Choose the equation of the line parallel to the x axis

What is the sin inverse of (sine 9pi/7)?

Sveta_85 [38]

For the answer to the question above, I've used my calculator to inverse your given equation because my brain can't do that.
The answer to your question is
In radians, arcsin(sin(9π) / 7) = 0.
i hope my answer helped you.

7 0

3 years ago

Given that ΔPQR is similar to ΔPTS, which statement MUST be true? A) m∠PST = m∠QPR B) m∠TPS = m∠RPQ C) m∠SPT = m∠PTS D) m∠PRQ =

galina1969 [7]

Answer:

The correct option is B.

Step-by-step explanation:

Two triangle are similar if their corresponding sides are in the same proportion or the corresponding angles are same.

It is given that the ΔPQR is similar to ΔPTS. It means all corresponding angles are same.

$\angle R=\angle S$

$\angle Q=\angle T$

$\angle P=\angle P$

Angle P can be defined as

$\angle QPR=\angle TPS$

$\angle RPQ=\angle TPS$

Therefore option B is correct.

$\angle PST\neq\angle QPR$

$\angle SPT\neq\angle PTS$

$\angle PRQ\neq\angle PTS$

Therefore option A, C and D are incorrect.

7 0

3 years ago

Read 2 more answers

The library charges a fine

Alex787 [66]

.55 when rounded up but this is the actual answer: 0.548461538461538

7 0

2 years ago

Please help me i need answers ASAP!!

mafiozo [28]

Answer:

I believe the first one is A

Step-by-step explanation:

4 0

3 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.

8 0

3 years ago