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You are planting 47 flowers but want them to go into 6 rows . How many flowers will be in each row ?

Darina [25.2K]

Answer:

<u><em /></u> $7$ $flowers$ $with$ $4$ $flowers$ $left$ $over$

Step-by-step explanation

$47$ ÷ $6$

4 0

3 years ago

Find the value of 8!<br> O A. 56<br> O B. 16,777,216<br> O c. 40,320<br> O D. 64

Ksivusya [100]

8! is 40,320

Step-by-step explanation:

Step 1: Find 8! (8 factorial).

Factorial of 8 is the product of 8 and all the integers below 8.

⇒ 8! = 8×7×6×5×4×3×2×1

= 40,320

8 0

3 years ago

nickel has an approximate diameter of 20 mm and an approximate thickness of 2 mm what is the approximate volume of a stack of 40

ziro4ka [17]

Answer:

The approximate volume of a stack of 40 nickels is 25132.741 cubic milimeters.

Step-by-step explanation:

A nickel can be modelled by the formula of a right cylinder, the approximate volume of the stack of nickels ( $V_{net}$ ), measured in cubic milimeters, is the product of the number of elements ( $n$ ), no unit, and the volume of each element ( $V$ ), measured in cubic milimeters. That is:

$V_{net} = n\cdot V$ (1)

By volume equation of right cylinder, we have the following formula:

$V_{net} = \frac{\pi}{4}\cdot n \cdot D^{2}\cdot h$ (2)

Where:

$n$ - Amount of nickels, no unit.

$D$ - Diameter of nickel, measured in milimeters.

$h$ - Thickness of nickel, measured in milimeters.

If we know that $n = 40$ , $D = 20\,mm$ and $h = 2\,mm$ , then the net volume of the stack of nickels is:

$V_{net} = \frac{\pi}{4}\cdot (40)\cdot (20\,mm)^{2}\cdot (2\,mm)$

$V_{net} \approx 25132.741\,mm^{3}$

The approximate volume of a stack of 40 nickels is 25132.741 cubic milimeters.

8 0

2 years ago

Can any one pls help me

charle [14.2K]

Start off with the y-intercept and place it on the y axis. Then do rise/run to place the slope

8 0

3 years ago

Read 2 more answers

In a trapezoid the lengths of bases are 11 and 18. The lengths of legs are 3 and 7. The extensions of the legs meet at some poin

FrozenT [24]

Answer: The length of segments between this point and the vertices of greater base are $7\frac{5}{7}$ and 18.

Step-by-step explanation:

Let ABCD is the trapezoid, ( shown in below diagram)

In which AB is the greater base and AB = 18 DC= 11, AD= 3 and BC = 7

Let P is the point where The extended legs meet,

So, according to the question, we have to find out : AP and BP

In Δ APB and Δ DPC,

∠ DPC ≅ ∠APB ( reflexive)

∠ PDC ≅ ∠ PAB ( By alternative interior angle theorem)

And, ∠ PCD ≅ ∠ PBA ( By alternative interior angle theorem)

Therefore, By AAA similarity postulate,

$\triangle APB\sim \triangle D PC$

Let, DP =x

⇒ $\frac{3+x}{18} = \frac{x}{11}$

⇒ 33 +11x = 18x

⇒ x = 33/7= $4\frac{5}{7}$

Thus, PD= $4\frac{5}{7}$

But, AP= PD + DA

AP= $4\frac{5}{7}+3 =7\frac{5}{7}$

Now, let PC =y,

⇒ $\frac{7+y}{18} = \frac{y}{11}$

⇒ 77 + 11y = 18y

⇒ y = 77/7 = 11

Thus, PC= 11

But, PB= PC + CB

PB= 11+7 = 18

7 0

2 years ago