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konstantin123 [22]
1 year ago
13

Thrifty supply store normally sells candy bars for .59cents

Mathematics
1 answer:
Assoli18 [71]1 year ago
7 0

Answer:

well that is a pretty good deal i might have to go there

Step-by-step explanation:

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10X3 tens in standard form:
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The average price for a gallon of gasoline in the United States is and in Russia it is . Assume these averages are the populatio
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This question is incomplete

Complete Question

The average price for a gallon of gasoline in the United States is $3.73 and in Russia it is $3.40 (Bloomberg Businessweek, March 5–March 11, 2012). Assume these averages are the population means in the two countries and that the probability distributions are normally distributed with a standard deviation of $.25 in the United States and a standard deviation of $.20 in Russia.

a. What is the probability that a randomly selected gas station in the United States charges less than $3.50 per gallon?(to 4 decimals)

Answer:

0.1788

Step-by-step explanation:

We solve this question using z score formula.

z = (x-μ)/σ,

where x is the raw score

μ is the population mean

σ is the population standard deviation.

a. What is the probability that a randomly selected gas station in the United States charges less than $3.50 per gallon?(to 4 decimals)

For the United States

x is the raw score = $3.50

μ is the population mean = Average price for a gallon of gasoline in the United States is $3.73

σ is the population standard deviation = standard deviation of $.25 in the United States = $0.25

z score = $3.50 - $3.73/$0.25

=-0.92

Determining the Probability value from Z-Table:

P(x ≤ 3.50) = P(x < 3.5) = P(z = -0.92)

= 0.17879

Approximately to 4 decimal places = 0.1788

Therefore, the probability that a randomly selected gas station in the United States charges less than $3.50 per gallon is 0.1788

5 0
1 year ago
destiny sells pencil cases for $5 each and mechanical pencils for $2 each at the school supply store. she sells p pencil cases a
Andrew [12]

Answer:

Step-by-step explanation:

5p + 2(p + 4) .......because pencil cases (p) sell for 5 bucks a piece....and mechanical pencils (p + 4), sell for 2 bucks a piece. She is basically selling 4 more mechanical pencils then she is pencil cases.

8 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
1 year ago
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