What is the measure of the angle? Enter the answer in the box. degrees

A kitchen sink faucet streams 0.5 gallons of water in 10 seconds. At this rate, how much water do you predict it will stream in

AVprozaik [17]

Answer:

3 Gallons of water

Step-by-step explanation:

Because 1 gallon in 20 seconds and 60 seconds in one minute

3 0

3 years ago

You have a cookie recipe that calls for 1/3 cup of sugar. You want to make 3 times as many cookies by tripling the recipe. How m

Rufina [12.5K]

Answer:

You will need 1 cup of sugar

Step-by-step explanation:

1/3*3/1=3/3=1

4 0

3 years ago

PLEASE HELP! What is the length of side s of the square shown below!w?

Aloiza [94]

Answer: 2

Step-by-step explanation: because the x in the middle has to be the same

4 0

3 years ago

Read 2 more answers

Please answer ASAP! An explanation is a MUST REQUIREMENT in order to receive points and the Brainliest answer. Thank you.

galben [10]

Remmber pemdas or gemdas or bodmas or whatever

exponential laws
(x^m)^n=x^(mn)

ok so

(3^2)(2a)^(3/2))^2=
(3^2)(2a)^((3/2)*2)=
(3^2)(2a))^(6/2)=
(3^2)(2a)^3=
(3^2)(8a^3)^3=
(9)(8a^3)=
72a^3

2nd one from left

3 0

3 years ago

A student is getting ready to take an important oral examination and is concerned about the possibility of having an "on" day or

Tamiku [17]

Answer:

The students should request an examination with 5 examiners.

Step-by-step explanation:

Let X denote the event that the student has an “on” day, and let Y denote the

denote the event that he passes the examination. Then,

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

The events ( $Y|X$ ) follows a Binomial distribution with probability of success 0.80 and the events ( $Y|X^{c}$ ) follows a Binomial distribution with probability of success 0.40.

It is provided that the student believes that he is twice as likely to have an off day as he is to have an on day. Then,

$P(X)=2\cdot P(X^{c})$

Then,

$P(X)+P(X^{c})=1$

⇒

$2P(X^{c})+P(X^{c})=1\\\\3P(X^{c})=1\\\\P(X^{c})=\frac{1}{3}$

Then,

$P(X)=1-P(X^{c})\\=1-\frac{1}{3}\\=\frac{2}{3}$

Compute the probability that the students passes if request an examination with 3 examiners as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

$=[\sum\limits^{3}_{x=2}{{3\choose x}(0.80)^{x}(1-0.80)^{3-x}}]\times\frac{2}{3}+[\sum\limits^{3}_{x=2}{{3\choose x}(0.40)^{3}(1-0.40)^{3-x}}]\times\frac{1}{3}$

$=0.715$

The probability that the students passes if request an examination with 3 examiners is 0.715.

Compute the probability that the students passes if request an examination with 5 examiners as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

$=[\sum\limits^{5}_{x=3}{{5\choose x}(0.80)^{x}(1-0.80)^{5-x}}]\times\frac{2}{3}+[\sum\limits^{5}_{x=3}{{5\choose x}(0.40)^{x}(1-0.40)^{5-x}}]\times\frac{1}{3}$

$=0.734$

The probability that the students passes if request an examination with 5 examiners is 0.734.

As the probability of passing is more in case of 5 examiners, the students should request an examination with 5 examiners.

8 0

3 years ago