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skad [1K]
3 years ago
15

Please help me someone? Will give a thanks and award

Mathematics
1 answer:
Setler [38]3 years ago
3 0
The probabilities of knocking down zero, one, and two pins are 0.175, 0.189, and 0.264, respectively.  The total probability of knocking down at most two is thus 0.175+0.189+0.264=0.628, or 62.8%.
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HELPPP <br> 8g + 25 = 97<br> Solve for G
RoseWind [281]

Answer: 120 g

Step-by-step explanation:

4 0
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 <em>X</em> is 2.6 and 1.2 respectively.

(b) The mean and standard deviation of <em>T</em> are 390 and 180 respectively.

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

Step-by-step explanation:

(a)

The random variable <em>X</em> 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 <em>X</em> is 2.6 and 1.2 respectively.

(b)

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

That is, <em>T</em> = 150 <em>X</em>

Compute the mean of <em>T</em> as follows:

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

Compute the standard deviation of <em>T</em> 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 <em>T</em> 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 <em>T</em> = 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 <em>T</em> 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.

8 0
3 years ago
Write 863.141 in expanded form
Mashutka [201]
Eight hundred sixty three and 1 hundred 41 thousandths
6 0
3 years ago
Read 2 more answers
12/18/2020
larisa86 [58]

Answer:

,twenty five divided by five

4 0
2 years ago
(-9.506)(18.78) =<br> wha does it equal?
notka56 [123]

Answer:

-178.52268

Step-by-step explanation:

This is the answer..

Plss mark me as brainliest.

5 0
3 years ago
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