Answer:
V = 672 cm³ (Option 3)
Step-by-step explanation:
Volume of Rectangular Prism =
<u><em>Where L = 14 cm, W = 6 cm and H = 8 cm</em></u>
V = 14 * 6 * 8
V = 672 cm³
In if 5/7 of students are wearing jeans today and there are about 27 students in the class how many students are wearing the jea
1 answer:
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Answer:
The discontinuity is x = 4
There no holes
The equation of the vertical asymptote is x = 4
The x intercept is 2
The equation of the horizontal asymptote is y = 1
Step-by-step explanation:
* Lets explain the problem
∵
- To find the point of discontinuity put the denominator = 0 and find
the value of x
∵ The denominator is x - 4
∵ x - 4 = 0 ⇒ add 4 to both sides
∴ x = 4
* The discontinuity is x = 4
- A hole occurs when a number is both a zero of the numerator
and denominator
∵ The numerator is x - 2
∵ x - 2 = 0 ⇒ add 2 to both sides
∴ x = 2
∵ The denominator is x - 4
∵ x - 4 = 0 ⇒ add 4 to both sides
∴ x = 4
∵ There is no common number makes the numerator and denominator
equal to 0
∴ There no holes
- Vertical asymptotes are vertical lines which correspond to the zeroes
of the denominator of the function
∵ The zero of the denominator is x = 4
∴ The equation of the vertical asymptote is x = 4
- x- intercept is the values of x which make f(x) = 0, means the
intersection points between the graph and the x-axis
∵ f(x) = 0
∴ ⇒ by using cross multiplication
∴ x - 2 = 0 ⇒ add 2 to both sides
∴ x = 2
* The x intercept is 2
- If the highest power of the numerator = the highest power of the
denominator, then the equation of the horizontal asymptote is
y = The leading coeff. of numerator/leading coeff. of denominator
∵ The numerator is x - 2
∵ The denominator is x - 4
∵ The leading coefficient of the numerator is 1
∵ The leading coefficient of the denominator is 1
∴ y = 1/1 = 1
* The equation of the horizontal asymptote is y = 1
Since the dice are fair and the rolling are independent, each single outcome has probability 1/15. Every time we choose
We have and
, because the dice are fair.
Now we use the assumption of independence to claim that
Now, we simply have to count in how many ways we can obtain every possible outcome for the sum. Consider the attached table: we can see that we can obtain:
- 2 in a unique way (1+1)
- 3 in two possible ways (1+2, 2+1)
- 4 in three possible ways
- 5 in three possible ways
- 6 in three possible ways
- 7 in two possible ways
- 8 in a unique way
This implies that the probabilities of the outcomes of are the number of possible ways divided by 15: we can obtain 2 and 8 with probability 1/15, 3 and 7 with probability 2/15, and 4, 5 and 6 with probabilities 3/15=1/5
