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2 years ago

Two numbers are randomly selected on a number line numbered from 1 to 9. Match each scenario to its probability.

Mathematics

2 answers:

Talja [164]2 years ago

5 0

Answer:

1/9; 4/9; 1/12; 1/6

Step-by-step explanation:

the probability that both numbers are greater than 6 if the same number can be chosen twice--> 3/9 * 3/9 = 1/9

the probability that both numbers are less than 7 if the same number can be chosen twice --> 6/9 * 6/9 = 4/9

the probability that both numbers are odd numbers less than 6 if the same numbers cannot be chosen twice --> 3/9 * 2/8 = 1/12

the probability that both numbers are even numbers if the same numbers cannot be chosen twice --> 4/9 * 3/8 = 1/6

maxonik [38]2 years ago

4 0

Answer:

1/9; 4/9; 1/12; 1/6

Step-by-step explanation:

greater than 6 not chosen twice = 1/12

greater than 6 can be twice = 1/9

Less than 7 and can be chosen twice = 4/9

both are even but cannot be chosen twice =1/6

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2 years ago

A consensus forecast is the average of a large number of individual analysts' forecasts. Suppose the individual forecasts for a

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Answer:

a) 0.954

b) 0.937

c) 0.891

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 6 percent

Standard Deviation, σ = 1.3 percent

We are given that the distribution of particular interest rate is a bell shaped distribution that is a normal distribution.

Formula:

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

a) P(At least 3.8 percent.)

$P(x \geq 3.8)$

$P( x \geq 3.8) = P( z \geq \displaystyle\frac{3.8 - 6}{1.3}) = P(z > -1.69)$

$= 1 - P(z < -1.69)$

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

$P(x \geq 3.8) = 1 - 0.046 = 0.954 = 95.4\%$

b) P(At most 8 percent)

$P(x \leq 8) = P(z \leq \displaystyle\frac{8-6}{1.3}) = P(z \leq 1.53)$

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$P( x \leq 8) =0.937= 93.7\%$

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$P(3.8 \leq x \leq 8) = 89.1\%$

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2 years ago