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jarptica [38.1K]
2 years ago
12

Each of the 10 students in the baking club made 2 chocolate cakes for a fundraiser. They all used the same recipe, using C cups

of flour in total.
Write an expression the represents the amount of flour required for one cake.
Mathematics
1 answer:
8090 [49]2 years ago
7 0

Answer:

The expression is C/20 cups of flour in one cake

Step-by-step explanation:

* Lets explain how to solve the problem

- There are 10 students

- Each one made 2 chocolate cakes

- They all used the same recipe

- The total amount of flour is C cups

* Lets change these information to equation

∵ There are 10 students

∵ Each student made 2 cakes

∴ The total number of cakes = 10 × 2 = 20 cakes

∵ The total amount of flour for the twenty cakes is C cups

- To find the amount of flour in each cake divide the total amount

  of flour by the total number of the cakes

∴ The amount of flour in each cake = C/20

* The expression is C/20 cups of flour in one cake

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During the 2015-16 NBA season, J.J. Redick of the Los Angeles Clippers had a free throw shooting percentage of 0.901 . Assume th
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Step-by-step explanation:

Given :  J.J. Redick of the Los Angeles Clippers had a free throw shooting percentage of 0.901 .

We assume that,

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So it is a binomial distribution .

Using binomial probability formula, the probability of getting success in x trials :

P(X=x)^nC_xp^x(1-p)^{n-x}

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Let x be binomial variable that represents the number of a=makes.

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The probability that he makes at least 13 of them will be :-

P(x\geq13)=P(x=13)+P(x=14)

=^{14}C_{13}(0.901)^{13}(1-0.901)^1+^{14}C_{14}(0.901)^{14}(1-0.901)^0\\\\=(14)(0.901)^{13}(0.099)+(1)(0.901)^{14}\ \ [\because\ ^nC_n=1\ \&\ ^nC_{n-1}=n ]\\\\\approx0.3574+0.2324=0.5898

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2 years ago
Two cross sections of a right hexagonal pyramid are obtained by cutting the pyramid with planes parallel to the hexagonal base.
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Answer:

The larger cross section is 24 meters away from the apex.

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Now we are given the areas of the two cross sections of the right hexagonal pyramid:A_1=216\:ft^2\: \:\:\:A_2=486\:ft^2

From these areas we find the radius of the hexagons:

r_1=\sqrt{\frac{2A_1}{3\sqrt{3} } } =\sqrt{\frac{2*216}{3\sqrt{3} } }=\boxed{9.12ft}

r_2=\sqrt{\frac{2A_2}{3\sqrt{3} } } =\sqrt{\frac{2*486}{3\sqrt{3} } }=\boxed{13.68ft}

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Therefore we have:

\frac{H-8}{r_1} =\frac{H}{r_2}

We put in the numerical values of r_1, r_2 and solve for H:

\boxed{H=\frac{8r_2}{r_2-r_1} =\frac{8*13.677}{13.68-9.12} =24\:feet.}

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