If the parent function f(x) = |x| is transformed to g(x) = |x| + 4, what transformation occurs from f(x) to g(x)?

Conner and Jana are multiplying (3^5 6^8)(3^9 6^10)

LekaFEV [45]

Answer:

None of them are correct. They only thought of combining like terms, which isn't the case.

Step-by-step explanation:

(3^5 6^8)(3^9 6^10) =

(3^5+9) + (3^5 6^10) + (6^8 3^9) + (6^8+10) =

3^14 + (3^5 6^10) + (6^8 3^9) + 6^18

8 0

3 years ago

Please answer quickly!!!!! Will give the brainiest There is a pic attached.

Alexeev081 [22]

Answer:

A. 2/3

Step-by-step explanation:

There are 12 tiles in total, and 8 tiles with numbers that are greater than 4.

Express this as a fraction:

8/12

Simplify:

2/3

Hope this helps

5 0

3 years ago

Solved and graph each inequality X+7>11

Nina [5.8K]

Answer:

$x>4$

Step-by-step explanation:

Given inequality:

$x+7>11$

We need to solve the given inequality for solutions of $x$ .

In order to solve for the inequality, we will isolate $x$ in one side of the in-equation.

We have,

$x+7>11$

Subtracting both sides by 7.

$x+7-7>11-7$

$x>4$

Thus, the solution of the given inequality is all real numbers greater than 4.

The solution in interval form can be written as (4,∞).

The graph for the inequality is shown below.

6 0

3 years ago

M&M plain candies come in various colors. According to the M&M/Mars Department of Consumer Affairs, the distribution of

lyudmila [28]

Answer:

A) 0.12. Yes. Choosing a green and blue M&M is possible

B) 0.43. Yes. Choosing a yellow and red M&M is possible

C) 0.78

Step-by-step explanation:

First of all, the summation of the distribution of all colours is;

Σ(all colors ) = 22% + 20% + 23% + 10% + 6% + 6% + 13% = 100%, or 1.

Thus;

a) P(green candy or blue candy) is;

P(GREEN ∪ BLUE) = P(G) + P(BL)

P(GREEN ∪ BLUE) = 6%+6%

P(GREEN ∪ BLUE) = 12% or 0.12

Now, due to the fact that we have to choose ONE candy and only ONE candy at random, then they are mutually exclusive: Yes. Choosing a green and blue M&M is possible

b)P(yellow candy or red candy is;

P(YELLOW ∪ RED) = P(Y) + P(R)

P(YELLOW ∪ RED) = 20% + 23% = 43% or 0.43

Yes. Choosing a yellow and red M&M is possible

c) P(NOT PURPLE)

the probability of having a purple is;

P(PURPLE) = 22% or 0.22

So, the Probability of NOT having a PURPLE is 1 - 0.22 = 0.78

4 0

3 years ago

What is the probability of getting zero heads in six tosses? What is the probability of getting exactly two heads in six tosses?

alexira [117]

Answer:

The probability of getting zero heads in six tosses is 0.015625.

The probability of getting exactly two heads in six tosses is 0.234375.

Step-by-step explanation:

Using the Minitab software for computing the desired probabilities.

We take n=6 and p=1/2 because we have six tosses and in binomial distribution the probability of success p remains constant in each trial whereas the probability of success in this case is getting heads.

When a coin is tossed then there are two possible outcomes head or tail.

So,

p= P(heads)=1/2=0.5

The probability of getting zero heads in six tosses is computed by considering the following steps:

In Minitab

Calc >Probability Distributions > Binomial

Select n=6 and p=0.5 and select input constant=0 and bubble the probability and by clicking OK we get the following output:

Probability Density Function

Binomial with n = 6 and p = 0.5

x P( X = x )

0 0.015625

So, the probability of getting zero heads in six tosses is 0.015625.

The probability of getting exactly two heads in six tosses is computed by considering the following steps :

In Minitab

Calc >Probability Distributions > Binomial

Select n=6 and p=0.5 and select input constant=2 and bubble the probability and by clicking OK we get the following output:

Probability Density Function

Binomial with n = 6 and p = 0.5

x P( X = x )

2 0.234375

So, the probability of getting exactly two heads in six tosses is 0.234375.

7 0

4 years ago