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2 years ago

The lunch at Dion's Place has choice of ham, turkey, or roast beef on rye or white bread white juice, milk, or tea.

Mathematics

1 answer:

spayn [35]2 years ago

6 0

Total combinations are found by multiplying the quantities of each part by each other.

3 kinds of meat x 2 types of bread x 3 types of drinks

A) total combinations = 3 x 2 x 3 = 18

B) there are 18 different combinations, ham in rye with tea is 1 combination.

The probability would be 1/18

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3 years ago

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kodGreya [7K]

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Multiple Choice What is the sum of five-sixths + start fraction 2 over 3 end fraction? A. one and one-third B. One and one half

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Answer:

Step-by-step explanation:

$\frac{5}{6} +\frac{2}{3} \\\\=\frac{5}{6}+\frac{2*2}{2*3} \\\\=\frac{5}{6}+\frac{4}{6} \\\\=\frac{5+4}{6} \\\\=\frac{9}{6} \\\\=\frac{3*3}{3*2} \\\\=\frac{3}{2}$

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3 years ago

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3 years ago

Read 2 more answers

According to records in a large hospital, the birth weights of newborns have a symmetric and bell-shaped relative frequency dist

Kamila [148]

Answer:

15.9% of babies are born with birth weight under 6.3 pounds.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 6.8 pounds

Standard Deviation, σ = 0.5

We are given that the distribution of birth weights is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(birth weight under 6.3 pounds)

P(x < 6.3)

$P( x < 6.3) = P( z < \displaystyle\frac{6.3 - 6.8}{0.5}) = P(z < -1)$

Calculation the value from standard normal z table, we have,

$P(x < -1) = 0.159 = 15.9\%$

15.9% of babies are born with birth weight under 6.3 pounds.

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3 years ago