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Ksju [112]
2 years ago
5

A store gets a bucket of eggs for $16.20 and sells a dozen eggs for $3.50. If there are 450 eggs in the bucket, by what percent

has the store increased the price per dozen eggs?
Please help every other answer when I searched up was wrong.
Mathematics
1 answer:
Murljashka [212]2 years ago
3 0

Answer:

710%

Step-by-step explanation:

Find each unit price.

16.20 $/bucket × (1 bucket / 450 eggs) = 0.036 $/egg

3.50 $/dozen × (1 dozen / 12 eggs) = 0.292 $/egg

The percent markup is:

(0.292 − 0.036) / 0.036 × 100% ≈ 710%

The store marks up the price by approximately 710%.

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Two functions are defined as shown. f(x) =-1/2 x – 2 g(x) = –1 Which graph shows the input value for which f(x) = g(x)?
Natalka [10]

Answer:

See attachment.

Step-by-step explanation:

The given functions are:

f(x) =  -  \frac{1}{2}x - 2

and g(x)=-1

To find the x-value of the point of intersection of the two functions, we equate the two functions and solve for x.

-  \frac{1}{2}x - 2 = -1

x + 4 = 2

x = 2 - 4 =  - 2

The graph that shows the input value for which f(x)=g(x) is the graph which shows the point of intersection of f(x) and g(x) to be at x=-2.

4 0
2 years ago
Read 2 more answers
Paula has a dog that weighs 3 times as much as Carla's dog
ivanzaharov [21]

Answer:

The weight of Paula's dog is 36 pounds.

Step-by-step explanation:

SO we would draw one box for the weight of Carla's dog. Then we would draw 3 boxes for the weight of Paula's dog.

To find how much does Paula's dog weigh we divide the total weight of the dogs by the total number of boxes.

So then you would calculate 48 divided by 4.

The weight of Carla's dog is 12 pounds.

Therefore the weight of Paula's dog is 3x12=3x10+3x2=30+6=36 pounds.

4 0
2 years ago
Researchers have a sample size of 24 , and they are using the Student's t-distribution. What are the degrees of freedom and how
DerKrebs [107]

Answer:

For this case assuming that the random variable is X

df = n-1

And replacing n = 24 we got:

df = 24-1=23

And we notate the distribution we got: X \sim t_{n-1}= t_{23}

Step-by-step explanation:

Previous concepts

The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".

The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.  

The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."

Solution to the problem

For this case assuming that the random variable is X

df = n-1

And replacing n = 24 we got:

df = 24-1=23

And we notate the distribution we got: X \sim t_{n-1}= t_{23}

7 0
2 years ago
A really bad carton of eggs contains spoiled eggs. An unsuspecting chef picks eggs at random for his ""Mega-Omelet Surprise."" F
Dima020 [189]

Answer:

(a) The probability that of the 5 eggs selected exactly 5 are unspoiled is 0.0531.

(b) The probability that of the 5 eggs selected 2 or less are unspoiled is 0.3959.

(c) The probability that of the 5 eggs selected more than 1 are unspoiled is 0.8747.

Step-by-step explanation:

The complete question is:

A really bad carton of 18 eggs contains 8 spoiled eggs. An unsuspecting chef picks 5 eggs at random for his “Mega-Omelet Surprise.” Find the probability that the number of unspoiled eggs among the 5 selected is

(a) exactly 5

(b) 2 or fewer

(c) more than 1.

Let <em>X</em> = number of unspoiled eggs in the bad carton of eggs.

Of the 18 eggs in the bad carton of eggs, 8 were spoiled eggs.

The probability of selecting an unspoiled egg is:

P(X)=p=\frac{10}{18}=0.556

A randomly selected egg is unspoiled or not is independent of the others.

It is provided that a chef picks 5 eggs at random.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 5 and <em>p</em> = 0.556.

The success is defined as the selection of an unspoiled egg.

The probability mass function of <em>X</em> is given by:

P(X=x)={5\choose x}(0.556)^{x}(1-0.556)^{5-x};\ x=0,1,2,3...

(a)

Compute the probability that of the 5 eggs selected exactly 5 are unspoiled as follows:

P(X=5)={5\choose 5}(0.556)^{5}(1-0.556)^{5-5}\\=1\times 0.05313\times 1\\=0.0531

Thus, the probability that of the 5 eggs selected exactly 5 are unspoiled is 0.0531.

(b)

Compute the probability that of the 5 eggs selected 2 or less are unspoiled as follows:

P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)

              =\sum\imits^{2}_{x=0}{{5\choose 5}(0.556)^{5}(1-0.556)^{5-5}}\\=0.0173+0.1080+0.2706\\=0.3959

Thus, the probability that of the 5 eggs selected 2 or less are unspoiled is 0.3959.

(c)

Compute the probability that of the 5 eggs selected more than 1 are unspoiled as follows:

P (X > 1) = 1 - P (X ≤ 1)

              = 1 - P (X = 0) - P (X = 1)

              =1-\sum\limits^{1}_{x=0}{{5\choose 5}(0.556)^{5}(1-0.556)^{5-5}}\\=1-0.0173-0.1080\\=0.8747

Thus, the probability that of the 5 eggs selected more than 1 are unspoiled is 0.8747.

6 0
2 years ago
What is the quotient of -8a^8b^-2/ 10a^-4b^-10 in simplified form?
postnew [5]

Here are a few rules with exponents that you need to know:

  1. Dividing exponents of the same base: \frac{x^m}{x^n}=x^{m-n}

For this, divide:

\frac{-8a^8b^{-2}}{10a^{-4}b^{-10}}=\frac{-8}{10}a^{8-(-4)}b^{-2-(-10)}=-\frac{4}{5}a^{12}b^8

<u>Your final answer is -\frac{4}{5}a^{12}b^8</u>

7 0
2 years ago
Read 2 more answers
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