I need help with this question

Jane’s favorite fruit punch consists of pear, pineapple, and plum juices in the ratio 5:2:3.

Vlad [161]

5:2:3
5+2+3=10

If she has double of plum juice the ratio will be

10:4:6
10+4+6=20

So she can make 20 cups of fruit punch.

3 0

2 years ago

Find value of y and x

Vika [28.1K]

Since the triangle is isosceles the measure of angle B is also 47.
since you have dropped a perpendicular bisector the measure of angle x is 90. and since the angles must add to 180 you get 47+47+a=180 and a= 86. since y is half this we have y is 43

if you like my answer please rate as brainliest! thx

4 0

2 years ago

1. (SLO) Simplify the expression: 6(b-7)

Sav [38]

1 → B

multiply each term in the parenthesis by the 6 outside

6(b-7) = 6b - 42

2 → D

multiply each term in the parenthesis by the 3 outside

3(4x+6y+2) = 12x+18y+6

3 → A

the value outside the parenthesis is -1, hence each term inside is multiplied by -1

-1(6x+5y) = -6x -5y

3 0

2 years ago

Read 2 more answers

What additional information would you need to prove that ΔABC ≅ ΔDEF by ASA

aleksandr82 [10.1K]

What do you mean? You just need to prove there are two angles and the side between these two angles are the same in each triangle. One of the possible combinations could be ∠BAC=∠FDE, ∠ABC= ∠DEF, and AB=DE.

3 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 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

2 years ago