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Anna007 [38]
2 years ago
8

Billy visits a local restaurant. He has 3 meat choices, 10 vegetable choices, 2 bread choices, 6 dessert choices, and 5 drink ch

oices. How many different meals can he possibly have if he chooses one each of meat, vegetable, bread, dessert, and drink?
Mathematics
1 answer:
Galina-37 [17]2 years ago
8 0

There are 1800 different meals can he possibly have if he chooses one each of meat, vegetable, bread, dessert, and drink.

What is Fundamental principle of counting?

The Fundamental Counting Principle (also called the counting rule) is a way to figure out the number of outcomes in a probability problem. Basically, you multiply the events together to get the total number of outcomes.

3 meat choices, 10 veg choices, 2 bread choices, 6 dessert choices.

According to the fundamental principle of counting,

Total number of choices = 3 x 10 x 2 x 6 x 5 choices

                                         = 1800 choices

Thus, There are 1800 different meals can he possibly have if he chooses one each of meat, vegetable, bread, dessert, and drink.

Learn more about fundamental principle of counting from:

brainly.com/question/4747487

#SPJ1

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Brainliest!!!!!!
Nataly [62]

Answer:

2x² - 12x + 13

Step-by-step explanation:

Given following:

  • k(x) = 2x² - 5
  • p(x) = x - 3

Start solving from right to left across the function.

Steps:

  • k[p(x)]
  • k[x - 3]
  • 2(x - 3)² - 5
  • 2(x² - 6x + 9) - 5
  • 2x² - 12x + 18 - 5
  • 2x² - 12x + 13
4 0
2 years ago
Read 2 more answers
The demand for a daily newspaper at a newsstand at a busy intersection is known to be normally distributed with a mean of 150 an
AURORKA [14]

Answer:

171 newspapers.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 150, \sigma = 25

How many newspapers should the newsstand operator order to ensure that he runs short on no more than 20% of days

The number of newspapers must be on the 100-20 = 80th percentile. So this value if X when Z has a pvalue of 0.8. So X when Z = 0.84.

Z = \frac{X - \mu}{\sigma}

0.84 = \frac{X - 150}{25}

X - 150 = 0.84*25

X = 171

So 171 newspapers.

4 0
4 years ago
Which fraction is between 2/3 and 6/7
MrRa [10]

Answer:

For example, 16/21

Step-by-step explanation:

Are there choices?

There is an infinite number of numbers between any two numbers.

2/3 = 14/21

6/7 = 18/21

One example of a fraction between 2/3 and 6/7 is 16/21

16/21 is between 2/3 and 6/7.

5 0
2 years ago
20 - (3 2/3 + 2 1/4 + 2 1/3) = ?
Alex777 [14]

Answer:

20-5 11/12= 9 11/12

Step-by-step explanation:

we do (3 2/3 + 2 1/4+ 2 1/3 )which is

           3 8/12.    2 3/12    2 4/12 (common denominator)

then add 3 8/12 and 2 3/12 and 2 1/3 to get 10 1/12 then we get 20- 10 1/12 to get a final answer of 9 11/12

5 0
3 years ago
7. In a state lottery, a player must choose 8 of the numbers from 1 to 40. The lottery commission then performs an experiment th
Vlad1618 [11]

Answer:a) P(8 of the players numbers are drawn)=1.3×10^-8

b) P(7 of the players number are drrawn)=3.33×10^-c) P(at least 6 of the players number were drawn)=1.84×10^-4

Step-by-step explanation:

Players has 8 combinations of numbers from 1-40. The outcome S contains all the combinations of 8 out of 40

a) P(8 of the players numbers are drawn)= 1/40/8= 1.3×10^-8

There are one in hundred million chances that the draw numbers are precisely the chosen ones.

b) Number of ways of drawing 78 selected numbers from 1-40=8×(40-7)

8×32

P(7 of the players number are drawn)=8×32/40 =3.33×10^-6.

There are approximately 300,000 chances that 7 of the players numbers are chosen

c) P(at least 6 players numbers are drawn)= 32/2×(8/6) ways to draw.

P(at least 6 players numbers are drawn)=P(all 8 chosen are drawn)+P(7 players numbers drawn)+P(6 chosen are drawn) = 1+ 8 x32/40/8 +[8\6 ×32/2]

P(at least 6 players numbers are drawn) = 1.84×10^-4.

There are approximately 5400chances that at least6 of the numbers drawn are chosen by the player.

5 0
4 years ago
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