Ask question

Ask question

All categories

3 years ago

10

A MLB pitcher throws a strike on his first pitch to a batter about 60% of the time. if during the next season this pitcher faces

300 batters, how many of those batters would we expect there to be a strike on the first pitch?

1 answer:

Brilliant_brown [7]3 years ago

6 0

This question in not on here please try again

You might be interested in

A bottle holds 5 pints of lemonade. How much is this in fluid ounces?

Art [367]

I think the answer is 80 fluid ounces

3 0

4 years ago

Read 2 more answers

Which element has has a stronger ability to attract an electron from another atom within a bond.

WITCHER [35]

Answer:

its sulfur because they have higher electronegativity such as n, o, f, (flourine)

3 0

3 years ago

What’s the value in 8 of 0.837?

Romashka-Z-Leto [24]

Tenths, my dude. Hope I was a help!!

5 0

3 years ago

Read 2 more answers

1/2b-7=5.5 slove for b

melamori03 [73]

Answer:

b = 25

Step-by-step explanation:

Given

$\frac{1}{2}$ b - 7 = 5.5 ( multiply through by 2 to clear the fraction )

b - 14 = 11 ( add 14 to both sides )

b = 25

6 0

3 years ago

Read 2 more answers

Digiron [165]

What I gather from the question is that $X$ has second moment $E(X^2)=81$ and variance $V(X) = 58$ , and you're asked to find the expectation and variance of the random variable $Y=2X+10$ .

From the given second moment and variance, we find the expectation of $X$ :

$V(X) = E(X^2) - E(X)^2 \implies E(X) = \sqrt{E(X^2) - V(X)} = \sqrt{23}$

Expectation is linear, so

$E(Y) = E(2X+10) = 2 E(X) + 10 = \boxed{2\sqrt{23} + 10}$

Using the same variance identity, we have

$V(Y) = V(2X+10) = E((2X+10)^2) - E(2X+10)^2$

and

$E((2X+10)^2) = E(4X^2 + 40X + 100) = 4E(X^2) + 40E(X) + 100 = 424 + 40\sqrt{23}$

so that

$V(Y) = V(2X+10) = (424 + 40\sqrt{23}) - (2\sqrt{23} + 10)^2 = \boxed{232}$

Alternatively, we can use the identity

$V(aX+b) = a^2 V(X) \implies V(2X+10) = 4V(X) = 232$

5 0

3 years ago