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2 years ago

5

Thomas bought two and one half dozen eggs costing 45.00. He sold each eggs for 2.00. Did he make a profit or a loss and by how m

uch?

1 answer:

Reil [10]2 years ago

5 0

Answer:

Step-by-step explanation:

2.5doz(12/doz)=30 eggs. So revenue minus cost when selling them for $2 each is

2(30)-45=15

Since 15 is positive he made a profit of $15 dollars.

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Orange M&M’s: The M&M’s web site says that 20% of milk chocolate M&M’s are orange. Let’s assume this is true and set

SOVA2 [1]

Answer:

The correct option is (A).

Step-by-step explanation:

Let <em>X</em> = number of orange milk chocolate M&M’s.

The proportion of orange milk chocolate M&M’s is, <em>p</em> = 0.20.

The number of candies in a small bag of milk chocolate M&M’s is, <em>n</em> = 55.

The event of an milk chocolate M&M being orange is independent of the other candies.

The random variable <em>X</em> follows a Binomial distribution with parameter <em>n</em> = 55 and <em>p</em> = 0.20.

The expected value of a Binomial random variable is:

$E(X)=np$

Compute the expected number of orange milk chocolate M&M’s in a bag of 55 candies as follows:

$E(X)=np$

$=55\times 0.20\\=11$

It is provided that in a randomly selected bag of milk chocolate M&M's there were 14 orange ones, i.e. the proportion of orange milk chocolate M&M's in a random bag was 25.5%.

This proportion is not surprising.

This is because the average number of orange milk chocolate M&M’s in a bag of 55 candies is expected to be 11. So, if a bag has 14 orange milk chocolate M&M’s it is not unusual at all.

All unusual events have a very low probability, i.e. less than 0.05.

Compute the probability of P (X ≥ 14) as follows:

$P(X\geq 14)=\sum\limits^{55}_{x=14}{{55\choose x}0.20^{x}(1-0.20)^{55-x}}$

$=0.1968$

The probability of having 14 or more orange candies in a bag of milk chocolate M&M’s is 0.1968.

This probability is quite larger than 0.05.

Thus, the correct option is (A).

4 0

3 years ago

3.14 x 6.25 step by step

katen-ka-za [31]

Answer:

Step-by-step explanation:

3.14x6.25

19.625

8 0

2 years ago

sing traditional methods, it takes 108108 hours to receive a basic driving license. A new license training method using Computer

solniwko [45]

Answer:

$z_{stat} = -1.50$

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 108

Sample mean, $\bar{x}$ = 107

Sample size, n = 110

Alpha, α = 0.02

Population standard deviation, σ = 7

Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$z_{stat} = \displaystyle\frac{107 - 108}{\frac{7}{\sqrt{110}} } = -1.50$

The test statistic is -1.50

8 0

2 years ago

A transformer increases the input voltage from 12V to 12,000V. What is the ratio of the number of wire turns on the primary coil

erma4kov [3.2K]

Answer:

$\frac{n_p}{n_s}=\frac{1}{1000}$

Step-by-step explanation:

Given voltage of primary coil is $V_p=12V$

And voltage of secondary coil is $V_s=12,000V$

A transformer is a device that contains primary and secondary coil. These coils are mounted on soft iron. The change in voltage depends upon the number of turns of primary coil $(n_p)$ and secondary coil $(n_s)$ .

Following equation represents the relation between them.

$\frac{V_s}{V_p} =\frac{n_s}{n_p}\\\\\frac{12,000}{12}=\frac{n_s}{n_p}\\\\1000=\frac{n_s}{n_p}$

Then, ratio of the number of wire turns on the primary coil to the number of turns on the secondary coil is

$\frac{n_p}{n_s}=\frac{1}{1000}$

5 0

3 years ago

Alice has 1/5 as many miniature cars as Sylvester has Sylvester has 35 miniature cars how many miniature cars does alic have?

const2013 [10]

The answer is 7. Since Alice has 1/5 as many cars as Sylvester. You multiply 1/5 by how many cars Sylvester has. Sylvester has 35 cars, so you would do 35 times 1/5 or 35 divided by 5 (multiply by 1/5 and divided by 5 are the same thing) and you would get your answer of 7.

6 0

2 years ago