Jasmine ran 5 miles in 42.5 minutes. Caroline ran 3 miles in 26.4 minutes. Who ran at a faster pace?

You read online that the probability of being dealt four‑of‑a‑kind in a five‑card poker hand is 1 / 4165. Explain carefully what

Lapatulllka [165]

Answer:

There is about 4,164/4,165 chances of not getting getting a four of a kind. So, it is extremely unlikely or even borderline impossible in that situation to get a four of a kind.

<u>But in the long run, it can be increased only if you keep drawing. So, the awnser would have to be. D </u>

Step-by-step explanation:

A. It does mean that if you are dealt 4165 five‑card poker hands, one will be four‑of‑a‑kind.

B. It does not mean that all will be four‑of‑a‑kind. The probability is actually saying that only on the 4165 the poker hand will you get a four‑of‑a‑kind, not just on any of the 4165 poker hands.

C. The probability is actually saying that in the long run, with a large number of five‑card poker hands, the fraction in which you will be dealt a four‑of‑a‑kind is 1 / 4165.

D. The chance you will be dealt four‑of‑a‑kind is 1 / 4165 only on the first hand. This chance will then increase with each new hand you are dealt until you eventually win

6 0

2 years ago

Find the domain of the function y = 3 tan(23x)

solmaris [256]

Answer:

$\mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right) \; \right| k \in \mathbb{Z} \right\rbrace$ .

In other words, the $x$ in $f(x) = 3\, \tan(23\, x)$ could be any real number as long as $x \ne \displaystyle \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right)$ for all integer $k$ (including negative integers.)

Step-by-step explanation:

The tangent function $y = \tan(x)$ has a real value for real inputs $x$ as long as the input $x \ne \displaystyle k\, \pi + \frac{\pi}{2}$ for all integer $k$ .

Hence, the domain of the original tangent function is $\mathbb{R} \backslash \displaystyle \left\lbrace \left. \left(k\, \pi + \frac{\pi}{2}\right) \; \right| k \in \mathbb{Z} \right\rbrace$ .

On the other hand, in the function $f(x) = 3\, \tan(23\, x)$ , the input to the tangent function is replaced with $(23\, x)$ .

The transformed tangent function $\tan(23\, x)$ would have a real value as long as its input $(23\, x)$ ensures that $23\, x\ne \displaystyle k\, \pi + \frac{\pi}{2}$ for all integer $k$ .

In other words, $\tan(23\, x)$ would have a real value as long as $x\ne \displaystyle \frac{1}{23} \, \left(k\, \pi + \frac{\pi}{2}\right)$ .

Accordingly, the domain of $f(x) = 3\, \tan(23\, x)$ would be $\mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right) \; \right| k \in \mathbb{Z} \right\rbrace$ .

4 0

2 years ago

Determine whether the function below is an even function, an odd function, both, or neither.

kari74 [83]

All of the powers of x here are even, so the function f(x) is even. Note that 19 = 19x^0 = 19(1) = 19

6 0

3 years ago

Find the exact values of the remaining trigonometric functions of θ satisfying the given conditions. (if an answer is undefined,

natita [175]

The answer is undefined.

8 0

3 years ago

Prove Euler's identity using Euler's formula.<br> e^ix = cos x + i sin x

Korolek [52]

First list all the terms out.

e^ix = 1 + ix/1! + (ix)^2/2! + (ix)^3/3! ...

Then, we can expand them.

e^ix = 1 + ix/1! + i^2x^2/2! + i^3x^3/3!...

Then, we can use the rules of raising i to a power.

e^ix = 1 + ix - x^2/2! - ix^3/3!...

Then, we can sort all the real and imaginary terms.

e^ix = (1 - x^2/2!...) + i(x - x^3/3!...)

We can simplify this.

e^ix = cos x + i sin x

This is Euler's Formula.

What happens if we put in pi?

x = pi

e^i*pi = cos(pi) + i sin(pi)

cos(pi) = -1

i sin(pi) = 0

e^i*pi = -1 OR e^i*pi + 1 = 0

That is Euler's identity.

3 0

3 years ago