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3 years ago

10

You have a batting average of .250. This means that the probability of your getting a hit is 25%. If you have 520 attempts this

season, how many hits can you expect to get?

2 answers:

Ksju [112]3 years ago

6 0

Answer:

130 expected hits

Step-by-step explanation:

520 attempts times 0.25 is 130

Strike441 [17]3 years ago

3 0

Answer:

130 hits answer right I know because I got it right ...

Hopes it helps

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A 92 ft path cuts diagonally from Mateo St. to Cove St. It makes an angle of 68° with Cove St. and an angle of 34° with Mateo St

ohaa [14]

Answer:

41.802 feet

Step-by-step explanation:

Given that a 92 ft path cuts diagonally from Mateo St. to Cove St. It makes an angle of 68° with Cove St. and an angle of 34° with Mateo St.

We can visualize a triangle formed by the three streets One side is 92 feet.

Angle opposite 92 feet = $180-68-34 = 78$

Let x be length of Cove st and y length of Mateo st.

Using sine angle for triangles,

$\frac{92}{sin 78} =\frac{x}{sin 34} =\frac{y}{sin 68}$

Simplify to get

x=52.595

y=87.207

Distance saved= $x+y-92\\= 139.802-92\\=41.802$

7 0

3 years ago

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What is equivalent to 7(x - 3)?

Murljashka [212]

Simple....

What is equivalent to 7(x-3)?

Distributing....

7*x=7x

7*-3=-21

7x-21

Thus, your answer.

3 0

2 years ago

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Use the definition of Taylor series to find the Taylor series, centered at c, for the function. f(x) = sin x, c = 3π/4

anyanavicka [17]

Answer:

$\sin(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n$

Step-by-step explanation:

Given

$f(x) = \sin x\\$

$c = \frac{3\pi}{4}$

Required

Find the Taylor series

The Taylor series of a function is defines as:

$f(x) = f(c) + f'(c)(x -c) + \frac{f"(c)}{2!}(x-c)^2 + \frac{f"'(c)}{3!}(x-c)^3 + ........ + \frac{f*n(c)}{n!}(x-c)^n$

We have:

$c = \frac{3\pi}{4}$

$f(x) = \sin x\\$

$f(c) = \sin(c)$

$f(c) = \sin(\frac{3\pi}{4})$

This gives:

$f(c) = \frac{1}{\sqrt 2}$

We have:

$f(c) = \sin(\frac{3\pi}{4})$

Differentiate

$f'(c) = \cos(\frac{3\pi}{4})$

This gives:

$f'(c) = -\frac{1}{\sqrt 2}$

We have:

$f'(c) = \cos(\frac{3\pi}{4})$

Differentiate

$f"(c) = -\sin(\frac{3\pi}{4})$

This gives:

$f"(c) = -\frac{1}{\sqrt 2}$

We have:

$f"(c) = -\sin(\frac{3\pi}{4})$

Differentiate

$f"'(c) = -\cos(\frac{3\pi}{4})$

This gives:

$f"'(c) = - * -\frac{1}{\sqrt 2}$

$f"'(c) = \frac{1}{\sqrt 2}$

So, we have:

$f(c) = \frac{1}{\sqrt 2}$

$f'(c) = -\frac{1}{\sqrt 2}$

$f"(c) = -\frac{1}{\sqrt 2}$

$f"'(c) = \frac{1}{\sqrt 2}$

$f(x) = f(c) + f'(c)(x -c) + \frac{f"(c)}{2!}(x-c)^2 + \frac{f"'(c)}{3!}(x-c)^3 + ........ + \frac{f*n(c)}{n!}(x-c)^n$

becomes

$f(x) = \frac{1}{\sqrt 2} - \frac{1}{\sqrt 2}(x - \frac{3\pi}{4}) -\frac{1/\sqrt 2}{2!}(x - \frac{3\pi}{4})^2 +\frac{1/\sqrt 2}{3!}(x - \frac{3\pi}{4})^3 + ... +\frac{f^n(c)}{n!}(x - \frac{3\pi}{4})^n$

Rewrite as:

$f(x) = \frac{1}{\sqrt 2} + \frac{(-1)}{\sqrt 2}(x - \frac{3\pi}{4}) +\frac{(-1)/\sqrt 2}{2!}(x - \frac{3\pi}{4})^2 +\frac{(-1)^2/\sqrt 2}{3!}(x - \frac{3\pi}{4})^3 + ... +\frac{f^n(c)}{n!}(x - \frac{3\pi}{4})^n$

Generally, the expression becomes

$f(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n$

Hence:

$\sin(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n$

3 0

2 years ago

Ismail school is at the corner of high Street and second Avenue. The corner forms a right angle. How might he describe the way t

disa [49]

Answer:

Step-by-step explanation:

street never meet

4 0

3 years ago

472,558 to the nearest hundred thousand

Marrrta [24]

473,000 u round up to the next thousand

4 0

3 years ago

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