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kicyunya [14]
4 years ago
6

Given a polynomial function f(x), describe the effects on the y-intercept, regions where the graph is increasing and decreasing,

and the end behavior when the following changes are made. Make sure to account for even and odd functions.
A.) When f(x) becomes f(x) + 2
B.) When f(x) becomes - 1/2 • f(x)
Mathematics
1 answer:
Jobisdone [24]4 years ago
7 0
A.  When f(x) becomes g(x) = f(x) + 2, the graph of f(x) is shifted upward 2 units.

B  The graph of f(x) is reflected in the x-axis and then vertically compressed by a factor of 1/2.
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A gambler playing blackjack in a casino using four decks noticed that 20% of the cards that had been dealt were jacks, queens, k
Ksenya-84 [330]

Answer:

x = 162 cards need to be dealt with

Step-by-step explanation:

Given:

- Estimation of likelyhood of next card to be queen, a king, or an ace:

                                  ( 64 - 0.2*x ) / ( 208 - x )

Find:

For what values of x is this likelihood greater than 70%?

Solution:

- Use the likely-hood estimation relation and set up the inequality as follows:

                                 ( 64 - 0.2*x ) / ( 208 - x ) > 0.7

                                 ( 64 - 0.2*x ) > 0.7*( 208 - x )

                                   64 - 0.2*x > 145.6 - 0.7*x

                                          0.5*x > 81.6

                                                 x > 163.2

- It would require 162 cards to be dealt with for the probability of next card being queen, a king, or an ace is greater than 70%

6 0
3 years ago
Please help!!! i’ll give brainliest.
S_A_V [24]

Answer:

quadrilateral ABCD is not congruent to quadrilateral KLMN. quadrilateral ABCD cannot be mapped onto quadrilateral KLMN through a series of rotations, reflections or translations.

7 0
3 years ago
The weight of the chocolate and Hershey Kisses are normally distributed with a mean of 4.5338 G and a standard deviation of 0.10
Salsk061 [2.6K]

For the bell-shaped graph of the normal distribution of weights of Hershey kisses, the area under the curve is 1, the value of the median and mode both is 4.5338 G and the value of variance is 0.0108.

In the given question,

The weight of the chocolate and Hershey Kisses are normally distributed with a mean of 4.5338 G and a standard deviation of 0.1039 G.

We have to find the answer of many question we solve the question one by one.

From the question;

Mean(μ) = 4.5338 G

Standard Deviation(σ) = 0.1039 G

(a) We have to find for the bell-shaped graph of the normal distribution of weights of Hershey kisses what is the area under the curve.

As we know that when the mean is 0 and a standard deviation is 1 then it is known as normal distribution.

So area under the bell shaped curve will be

\int\limits^{\infty}_{-\infty} {f(x)} \, dx= 1

This shows that that the total area of under the curve.

(b) We have to find the median.

In the normal distribution mean, median both are same. So the value of median equal to the value of mean.

As we know that the value of mean is 4.5338 G.

So the value of median is also 4.5338 G.

(c) We have to find the mode.

In the normal distribution mean, mode both are same. So the value of mode equal to the value of mean.

As we know that the value of mean is 4.5338 G.

So the value of mode is also 4.5338 G.

(d) we have to find the value of variance.

The value of variance is equal to the square of standard deviation.

So Variance = (0.1039)^2

Variance = 0.0108

Hence, the value of variance is 0.0108.

To learn more about normally distribution link is here

brainly.com/question/15103234

#SPJ1

3 0
1 year ago
Can someone help me with this? Thanks a bunch! ;)
dexar [7]

Answer:

idk, can I?

Step-by-step explanation:

5 0
3 years ago
What is the solution of a system definded by y=-x+5 and 5x+2y=14?
Digiron [165]
The answer to the question

3 0
3 years ago
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