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Softa [21]
3 years ago
7

In how many ways can 6 different kinds of pizza be listed on the menu?

Mathematics
1 answer:
vredina [299]3 years ago
5 0

Answer: 720

Step-by-step explanation:

Since we are to list 6 different kinds of Pizza and no condition is attached , this arrangement could be done in 6! ways.

That is

6! = 720

This means that the different kind of pizza could be listed in 720 ways

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Would you please help me with this quiz answer, I will really appreciate if you do. ​
laila [671]

Answer:

a . domain 5,0,7,9,0

range -2,-2,-4,8,2

b. domain 2,4,8,9

range 1,2,4,11

Step-by-step explanation:

<h3>a is not a function</h3>

because function is a relationship in which each domain element occurs only once.

<h3>b is a function</h3>
6 0
3 years ago
Find g(-1)).<br> g[(-1)) =
Mashutka [201]

Answer:

Step-by-step explanation:

Incomplete question.

5 0
3 years ago
You are dealt one card from a standard 52-card deck. Find the probability of being dealt an ace or a 6.
Ainat [17]

Answer:

2/13

Step-by-step explanation:

Since there's 4 of each card in the deck this means that there're 4 sixes and 4 aces, so 8 cards in a 52 card deck gives the problem 8/52 which equals 2/13

4 0
3 years ago
I don't understand no this at all​
Katena32 [7]

Answer:

Just get the perimeter of the smaller rectangle then times it by 4

Step-by-step explanation:

i tried to use a calculator but i cant do the perimeter of the smaller triangle because i dont know what x is sorry ill try to figure it out.

8 0
3 years ago
Customers arrivals at a checkout counter in a department store per hour have a Poisson distribution with parameter λ = 7. Calcul
IgorLugansk [536]

Answer:

(1)14.9% (2) 2.96% (3) 97.04%

Step-by-step explanation:

Formula for Poisson distribution: P(k) = \frac{\lambda^ke^{-k}}{k!} where k is a number of guests coming in at a particular hour period.

(1) We can substitute k = 7 and \lambda = 7 into the formula:

P(k=7) = \frac{7^7e^{-7}}{7!}

P(k=7) = \frac{823543*0.000911882}{5040} = 0.149 = 14.9\%

(2)To calculate the probability of maximum 2 customers, we can add up the probability of 0, 1, and 2 customers coming in at a random hours

P(k\leq2) = P(k=0)+P(k=1)+P(k=2)

P(k\leq2) = \frac{7^0e^{-7}}{0!} + \frac{7^1e^{-7}}{1!} + \frac{7^2e^{-7}}{2!}

P(k \leq 2) = \frac{0.000911882}{1} + \frac{7*0.000911882}{1} + \frac{49*0.000911882}{2}

P(k\leq2) = 0.000911882+0.006383174+0.022341108 \approx 0.0296=2.96\%

(3) The probability of having at least 3 customers arriving at a random hour would be the probability of having more than 2 customers, which is the invert of probability of having no more than 2 customers. Therefore:

P(k\geq 3) = P(k>2) = 1 - P(k\leq2) = 1 - 0.0296 = 0.9704 = 97.04\%

4 0
3 years ago
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