Answer:
a) 0.00031
b) 0.0017
c) 0.31
d) 0.00018
Step-by-step explanation:
attached below is the detailed solution
Total number of 7-poker cards are 52P7 = 133784560
A) Determine the probabilities of Seven-card straight
probability of seven-card straight = 0.00031
B) Determine the probability of four cards of one rank and three of a different rank
P( four cards of one rank and three of different rank ) = 0.0017
C) Determine probability of three cards of one rank and two cards of each two different ranks
P( three cards one rank and two cards of two different ranks ) = 0.31
D) Determine probability of two cards of each of three different ranks and a card of a fourth rank
P ( two cards of each of three different ranks and a card of fourth rank ) = 0.00018
Answer: 5 = y
Step-by-step explanation:
Answer:
Hope I helped! Brainleist!
Step-by-step explanation:
you get $12/week
and you have $
your equation is
y=12x
now...
y = 384
so
384=12x
x= 32
Answer:
five times a number plus two
Step-by-step explanation:
Answer:

Domain: All Real Numbers
General Formulas and Concepts:
<u>Algebra I</u>
- Domain is the set of x-values that can be inputted into function f(x)
<u>Calculus</u>
The derivative of a constant is equal to 0
Basic Power Rule:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Chain Rule: ![\frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%5Bf%28g%28x%29%29%5D%20%3Df%27%28g%28x%29%29%20%5Ccdot%20g%27%28x%29)
Derivative: ![\frac{d}{dx} [ln(u)] = \frac{u'}{u}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%20%5Bln%28u%29%5D%20%3D%20%5Cfrac%7Bu%27%7D%7Bu%7D)
Step-by-step explanation:
<u>Step 1: Define</u>
f(x) = ln(2x² + 1)
<u>Step 2: Differentiate</u>
- Derivative ln(u) [Chain Rule/Basic Power]:

- Simplify:

- Multiply:

<u>Step 3: Domain</u>
We know that we would have issues in the denominator when we have a rational expression. However, we can see that the denominator would never equal 0.
Therefore, our domain would be all real numbers.
We can also graph the differential function to analyze the domain.