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2 years ago

11

Yasmien wants to give her friends 3/4 of a banana cake in her birthday party .She was able to bake 12 banana cake.How many peopl

e can share the banana cake?

1 answer:

Katen [24]2 years ago

8 0

Answer:

She can share it with 9 people

Step-by-step explanation:

75% of 12 =9

50% of 12= 6

25% of 12= 3

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On Monday 172 students went on a trip to the zoo all 4 buses were filled and 9 students had to travel in cars.How many students

il63 [147K]

Answer:

Three was 43 students in each bus

Step-by-step explanation:

4 0

3 years ago

Gina and Stewart are surf-fishing on the Atlantic coast, where both bluefish and pompano are

viktelen [127]

Using the normal distribution, it is found that due to the greater z-score, Stewart caught the longer fish relative to the same specie.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

To find who caught the longer fish relative to the same specie, we need to find out who had the higher z-score.

Stewart caught a bluefish with a length of 322 mm, hence the parameters are as follows:

$X = 322, \mu = 267, \sigma = 39$

The z-score is:

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

$Z = \frac{322 - 267}{39}$

Z = 1.41

Gina caught a pompano with a length of 176 mm, hence the parameters are as follows:

$X = 176, \mu = 155, \sigma = 28$

The z-score is:

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

$Z = \frac{176 - 155}{28}$

Z = 0.75.

Hence Stewart caught the longer fish relative to the same specie.

More can be learned about the normal distribution at brainly.com/question/4079902

#SPJ1

8 0

1 year ago

Suppose that 50% of all young adults prefer McDonald's to Burger King when asked to state a preference. A group of 12 young adul

ddd [48]

Answer:

a) 0.194 = 19.4% probability that more than 7 preferred McDonald's

b) 0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred McDonald's

c) 0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred Burger King

Step-by-step explanation:

For each young adult, there are only two possible outcomes. Either they prefer McDonalds, or they prefer burger king. The probability of an adult prefering McDonalds is independent from other adults. So we use the binomial probability distribution to solve this question.

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 combinations 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.

50% of all young adults prefer McDonald's to Burger King when asked to state a preference.

This means that $p = 0.5$

12 young adults were randomly selected

This means that $n = 12$

(a) What is the probability that more than 7 preferred McDonald's?

$P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)$

In which

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

$P(X = 8) = C_{12,8}.(0.5)^{8}.(0.5)^{4} = 0.121$

$P(X = 9) = C_{12,9}.(0.5)^{9}.(0.5)^{3} = 0.054$

$P(X = 10) = C_{12,10}.(0.5)^{10}.(0.5)^{2} = 0.016$

$P(X = 11) = C_{12,11}.(0.5)^{11}.(0.5)^{1} = 0.003$

$P(X = 12) = C_{12,12}.(0.5)^{12}.(0.5)^{0} = 0.000$

$P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.121 + 0.054 + 0.016 + 0.003 + 0.000 = 0.194$

0.194 = 19.4% probability that more than 7 preferred McDonald's

(b) What is the probability that between 3 and 7 (inclusive) preferred McDonald's?

$P(3 \leq X \leq 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)$

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

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.054$

$P(X = 4) = C_{12,4}.(0.5)^{4}.(0.5)^{8} = 0.121$

$P(X = 5) = C_{12,5}.(0.5)^{5}.(0.5)^{7} = 0.193$

$P(X = 6) = C_{12,6}.(0.5)^{6}.(0.5)^{6} = 0.226$

$P(X = 7) = C_{12,7}.(0.5)^{7}.(0.5)^{5} = 0.193$

$P(3 \leq X \leq 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 0.054 + 0.121 + 0.193 + 0.226 + 0.193 = 0.787$

0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred McDonald's

(c) What is the probability that between 3 and 7 (inclusive) preferred Burger King?

Since $p = 1-p = 0.5$ , this is the same as b) above.

So

0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred Burger King

7 0

2 years ago

Create a data set with 10 numbers that has a mean of 32?

densk [106]

30,77,55,71,19,40,10,5,7,6

4 0

3 years ago

= 3x +15 <br> = 4x-10<br><br> What is the value of x?

Viefleur [7K]

Answer:

3x+15= 4x-10

3x-4x= -10-15

-x= -25(minus will be cut on both sides)

so x= 25

7 0

3 years ago