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2 years ago

Please answer! I got more coming soon. :))

Mathematics

2 answers:

Neporo4naja [7]2 years ago

5 0

Answer:

Area of circle = $\boxed{\sf{28.3 \: {in}^{2}}}$ .

Step-by-step explanation:

Here's the required formula to find the area of circle :

$\twoheadrightarrow{\pmb{\sf{A_{(Circle)} = \pi{r}^{2}}}}$

A = Area
π = 3.14
r = radius

Substituting all the given values in the formula to find the area of circle :

${\twoheadrightarrow{\sf{Area_{(Circle)} = \pi{r}^{2}}}}$

${\twoheadrightarrow{\sf{Area_{(Circle)} = 3.14{(3)}^{2}}}}$

${\twoheadrightarrow{\sf{Area_{(Circle)} = 3.14{(3 \times 3)}}}}$

${\twoheadrightarrow{\sf{Area_{(Circle)} = 3.14{(9)}}}}$

${\twoheadrightarrow{\sf{Area_{(Circle)} = 3.14 \times 9}}}$

${\twoheadrightarrow{\sf{Area_{(Circle)} = 28.26}}}$

${\twoheadrightarrow{\sf{Area_{(Circle)} \approx 28.3}}}$

$\star{\boxed{\sf{\red{Area_{(Circle)} \approx 28.3 \: {in}^{2}}}}}$

Hence, the area of circle is 28.3 in².

$\rule{300}{2.5}$

astra-53 [7]2 years ago

4 0

Answer:

28.3 square inches

Step-by-step explanation:

Use the area of a circle to determine its area:

$A=\pi r^2\\\\A=\pi(3)^2\\\\A=9\pi\\\\A=9(3.14)\\\\A=28.26\\\\A\approx28.3$

Therefore, the area of a circle with a radius of 3 inches is 28.3 square inches

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3 years ago

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2 years ago

4. A small high school holds its graduation ceremony in the gym. Because of seating constraints, students are limited to a maxim

Ad libitum [116K]

Answer:

(a) The mean and standard deviation of X is 2.6 and 1.2 respectively.

(b) The mean and standard deviation of T are 390 and 180 respectively.

(c) The distribution of T is N (390, 180²). The probability that all students’ requests can be accommodated is 0.7291.

Step-by-step explanation:

(a)

The random variable X is defined as the number of tickets requested by a randomly selected graduating student.

The probability distribution of the number of tickets wanted by the students for the graduation ceremony is as follows:

X P (X)

0 0.05

1 0.15

2 0.25

3 0.25

4 0.30

The formula to compute the mean is:

$\mu=\sum x\cdot P(X)$

Compute the mean number of tickets requested by a student as follows:

$\mu=\sum x\cdot P(X)\\=(0\times 0.05)+(1\times 0.15)+(2\times 0.25)+(3\times 0.25)+(4\times 0.30)\\=2.6$

The formula of standard deviation of the number of tickets requested by a student as follows:

$\sigma=\sqrt{E(X^{2})-\mu^{2}}$

Compute the standard deviation as follows:

$\sigma=\sqrt{E(X^{2})-\mu^{2}}\\=\sqrt{[(0^{2}\times 0.05)+(1^{2}\times 0.15)+(2^{2}\times 0.25)+(3^{2}\times 0.25)+(4^{2}\times 0.30)]-(2.6)^{2}}\\=\sqrt{1.44}\\=1.2$

Thus, the mean and standard deviation of X is 2.6 and 1.2 respectively.

(b)

The random variable T is defined as the total number of tickets requested by the 150 students graduating this year.

That is, T = 150 X

Compute the mean of T as follows:

$\mu=E(T)\\=E(150\cdot X)\\=150\times E(X)\\=150\times 2.6\\=390$

Compute the standard deviation of T as follows:

$\sigma=SD(T)\\=SD(150\cdot X)\\=\sqrt{V(150\cdot X)}\\=\sqrt{150^{2}}\times SD(X)\\=150\times 1.2\\=180$

Thus, the mean and standard deviation of T are 390 and 180 respectively.

(c)

The maximum number of seats at the gym is, 500.

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sum of values of X, i.e ∑X, will be approximately normally distributed.

Here T = total number of seats requested.

Then, the mean of the distribution of the sum of values of X is given by,

$\mu_{T}=n\times \mu_{X}=390$

And the standard deviation of the distribution of the sum of values of X is given by,

$\sigma_{T}=n\times \sigma_{X}=180$

So, the distribution of T is N (390, 180²).

Compute the probability that all students’ requests can be accommodated, i.e. less than 500 seats were requested as follows:

$P(T$

Thus, the probability that all students’ requests can be accommodated is 0.7291.

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3 years ago