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3 years ago

I need help to get to 8.9 I need help to figure out the numbers for the empty boxes.

Mathematics

1 answer:

Rainbow [258]3 years ago

7 0

Answer:

Bottom row from left to right we have 0.4 0.3 0.3 0.1 0.5 0.5 Second row from right to left 0.7 0.6 0.4 0.6 1 On the Third row from right to left 1.3 1 1 1.6 On the Fourth row 2.3 2 2.6 On the Fith row 4.3 and 4.6 The last Row also the highest 8.9

Step-by-step explanation:

you take two numbers right and left add them and put the answer in the box above and in between

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Usimov [2.4K]

Answer:

a) $f(x) = \frac{1}{25.9}$

b) 0.2046 = 20.46% probability the driving distance for one of these golfers is less than 290 yards

Step-by-step explanation:

Uniform probability distribution:

An uniform distribution has two bounds, a and b.

The probability of finding a value of at lower than x is:

$P(X < x) = \frac{x - a}{b - a}$

The probability of finding a value between c and d is:

$P(c \leq X \leq d) = \frac{d - c}{b - a}$

The probability of finding a value above x is:

$P(X > x) = \frac{b - x}{b - a}$

The probability density function of the uniform distribution is:

$f(x) = \frac{1}{b-a}$

The driving distance for the top 100 golfers on the PGA tour is between 284.7 and 310.6 yards.

This means that $a = 284.7, b = 310.6$ .

a. Give a mathematical expression for the probability density function of driving distance.

$f(x) = \frac{1}{b-a} = \frac{1}{310.6-284.7} = \frac{1}{25.9}$

b. What is the probability the driving distance for one of these golfers is less than 290 yards?

$P(X < 290) = \frac{290 - 284.7}{310.6-284.7} = 0.2046$

0.2046 = 20.46% probability the driving distance for one of these golfers is less than 290 yards

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2 years ago

The number of students in the tutoring center was recorded for 47 randomly selected times. The data is summarized in the frequen

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Using the information given, it is found that the class width for this frequency distribution table is of 1.

In this problem, these following classes are given:

0 – 1 14

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A similar problem is given at brainly.com/question/24701109

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3 years ago

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Answer:

Following are the solution to the question:

Step-by-step explanation:

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The number of accidents on a certain section of I-40 averages 4 accidents per weekday independent across weekdays. Assuming the

AysviL [449]

Answer:

1.83% probability there are no car accidents on that stretch on Monday

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

The number of accidents on a certain section of I-40 averages 4 accidents per weekday independent across weekdays.

This means that $\mu = 4$

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This is P(X = 0).

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-4}*(4)^{0}}{(0)!} = 0.0183$

1.83% probability there are no car accidents on that stretch on Monday

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3 years ago