Choose the imensions that represen this rectlangl

Each base in this right prism-like figure is a quarter of a circle with a radius of 2m.

NISA [10]

Answer:

$volume = 10\pi m^3$

Step-by-step explanation:

Volume of a prism = Base Area x height

Base is a quarter of a circle.

$Area \ of \ circle = \pi r^2\\\\r = 2m\\\\Area \ of \ quarter \ of \ a \ circle = \frac{1}{4} \pi r^2 = \frac{1}{4}\pi \times 4 = \pi m^2$

Base Area = π square meters.

Therefore ,

$Volume = Base Area \times height = \pi \times 10 = 10\pi m^3$

7 0

3 years ago

A dance class consists of 22 students, of which 10 are women and 12 are men. if 5 men and 5 women are to be chosen and then pair

OverLord2011 [107]

solution:

I choose 5 women from a pool of 10 in 10C2 ways.

I choose 5 men from a pool of 12 in 12C2 ways.

So total number of ways of choosing in 10C2 x 12C2. Now I need to arrange them in 5 pairs. This is where I have a different solution. The solution says that there are 5! ways to arrange them in pairs.

But I cant seem to understand why? My reasoning is that for first pair position I need to choose 1 man from 5 and 1 woman from 5. So for the first position I have 5 x 5 choices (5 for man and 5 for woman). Similarly for the second position I have 4 x 4 choices and so on. Hence the total ways are 5! x 5!

So I calculate the total ways as 10C2 * 12C2 * 5! * 5!. Can anyone point the flaw in my reasoning for arranging the chosen men and women in pairs.

3 0

4 years ago

An equilateral triangle has perimeter P. The length of one side is p/3 which expression is equivalent to the length of one side?

LekaFEV [45]

Answer:

Your answer is C

Step-by-step explanation:

3 0

3 years ago

What is the approximate value of b, rounded to the nearest tenth?

ozzi

we know that

m∠ZWY=m∠WXY --------> given problem

so

In the right triangle ZWY

$tan\ ZWY=\frac{ZY}{YW}=\frac{5}{6}$

In the right triangle XWY

$tan\ WXY=\frac{WY}{XY}=\frac{6}{a}$

$\frac{6}{a} =\frac{5}{6} \\ \\5*a=6*6 \\ \\a=\frac{36}{5}=7.2\ units$

Find the value of b

Applying the Pythagorean Theorem

In the right triangle XWY

$b^{2}=6^{2}+a^{2} \\b^{2}=6^{2}+7.2^{2}\\b^{2}=87.84\\ b= 9.4\ units$

therefore

the answer is the option

D) 9.4 units

4 0

3 years ago

Read 2 more answers

Suppose that the weight of seedless watermelons is normally distributed with mean 6.2 kg. and standard deviation 1.5 kg. Let X b

Oksi-84 [34.3K]

Answer:

A.

X ~ N(6.2kg, 2.25kg²)

B. What is the median seedless watermelon weight?____ kg.

6.2 kg

C. What is the Z-score for a seedless watermelon weighing 8 kg?

1.2

D. What is the probability that a randomly selected watermelon will weigh more than 7 kg?

0.2981

E. What is the probability that a randomly selected seedless watermelon will weigh between 4 and 5 kg?

0.1411

F. The 80th percentile for the weight of seedless watermelons is _____ kg.

7.5 kg

Explanation:

A. X ~ N(___ , ____ )

The distribution of a random variable in a sample extracted from a population that follows a normal distribution is represented by the notation:

X ~ N(μ, σ²)

Where:

X is the random variable
N stands for normal distribution function
μ is the median of the population
σ² is the variance of the population

Here, you have:

μ = 6.2 kg

σ² = (1.5kg)² = 2.25 kg²

X ~ N(6.2kg, 2.25kg²)

B. What is the median seedless watermelon weight?____ kg.

The median of a random variable that follows a normal distribution is equal to the mean, thus it is 6.2 kg.

C. What is the Z-score for a seedless watermelon weighing 8 kg?

The z-score is the standardized value of the random variable. It measures how far away is the variable from the mean.

It is calcuated with the formula:

$Z-score(X=x)=\dfrac{x-\mu}{\sigma}$

Thus for X=8:

$Z(X=8)=\dfrac{8-6.2}{1.5}=1.2$

D. What is the probability that a randomly selected watermelon will weigh more than 7 kg?

You want P(X>7)

You must use the tables for the standardized normal distribution.

Find the corresponding Z-score for X = 7

$Z(X=7)=\dfrac{7-6.2}{1.5}\approx0.53$

You must use a table for the standardized normal distribution which gives the cumulative distribution or area under the curve of the standard normal distribution and find P(Z>0.53).

That is the area to the right of Z=0.53. The table shows 0.2981.

Thus, the probability that a randomly selected seedless watermelon will weigh more than 7 kg is 0.2981

E. What is the probability that a randomly selected seedless watermelon will weigh between 4 and 5 kg?

For this case you must find the Z-scores for X=4 and X=5 and then find the area under the curve of the standardized normal distribution between those two Z-scores.

Z(X=4) = (4 - 6.2)/1.5 ≈ -1.47

Z(X=5) = (5 - 6.2)/1.5 = -0.8

In the table the area to the right of Z = - 1.47 is 1 - 0.0708 = 0.9292

And the area to the right of Z = - 0.8 is 1 - 0.2119 = 0.7881

Thus, the area in between is the difference 0.9292 - 0.7881 = 0.1411.

F. The 80th percentile for the weight of seedless watermelons is _____ kg.

The 80th percentile is the weigh of the top 20% seedless watermelons: this is 80% of the weighs are below that weight.

You must find the Z-score for which the area under the curve is less than 0.80.

The area less than 0.80 is 1 less the area that is less than 0.20.

From the table, the Zscore that defines the area less than 0.20 is 0.845 (interpolating).

Thus, the 80th percentile is the X value that makes the Z-score greater than or equal to 0.845:

Z ≥ 0.845

(X - 6.2)/1.5 ≥ 0.845

X ≥ 0.845 × 1.5 + 6.2

X ≥ 7.4675

X ≥ 7.5 kg ← answer

7 0

3 years ago