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2 years ago

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What is the total amount of sap trees produced that day? Please tell me ASAP(as soon as possible) because this lesson is due for

a grade tomorrow and I’m stuck on this question. Please tell me now because I have to go, Thanks!

2 answers:

ArbitrLikvidat [17]2 years ago

6 0

Answer:

5 gallons

Step-by-step explanation:

Number of x's on 1/4 = 3

So, amount of sap trees produce that day = 3/4

Number of x's on 3/8 = 2

So, amount of sap trees produce that day = 2 × 3/8 = 3/4

Number of x's on 5/8 = 4

So, amount of sap trees produce that day = 4 ×5/8 = 5/2

Number of x's on 1 = 1

So, amount of sap trees produce that day = 1 × 1 =1

Therefore,

total amount of sap the trees produce = amount of sap trees produce based on number of x's on 1/4 + amount of sap trees produce based on number of x's on 3/8 + amount of sap trees based on number of x's on 5/8 + amount of sap trees based on number of x's on 1 = $\frac{5}{2}+\frac{3}{4}+\frac{3}{4}+1=\frac{10+3+3+4}{4}=\frac{20}{4}=5\,\,gallons$

gogolik [260]2 years ago

3 0

Answer: 5 gallons

Step-by-step explanation:

There are 3 x's on the graph for 1/4, so 1/4 times 3=3/4

There are 2 x's on the graph for 3/8, so 3/8 times 2=3/4

There are 4 x's on the graph for 5/8, so 5/8 times 4=5/2

There is only 1 x for 1, so it will remain 1.

So if you add these up,

3/4+3/4+5/2+1

6/4+10/4+1

16/4+1

4+1

There is 5 gallons of sap produced that day.

Hope this helped!

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Answer:

The ticket price that maximizes revenue is $50.

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Step-by-step explanation:

We have to write a function that describes the revenue of the airline.

We know one point of this function: when the price is $30, the amount of passengers is 7000.

We also know that for an increase of $1 in the ticket price, the amount of passengers will decrease by 100.

Then, we can write the revenue as the multiplication of price and passengers:

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where x is the variation in the price of the ticket.

Then, if we derive R in function of x, and equal to 0, we will have the value of x that maximizes the revenue.

$R(x)=(30+x)(7000-100x)=30\cdot7000-30\cdot100x+7000x-100x^2\\\\R(x)=-100x^2+(7000-3000)x+210000\\\\R(x)=-100x^2+4000x+210000\\\\\\\dfrac{dR}{dx}=100(-2x)+4000=0\\\\\\200x=4000\\\\x=4000/200=20$

We know that the increment in price (from the $30 level) that maximizes the revenue is $20, so the price should be:

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The maximum monthly revenue is:

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2 years ago

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2 years ago

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The second question:

Consider the division expression $7\frac{1}{2} / 2$ . Select all multiplication equations that correspond to this division expression.

$2 * ? = 7\frac{1}{2}$ $7\frac{1}{2} * ?= 2$ $? * 2 = 7\frac{1}{2}$

$2* 7\frac{1}{2} = ?$ $? * 7\frac{1}{2} = 2$

Answer:

1. See Explanation

2. $2 * ? = 7\frac{1}{2}$ and $? * 2 = 7\frac{1}{2}$

Step-by-step explanation:

Solving (a):

Given

$Students = 60$

$Group = Equal\ Sized$

Required

Interpret $\frac{60}{5}$ in 2 ways

<u>Interpretation 1:</u> Number of groups if there are 5 students in each

<u>Interpretation 2:</u> Number of students in each group if there are 5 groups

<u>Solving the quotient</u>

$Quotient = \frac{60}{5}$

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<u>For Interpretation 1:</u>

The quotient means: 12 groups

<u>For Interpretation 2:</u>

The quotient means: 12 students

Solving (b):

Given

$7\frac{1}{2} / 2$

Required

Select all equivalent multiplication equations

Let ? be the quotient of t $7\frac{1}{2} / 2$

So, we have:

$7\frac{1}{2}/2 = ?$

Multiply through by 2

$2 * 7\frac{1}{2}/2 = ? * 2$

$7\frac{1}{2} = ? * 2$

Rewrite as:

$? * 2 = 7\frac{1}{2}$ --- This is 1 equivalent expression

Apply commutative law of addition:

$2 * ? = 7\frac{1}{2}$ --- This is another equivalent expression

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2 years ago

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Answer:

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Binomial probability distribution

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$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

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This means that 100 - 97.4 = 2.6% = 0.026 did not finish, which means that $p = 0.026$

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This is:

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4 0

2 years ago