If the graph of f(x) = 4x is shifted down, then what would be the equation of the new graph?

Suppose that, when taking a random sample of three students' GPAs, you get a sample mean of 3.90. This sample mean is far high

vesna_86 [32]

Answer:

Your sample most likely is biased, but simply getting a high number does not 100 percent mean your answer was biased. It could be a pure coincidence that you picked students with high GPAs. However, it is likely that your answer is biased. You may have sampled kids that are in advanced classes. You may have sampled students that were in some sort of honors club. You may have sampled students that were all hanging out in the library, a place that most people come to study and prepare for classes. So again, while your higher than normal answer could have been a coincidence, you most likely had a biased sample.

Step-by-step explanation:

6 0

3 years ago

. The fourth grade class is donating 100 clothing items to charity. 63 hundredths of the items are sweaters and 0.20 of the item

inessss [21]

Answer: 17/100

Step-by-step explanation:

63/100 is sweaters and 20/100 is jeans and when you add them together you get 83/100. 100/100-83/100=17/100

6 0

2 years ago

38F hotter or 2C (Tempture)

viva [34]

38* F is hotter. 2*C is only 35.6*

4 0

3 years ago

Read 2 more answers

Find three consecutive odd integers such that

Anuta_ua [19.1K]

Answer:

17, 19, 21

Step-by-step explanation:

x = first number

x+2 = second number

x+4 = third number

4x + 3(x+2) + 2(x+4) = 167

simplify:

4x + 3x + 6 + 2x + 8 = 167

combine like terms:

9x + 14 = 167

subtract 14 from each side of the equation:

9x = 153

divide both sides by 9:

x = 17

7 0

3 years ago

Suppose a certain computer virus can enter a system through an email or through a webpage. There is a 40% chance of receiving th

DedPeter [7]

Answer:

P = 0.42

Step-by-step explanation:

This probability problem can be solved by building a Venn like diagram for each probability.

I say that we have two sets:

-Set A, that is the probability of receiving this virus through the email.

-Set B, that is the probability of receiving it through the webpage.

The most important information in these kind of problems is the intersection. That is, that he virus enters the system simultaneously by both email and webpage with a probability of 0.17. It means that $A \cap B$ = 0.17.

By email only

The problem states that there is a 40 chance of receiving it through the email. It means that we have the following equation:

$A + (A \cap B) = 0.40$

$A + 0.17 = 0.40$

$A = 0.23$

where A is the probability that the system receives the virus just through the email.

The problem states that there is a 40% chance of receiving it through the email. 23% just through email and 17% by both the email and the webpage.

By webpage only

There is a 35% chance of receiving it through the webpage. With this information, we have the following equation:

$B + (A \cap B) = 0.35$

$B + 0.17 = 0.35$

$B = 0.18$

where B is the probability that the system receives the virus just through the webpage.

The problem states that there is a 35% chance of receiving it through the webpage. 18% just through the webpage and 17% by both the email and the webpage.

What is the probability that the virus does not enter the system at all?

So, we have the following probabilities.

- The virus does not enter the system: P

- The virus enters the system just by email: 23% = 0.23

- The virus enters the system just by webpage: 18% = 0.18

- The virus enters the system both by email and by the webpage: 17% = 0.17.

The sum of the probabilities is 100% = 1. So:

P + 0.23 + 0.18 + 0.17 = 1

P = 1 - 0.58

P = 0.42

There is a probability of 42% that the virus does not enter the system at all.

5 0

3 years ago