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gregori [183]
3 years ago
6

Which term, when added to the given polynomial, will change the end behavior of the graph?

Mathematics
2 answers:
olga_2 [115]3 years ago
4 0

Answer:

<u>A) -6x9</u> will change the end behavior of the graph.

<u>OAmalOHopeO</u>

xxTIMURxx [149]3 years ago
3 0

a. –6x9 when added to the given polynomial, will change the end behavior of the graph.

hope it helps :)

please mark brainliest!

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VikaD [51]

Answer:

Thanks I guess

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
2,10,18,26 find the 61st term
saw5 [17]

Answer:

482

Step-by-step explanation:

We can see that the numbers shown resemble an arithmetic sequence because they have a common difference. The formula for the nth term of an arithmetic sequence is:

a_{n} =a_{1} +(n-1)d

Where a_{1} is the first term, a_{n} is the nth term, and d is the common difference. To find the 61st term, all we need is the first term and the common difference. By looking at what given, we can say the first term is 2. Now, to find the common difference, we find the difference of a term from the term before it. In this case we can do 10-2, which is 8 , or the common difference. Since we have everything we need, it can be plugged into the equation:

a_{61} =2 + (61 - 1)*8\\a_{61} = 2 + 60*8\\a_{61} = 2 + 480\\a_{61} = 482

So, the 61st term is 482.

4 0
3 years ago
I have 100 items of product in stock. The probability mass function for the product's demand D is P(D=90)=P(D=100)=P(D=110)=1/3.
masya89 [10]

Answer:

The probability mass function for the items sold is

P_X(k) = \left \{ {\frac{1}{3} \, \, \, {k=90} \atop \, \frac{2}{3} \, \, \, {k=100}} \right.

The mean is 96.667

The variance is 22.222

b) The probability mass function for the unfilled demand due to lack of stock is

P_Y(k) = \left \{ {\frac{2}{3} \, \, \, {k=0} \atop \, \frac{1}{3} \, \, \, {k=10}} \right.

The mean is 3.333

The variance is 33.333

Step-by-step explanation:

If the demand is higher than 100, then you will sell 100 items only. Thus, there is a probability of 1/3+1/3 = 2/3 that you will sell 100 items, while there is a probability of 1/3 that you will sell 90.

The probability mass function for the items sold is

P_X(k) = \left \{ {\frac{1}{3} \, \, \, {k=90} \atop \, \frac{2}{3} \, \, \, {k=100}} \right.

The mean is 1/3 * 90 + 2/3 * 100 = 290/3 = 96.667

The variance is V(X) = E(X²)-E(X)² = (1/3*90² + 2/3*100²) - (290/3)² = 200/9 = 22.222

b) If order to be unfilled demand, you need to have a demand of 110, which happens with probability 1/3. In that case, the value of the variable, lets call it Y, that counts the amount of unfilled demand due to lack of stock is 110-100 = 10. In any other case, the value of Y is 0, which would happen with probability 1-1/3 = 2/3. Thus

P_Y(k) = \left \{ {\frac{2}{3} \, \, \, {k=0} \atop \, \frac{1}{3} \, \, \, {k=10}} \right.

The mean is 2/3 * 0 + 1/3 * 10 = 10/3 = 3.333

The variance is 2/3*0² + 1/3*10² = 100/3 = 33.333

4 0
3 years ago
Do anyone know how to solve this? need the answer fast.
Nikitich [7]
The value of g^-1(-3) is the value of x that makes g(x) = -3. To find it, we can solve
.. (x +4)/(2x -5) = -3
.. x +4 = -3(2x -5)
.. 7x = 11
.. x = 11/7

The desired value is
.. (3/11)*g^-1(-3)
.. = (3/11)*(11/7)
.. = 3/7

\frac{3}{11} g^{-1}(-3) = \frac{3}{7}
6 0
3 years ago
2. f(x)=4x+1 and ​g(x)=x2−4x−5​, find ​(g◦​f)(4​).
Goryan [66]

Answer:

(g · f)(4) = 45

Step-by-step explanation:

f(x)=4x+1

g(x)=x² - 4x- 5

(g · f)(x) = 4(x² - 5) + 1

(g · f)(4) = 4(4² - 5) + 1

Following pemdas

(g · f)(4) = 4(16 - 5) + 1

(g · f)(4) = 4(11) + 1

(g · f)(4) = 44 + 1

(g · f)(4) = 45

8 0
2 years ago
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