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MrMuchimi
3 years ago
7

Round 23.5481 to the nearest thousandth

Mathematics
1 answer:
NemiM [27]3 years ago
5 0

Answer: 23.548

Step-by-step explanation: The thousandths place is 3 places to the right of the decimal point so in this problem, the digit in the rounding place is 8.

The rules of rounding says that if the digit to the right of the rounding place is greater than or equal to 5, we round down but if the digit to the right of the rounding place is less than 5, we round down.

Since the digit to the right of the rounding place, 1, is less than 5, we round down. This means that the digit in the rounding place which is 8 stays the same and we change all digits to the right of 8 to zero.

So we have 23<em>.</em>5480.

Finally, it's important to understand that when rounding decimals, we can drop any zeroes to the right of the decimal point as long as they're also to the right of the rounding place.

This means that 23.5481 rounded to the nearest thousandth is 23<em>.</em>548.

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A survey is taken among customers of a fast-food restaurant to determine preference for hamburger or chicken. Of 200 respondents
kicyunya [14]

Answer:

                          Hamburger       Chicken

Adults                    65                        60         125

children                 55                        20          75

                             120                        80         200

a)What is the probability that a randomly selected individual is an adult?

Total no. of adults = 125

Total no. of people  200

The probability that a randomly selected individual is an adult = \frac{125}{200}=0.625

b) What is the probability that a randomly selected individual is a child and prefers chicken?

No. of child prefers chicken = 20

The probability that a randomly selected individual is a child and prefers chicken= \frac{20}{200}=0.1

c)Given the person is a child, what is the probability that this child prefers a hamburger?

No. of children prefer hamburger = 55

No. of child = 75

The probability that this child prefers a hamburger= \frac{35}{75}=0.46

d) Assume we know that a person has ordered chicken, what is the probability that this individual is an adult?

No. of adults prefer chicken = 60

No. of total people like chicken = 80

A person has ordered chicken, the probability that this individual is an adult= \frac{60}{80}=0.75

3 0
2 years ago
All real numbers between −14 and 14 inclusive
Vladimir79 [104]

Answer:

-13, -12, -11, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13

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3 years ago
Read 2 more answers
A food-packaging apparatus underfills 10% of the containers. Find the probability that for any particular 10 containers the numb
Maksim231197 [3]

Answer:

a) P(X = 1) = 0.38742

b) P(X = 3) = 0.05740

c) P(X = 9) = 0.00000

d) P(X \geq 5) = 0.00163

Step-by-step explanation:

For each container, there are only two possible outcomes. Either it is undefilled, or it is not. This means that we can solve this problem using the binomial probability distribution.

Binomial probability distribution:

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

In this problem

There are 10 containers, so n = 10.

A food-packaging apparatus underfills 10% of the containers, so p = 0.1.

a) This is P(X = 1)

P(X = 1) = C_{10,1}.(0.1)^{1}.(0.9)^{9} = 0.38742

b) This is P(X = 3)

P(X = 3) = C_{10,3}.(0.1)^{3}.(0.9)^{7} = 0.05740

c) This is P(X = 9)

P(X = 9) = C_{10,9}.(0.1)^{9}.(0.9)^{1} = 0.00000

d) This is P(X \geq 5).

Either the number is lesser than five, or it is five or larger. The sum of the probabilities of each event is decimal 1. So:

P(X < 5) + P(X \geq 5) = 1

P(X \geq 5) = 1 - P(X < 5)

In which

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{10,0}.(0.1)^{0}.(0.9)^{10} = 0.34868

P(X = 1) = C_{10,1}.(0.1)^{1}.(0.9)^{9} = 0.38742

P(X = 2) = C_{10,2}.(0.1)^{2}.(0.9)^{8} = 0.1937

P(X = 3) = C_{10,3}.(0.1)^{3}.(0.9)^{7} = 0.05740

P(X = 4) = C_{10,4}.(0.1)^{1}.(0.9)^{9} = 0.38742

So

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.34868 + 0.38742 + 0.19371 + 0.05740 + 0.01116 = 0.99837

Finally

P(X \geq 5) = 1 - P(X < 5) = 1 - 0.99837 = 0.00163

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