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4 years ago

12

At middle ave school 1/3 of the students are on the track team.the track coach offered to buy a pizza or subs for everyone on th

e team.of the students for whom the coach bought good 3/5 ordered pizza.what fraction of the students at middle ave school atenpizza

1 answer:

Anvisha [2.4K]4 years ago

6 0

1/18 of the students ate pizza at middle ave school.

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When a wholesaler sold a product at $50 per unit, sales were 184 units per week. After a price increase of $5, however, the aver

nalin [4]

Answer:

$48 per unit

Step-by-step explanation:

Increasing the price by $5 reduces demand by 20 units, so the slope of the curve is -4 units per dollar. This lets us write a demand equation as ...

q = -4(p -50) +184

q = -4p + 384

q = 4(96 -p)

The revenue is the product of price and demand:

r = pq = 4p(96 -p)

This is the equation of a quadratic curve that opens downward and has zeros at p=0 and p=96. The vertex (maximum) will be halfway between the zeros, at ...

p = (0+96)/2 = 48

A price of $48 per unit will yield a maximum total revenue.

7 0

4 years ago

Use combining like terms and simplify this expression:6(x+4)+1

Yuliya22 [10]

Distribute the 6 into the (x+4). So it would be (6x+24)+1 which is 6x+25. that is the answer

6 0

3 years ago

Triangle ABC has AB=5, BC=7, and AC=9. D is on AC with BD=5. Find the length of DC

Amanda [17]

Answer:

DC = 4.5

Step-by-step explanation:

Since DC is on AC it is a segment bisector.

This divides AC exactly in half.

AC = 9 / 2 = 4.5

DC = 4.5

7 0

3 years ago

A colony of cats were introduced into a reserve. The number of cats present at timet(measured in years since the colony was intr

RUDIKE [14]

ANSWER:

a) 12 cats

b) 232

c) 246

d) 14

EXPLANATION:

Given the expression for number of cats C present at time t:

$C\text{ = }\frac{12.31}{0.05+0.56^t^{}}$

a) To find the number of cats initially on the reserve, let t = 0

Therefore, substitute 0 for t in the equation

$\begin{gathered} C\text{ = }\frac{12.31}{0.05+0.56^0} \\ \text{ = }\frac{12.31}{0.05\text{ + 1}} \\ \text{ = }\frac{12.31}{1.05} \\ =\text{ }11.72 \end{gathered}$

Number of cats initially on the reserve are approximately 12 cats

b) C(10):

$\begin{gathered} C(10)\text{ = }\frac{12.31}{0.05+0.56^{10}}\text{ = 232.12} \\ \end{gathered}$

C(10) = 232

Here, C(10) means that C is a function of 10. This means at time = 10 years

C)Using function notation to express the number of cats present after 17 years, we have:

$C(17)\text{ = }\frac{12.31}{0.05+0.56^{17}}$

$C(17)\text{ = }\frac{12.31}{0.05+0.56^{17}}\text{ = }245.94$

Therefore, number of cats present after 17 years are approximately 246 cats

C(17) = 246 cats

d) In this case, first find the number of cats present in the 10th year and subtract from the number of cats present in the 17th year.

$C(10)\text{ = }\frac{12.31}{0.05+0.56^{10}}=\text{ }232.12$

From question C above, we know C(17) = 246

Therefore, the increase in cat population to be expected from the 10th year to the 17th year is:

C(17) - C(10) = 246 - 232 = 14 cats

7 0

1 year ago

Solve. Write answer in simplest form.<br> 7 - 2/5 =

Arlecino [84]

Answer:

6 3/5

Step-by-step explanation:

7 - 2/5 = 6 3/5

Sorry I can’t really explain this. It just comes into my head

3 0

2 years ago

Read 2 more answers