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Drupady [299]
3 years ago
14

A local pizzeria offers 15 toppings for their pizzas and you can choose any 3 of them for one fixed price. How many different ty

pes of pizzas can you order with 3 toppings?
Mathematics
1 answer:
Shtirlitz [24]3 years ago
6 0

Answer:

  455 or 680, depending

Step-by-step explanation:

If we assume the three choices are different, then there are ...

  15C3 = 15·14·13/(3·2·1) = 35·13 = 455

ways to make the pizza.

___

If two or three of the topping choices can be the same, then there are an additional ...

  2(15C2) +15C1 = 2·105 +15 = 225

ways to make the pizza, for a total of ...

  455 + 225 = 680

different types of pizza.

__

There is a factor of 2 attached to the number of choices of 2 toppings, because you can have double anchovies and tomato, or double tomato and anchovies, for example, when your choice of two toppings is anchovies and tomato.

_____

nCk = n!/(k!(n-k)!)

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iragen [17]

a. Answer:  m = 2

<u>Step-by-step explanation:</u>

f(x) = x² - 4x + 1

f(m) = m² - 4m + 1 = -3

         m² - 4m + 4 = 0

         (m - 2)(m - 2) = 0

               m = 2

***************************************************

b. Answer:  k = 2 and k = 5

<u>Step-by-step explanation:</u>

f(x) = x² - 7x + 14

f(k) = k² - 7k + 14 = 4

        k² - 7k + 10 = 0

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3 years ago
The table and the graph each show a different relationship between the same two variables, x and y:
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1. y₁ = 70x
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3 years ago
34xy What are the factors of the expression? 3/4,
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Assuming that the sample mean carapace length is greater than 3.39 inches, what is the probability that the sample mean carapace
joja [24]

Answer:

The answer is "".

Step-by-step explanation:

Please find the complete question in the attached file.

We select a sample size n from the confidence interval with the mean \muand default \sigma, then the mean take seriously given as the straight line with a z score given by the confidence interval

\mu=3.87\\\\\sigma=2.01\\\\n=110\\\\

Using formula:

z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}  

The probability that perhaps the mean shells length of the sample is over 4.03 pounds is

P(X>4.03)=P(z>\frac{4.03-3.87}{\frac{2.01}{\sqrt{110}}})=P(z>0.8349)

Now, we utilize z to get the likelihood, and we use the Excel function for a more exact distribution

=\textup{NORM.S.DIST(0.8349,TRUE)}\\\\P(z

the required probability: P(z>0.8349)=1-P(z

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2 years ago
In the diagram the sphere touches each face of the cube at one point-the center of each face. if each side of the cube is 7cm, w
Amiraneli [1.4K]

Answer:

C

Step-by-step explanation:

Te unoccupied volume (V) of the cube is calculated as

V = volume of cube - volume of sphere

   = 7³ - \frac{4}{3} πr³ ( r is the radius and r = 7 ÷ 2 = 3.5 )

   = 343 - \frac{4}{3} π × 3,5³

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   = 163.4....

    ≈ 160 cm³ ( to the nearest 10 cm³ )

5 0
3 years ago
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