there are 325 students each class is limited to at most 30 students. how many classes need to be offered?

Solve by taking the square root of both sides.

horrorfan [7]

Answer:

a

Step-by-step explanation:

Given

3(x + 3)² - 81 = 0 ( add 81 to both sides )

3(x + 3)² = 81 ( divide both sides by 3 )

(x + 3)² = 27 ( take the square root of both sides )

x + 3 = ± $\sqrt{27}$ = ± 3 $\sqrt{3}$ ( subtract 3 from both sides )

x = - 3 ± 3 $\sqrt{3}$

x = - 3 - 3 $\sqrt{3}$ or x = - 3 + 3 $\sqrt{3}$ → a

8 0

3 years ago

Read 2 more answers

Prove that (4n+1)^2(4n-1)^2 is a multiple of 8 for all integers of n.<br><br> thanks.

azamat

Answer: i think (4n+1)^2(4n-1)^2 isnt a multiple of 8 for all integers of n because:

(4n + 1)²(4n - 1)²

= [(4n + 1)(4n - 1)]²

= (16n² - 1)²

= 16².n².n² - 2.16.n² + 1

= 8n²(32n² - 4) + 1

can see 8n²(32n² - 4) is a multiple of 8 but 1 isnt a multiple of 8

=> (4n + 1)²(4n - 1)² isnt a multiple of 8 for all integers of n.

Step-by-step explanation:

6 0

3 years ago

In an experiment, college students were given either four quarters or a $1 bill and they could either keep the money or spend it

gavmur [86]

Answer:

a) $P(A|B) = \frac{15/83}{44/83} =\frac{15}{44}=0.341$

b) $P(B|A) = \frac{29/83}{44/83} =\frac{29}{44}=0.659$

c) A. A student given a $1 bill is more likely to have kept the money.

Because the probability 0.659 is atmoslt two times greater than 0.341

Step-by-step explanation:

Assuming the following table:

Purchased Gum Kept the Money Total

Students Given 4 Quarters 25 14 39

Students Given $1 Bill 15 29 44

Total 40 43 83

a. find the probability of randomly selecting a student who spent the money, given that the student was given a $1 bill.

For this case let's define the following events

B= "student was given $1 Bill"

A="The student spent the money"

For this case we want this conditional probability:

$P(A|B) =\frac{P(A and B)}{P(B)}$

We have that $P(A)= \frac{40}{83} , P(B)= \frac{44}{83}, P(A and B)= \frac{15}{83}$

And if we replace we got:

$P(A|B) = \frac{15/83}{44/83} =\frac{15}{44}=0.341$

b. find the probability of randomly selecting a student who kept the money, given that the student was given a $1 bill.

For this case let's define the following events

B= "student was given $1 Bill"

A="The student kept the money"

For this case we want this conditional probability:

$P(A|B) =\frac{P(A and B)}{P(B)}$

We have that $P(A)= \frac{43}{83} , P(B)= \frac{44}{83}, P(A and B)= \frac{29}{83}$

And if we replace we got:

$P(B|A) = \frac{29/83}{44/83} =\frac{29}{44}=0.659$

c. what do the preceding results suggest?

For this case the best solution is:

A. A student given a $1 bill is more likely to have kept the money.

Because the probability 0.659 is atmoslt two times greater than 0.341

3 0

4 years ago

Normal Distribution. Cherry trees in a certain orchard have heights that are normally distributed with mu = 112 inches and sigma

Lubov Fominskaja [6]

Answer:

The probability that a randomly chosen tree is greater than 140 inches is 0.0228.

Step-by-step explanation:

Given : Cherry trees in a certain orchard have heights that are normally distributed with $\mu = 112$ inches and $\sigma = 14$ inches.

To find : What is the probability that a randomly chosen tree is greater than 140 inches?

Solution :

Mean - $\mu = 112$ inches

Standard deviation - $\sigma = 14$ inches

The z-score formula is given by, $Z=\frac{x-\mu}{\sigma}$

Now,

$P(X>140)=P(\frac{x-\mu}{\sigma}>\frac{140-\mu}{\sigma})$

$P(X>140)=P(Z>\frac{140-112}{14})$

$P(X>140)=P(Z>\frac{28}{14})$

$P(X>140)=P(Z>2)$

$P(X>140)=1-P(Z$

The Z-score value we get is from the Z-table,

$P(X>140)=1-0.9772$

$P(X>140)=0.0228$

Therefore, the probability that a randomly chosen tree is greater than 140 inches is 0.0228.

5 0

3 years ago

Malik, who is 5 feet tall, stands 15 feet from a mirror. In the mirror, he can see the reflection of the top of a tree, which he

Goryan [66]

Answer:

I don't get the question?? What is the question??

Step-by-step explanation:

4 0

3 years ago