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3 years ago

Hi. I need some help on 15 and 16! Thank you!

Mathematics

1 answer:

Alenkinab [10]3 years ago

7 0

Answer:

MN = 34, x = 7

Step-by-step explanation:

look at the file, cut off numbers are:

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An owl begins at rest on a tree branch searching for prey. Once the owl locates the prey it flies from its perch and four second

aleksklad [387]

Answer: 1.4 m/s

Explanation: the acceleration equation is a = change in velocity/change in time.

Plug the values in and you get 5.6/4

Divide and get 1.4 m/s

6 0

4 years ago

The driving distance for the top 100 golfers on the PGA tour is between 284.7 and 310.6 yards (according to GolfWeek). Assume th

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

3 0

3 years ago

List all the elements of the sample space for the following experiment: You spin a spinner with four equal sections labeled 1, 2

telo118 [61]

Answer:

Option B.

Step-by-step explanation:

When we spin a spinner the sample space for a spinner having four equal sections labeled 1, 2, 3, 4 will be {1, 2, 3, 4} and for tossing the dime is {H, T}.

Now if we toss a dime and spinning the spin is done simultaneously then sample space will have the elements {(1, H), (2, H), (3, H), (4, H), (1, T), (2, T), (3,T), (4, T)}

Therefore Option B. is the correct option.

4 0

4 years ago

What is 1 and 2/5× 7

OleMash [197]

7/5 x 7/1 = 49/5 or 9 4/5

3 0

3 years ago

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What is the volume of the cylinder shown below? A. 12157 units B. 3617 units C. 647 units D. 152 T units

Slav-nsk [51]

Answer: