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Ksju [112]
2 years ago
5

A store gets a bucket of eggs for $16.20 and sells a dozen eggs for $3.50. If there are 450 eggs in the bucket, by what percent

has the store increased the price per dozen eggs?
Please help every other answer when I searched up was wrong.
Mathematics
1 answer:
Murljashka [212]2 years ago
3 0

Answer:

710%

Step-by-step explanation:

Find each unit price.

16.20 $/bucket × (1 bucket / 450 eggs) = 0.036 $/egg

3.50 $/dozen × (1 dozen / 12 eggs) = 0.292 $/egg

The percent markup is:

(0.292 − 0.036) / 0.036 × 100% ≈ 710%

The store marks up the price by approximately 710%.

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The formula for the volume V of a cylinder is V = πr2h, where r is the radius of the base and h is the height of the cylinder. S
lana [24]

Answer:

Part A) h=V/(\pi r^{2})

Part B) h=4\ cm

Step-by-step explanation:

Part A)

we have the formula of the volume of a cylinder

V=\pi r^{2}h

Solve for h

That means ----> isolate the variable h

Divide both side by πr²

V/(\pi r^{2})=\pi r^{2}h/(\pi r^{2})

Simplify

V/(\pi r^{2})=h

Rewrite

h=V/(\pi r^{2})

Part B) Find the height of a cylinder with a volume of 36π cm3 and a base with a radius of 3 cm

we have

V=36\pi\ cm^3

r=3\ cm

substitute in the formula and solve for h

h=36\pi/(\pi 3^{2})

h=4\ cm

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2 years ago
In a class, 4/5 of the students have mathematical instruments. 1/4 of these students have lost their protractors. What fraction
Novay_Z [31]

Answer:

\frac{11}{20}

Step-by-step explanation:

\frac{4}{5} - \frac{1}{4}          <em>Subtract the students who don't have protractors from the students who have mathematical instruments.</em>

\frac{11}{20}

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2 years ago
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9 out of 10 people at a game are rooting for the home team. What is the probability that exactly 6 of 8 people sitting together
scZoUnD [109]

A random supporter roots the home team with probability 0.9, and the away team with probability 0.1.

Choosing 6 out of 8 supporters who root for the home team has probability

\displaystyle\binom{8}{6}\cdot 0.9^6\cdot 0.1^2 = 28\cdot0.531441\cdot 0.01=0.14

7 0
2 years ago
Find the smallest relation containing the relation {(1, 2), (1, 4), (3, 3), (4, 1)} that is:
professor190 [17]

Answer:

Remember, if B is a set, R is a relation in B and a is related with b (aRb or (a,b))

1. R is reflexive if for each element a∈B, aRa.

2. R is symmetric if satisfies that if aRb then bRa.

3. R is transitive if satisfies that if aRb and bRc then aRc.

Then, our set B is \{1,2,3,4\}.

a) We need to find a relation R reflexive and transitive that contain the relation R1=\{(1, 2), (1, 4), (3, 3), (4, 1)\}

Then, we need:

1. That 1R1, 2R2, 3R3, 4R4 to the relation be reflexive and,

2. Observe that

  • 1R4 and 4R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 4R1 and 1R2, then 4 must be related with 2.

Therefore \{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(4,1),(4,2)\} is the smallest relation containing the relation R1.

b) We need a new relation symmetric and transitive, then

  • since 1R2, then 2 must be related with 1.
  • since 1R4, 4 must be related with 1.

and the analysis for be transitive is the same that we did in a).

Observe that

  • 1R2 and 2R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 2R1 and 1R4, then 2 must be related with 4.
  • 4R1 and 1R2, then 4 must be related with 2.
  • 2R4 and 4R2, then 2 must be related with itself

Therefore, the smallest relation containing R1 that is symmetric and transitive is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

c) We need a new relation reflexive, symmetric and transitive containing R1.

For be reflexive

  • 1 must be related with 1,
  • 2 must be related with 2,
  • 3 must be related with 3,
  • 4 must be related with 4

For be symmetric

  • since 1R2, 2 must be related with 1,
  • since 1R4, 4 must be related with 1.

For be transitive

  • Since 4R1 and 1R2, 4 must be related with 2,
  • since 2R1 and 1R4, 2 must be related with 4.

Then, the smallest relation reflexive, symmetric and transitive containing R1 is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

5 0
1 year ago
Peter Johnson, the CFO of Homer Industries, Inc is trying to determine the Weighted Cost of Capital (WACC) based on two differen
Rzqust [24]

Answer:

Scenario 1 has WACC of 7.93%

Scenario 2 has WACC of 10.33%

Step-by-step explanation:

WACC=Ke*E/V+Kp*P/V+Kd(after tax)*D/V

Ke is the cost of equity =13%

Kp is the cost of preferred stock=10%

Kd(after tax) is the cost of debt of 8% adjusted for tax as below:

Kd(after tax )=Kd(before tax)*(1-t)

t is the tax rate of 30% or 0.3

Kd(after tax)=8%*(1-0.3)=5.60%

E is equity value,which is $1.8 million under scenario 1 and $3.8 under scenario 2

P is the value of preferred stock which is $1.2 million and $2.2 million respectively

D is the value of debt which is $5 million and $2 million

V=total value of capital structure=$8 million in both cases.

Scenario 1 WACC

WACC=13%*1.8/8+10%*1.2/8+5.6%*5/8=7.93%

Scenario 2 WACC

WACC=13%*3.8/8+10%*2.2/8+5.6%*2/8=10.33%

7 0
1 year ago
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