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kozerog [31]
3 years ago
9

A bakery sold 112 cupcakes in one day. The head baker predicted he would sell 92 cupcakes that day. What was the percent error o

f the baker's prediction?
Mathematics
1 answer:
lesya [120]3 years ago
8 0
22% I took 20 the number he was off and divided by the predicted number which is .217
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The chorus teacher plans to arrange the students in equal rows. Only the girls or boys will be in each row. What is the greatest
emmainna [20.7K]

Answer:

2

Step-by-step explanation:

68+48=116

116÷16=7.24 (wrong)

116÷12=9.7 (wrong)

116÷4=29 (lowest even)

116÷2=58 (highest even)

If I'm wrong, let me know

8 0
2 years ago
Can someone help me
serg [7]
Your answer is D. right
4 0
3 years ago
A representative from the National Football League's Marketing Division randomly selects people on a random street in Kansas Cit
Orlov [11]

Using the binomial distribution, we have that:

a) 0.1024 = 10.24% probability that the marketing representative must select 4 people to find one who attended the last home football game.

b) 0.2621 = 26.21% probability that the marketing representative must select more than 6 people to find one who attended the last home football game.

c) The expected number of people is 4, with a variance of 20.

For each person, there are only two possible outcomes. Either they attended a game, or they did not. The probability of a person attending a game is independent of any other person, which means that the binomial distribution is used.

Binomial probability distribution  

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

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

The parameters are:

  • p is the probability of a success on a single trial.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

The expected number of <u>trials before q successes</u> is given by:

E = \frac{q(1-p)}{p}

The variance is:

V = \frac{q(1-p)}{p^2}

In this problem, 0.2 probability of a finding a person who attended the last football game, thus p = 0.2.

Item a:

  • None of the first three attended, which is P(X = 0) when n = 3.
  • Fourth attended, with 0.2 probability.

Thus:

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

P(X = 0) = C_{3,0}.(0.2)^{0}.(0.8)^{3} = 0.512

0.2(0.512) = 0.1024

0.1024 = 10.24% probability that the marketing representative must select 4 people to find one who attended the last home football game.

Item b:

This is the probability that none of the first six went, which is P(X = 0) when n = 6.

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

P(X = 0) = C_{6,0}.(0.2)^{0}.(0.8)^{6} = 0.2621

0.2621 = 26.21% probability that the marketing representative must select more than 6 people to find one who attended the last home football game.

Item c:

  • One person, thus q = 1.

The expected value is:

E = \frac{q(1-p)}{p} = \frac{0.8}{0.2} = 4

The variance is:

V = \frac{0.8}{0.04} = 20

The expected number of people is 4, with a variance of 20.

A similar problem is given at brainly.com/question/24756209

3 0
1 year ago
A survey of 125 households was done to investigate whether households have laptops, tablets, or
riadik2000 [5.3K]

Answer:

3) 0.326

Step-by-step explanation:

Total Number of surveyed households =125

We want to determine the probability that a randomly selected household would have a laptop given that it does not  have a tablet.

Written in Probability notation: P(Laptop|No Tablet)

P(Laptop|No Tablet)=\dfrac{P(L\cap T')}{P(T')} \\=\dfrac{31/125}{95/125}\\=\dfrac{31}{95}\\\\=0.326

The correct option is C,

4 0
3 years ago
Which polygon is a pentagon? A 4-sided figure. A 6-sided figure. A 5-sided figure. A 4-sided figure.
kaheart [24]

Answer:

A Pentagon is a 5 sided figure

Step-by-step explanation:

I hope this helps

3 0
2 years ago
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