What is 11inches redistributed equally among the 12 months

QUESTION 1 The driving distance for the top 100 golfers on the PGA tour is between 284.7 and 310.6 yards.Assume that the driving

GarryVolchara [31]

Answer:

$P(X>300)$

And we can find this probability with the complement rule:

$P(X>300)= 1-P(X$

Step-by-step explanation:

For this case we define the random variable X ="driving distance for the top 100 golfers on the PGA tour" and we know that:

$X \sim Unif (a=284.7, b=310.6)$

And for this case the probability density function is given by:

$f(x) =\frac{1}{310.6 -284.7} =0.0386 , 284.7 \leq X \leq 310.6$

And the cumulative distribution function is given by:

$F(x) =\frac{x-284.7}{310.6-284.7} , 284.7 \leq X \leq 310.6$

And we want to find this probability:

$P(X>300)$

And we can find this probability with the complement rule:

$P(X>300)= 1-P(X$

6 0

3 years ago

Can someone help me out? (Show work on how to solve)

belka [17]

Answer:

dghggjj

Step-by-step explanation:

s gm knfc JC. jz. jz. .xx nkcmxd I nt☺️️

8 0

3 years ago

Read 2 more answers

According to a survey done at your school, about 42% of all the female students participate in 2 sport seasons. You randomly ask

Leni [432]

Answer:

Hi Christine

Step-by-step explanation:

Ms. Paik would be mad, wouldn't she?

5 0

3 years ago

Can anyone answer this word problem for me?

mel-nik [20]

What’s the word problem

8 0

3 years ago

Calculus piecewise function.

Kipish [7]

Part A

The notation $\lim_{x \to 2^{+}}f(x)$ means that we're approaching x = 2 from the right hand side (aka positive side). This is known as a right hand limit.

So we could start at say x = 2.5 and get closer to 2 by getting to x = 2.4 then to x = 2.3 then 2.2, 2.1, 2.01, 2.001, etc

We don't actually arrive at x = 2 itself. We simply move closer and closer.

Since we're on the positive or right hand side of 2, this means we go with the rule involving x > 2

Therefore f(x) = (x/2) + 1

Plug in x = 2 to find that...

f(x) = (x/2) + 1

f(2) = (2/2) + 1

f(2) = 2

This shows $\lim_{x \to 2^{+}}f(x) = 2$

Then for the left hand limit $\lim_{x \to 2^{-}}f(x)$ , we'll involve x < 2 and we go for the first piece. So,

f(x) = 3-x

f(2) = 3-2

f(2) = 1

Therefore, $\lim_{x \to 2^{-}}f(x) = 1$

===============================================================

Part B

Because $\lim_{x \to 2^{+}}f(x) \ne \lim_{x \to 2^{-}}f(x)$ this means that the limit $\lim_{x \to 2}f(x)$ does not exist.

If you are a visual learner, check out the graph below of the piecewise function. Notice the gap or disconnect at x = 2. This can be thought of as two roads that are disconnected. There's no way for a car to go from one road to the other. Because of this disconnect, the limit doesn't exist at x = 2.

===============================================================

Part C

You'll follow the same type of steps shown in part A.

However, keep in mind that x = 4 is above x = 2, so we'll deal with x > 2 only.

So you'd only involve the second piece f(x) = (x/2) + 1

You should find that f(4) = 3, and that both left and right hand limits equal this value. The left and right hand limits approach the same y value. The limit does exist here. There are no gaps to worry about when x = 4.

===============================================================

Part D

As mentioned earlier, since $\lim_{x \to 4^{+}}f(x) = \lim_{x \to 4^{-}}f(x) = 3$ , this means the limit $\lim_{x \to 4}f(x)$ does exist and it's equal to 3.

As x gets closer and closer to 4, the y values are approaching 3. This applies to both directions.

4 0

1 year ago