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aivan3 [116]
2 years ago
12

Find the value of X 4.5 8.9 6.5 X 13 b 11 can someone answer

Mathematics
1 answer:
Lorico [155]2 years ago
5 0
11, add six and a half to four and a half.
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If you have 49 questions on a test what score must you get to pass the test?​
arsen [322]

Well, on every test I have at a school and I believe most schools do this. You need at leasr 80% to pass. But to pass you'd most likely need 80% or 70%. But if you get an 80% you'll be sure to pass! I hope this helps you!

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
The system of equations y = 2x + 5 and y = –3x – 15 is shown on the graph below.According to the graph, what is the solution to
gtnhenbr [62]

Answer:

(-4, -3)

Step-by-step explanation:

The graph of both lines will be shown below.

To find the solution, we need to find the point on the graph that both lines intersect on.

As we can see, both lines intersect on the point (-4, -3). Thus, that is the solution.

Best of Luck!

6 0
3 years ago
Read 2 more answers
Can someone please fill in blank ASAP
Annette [7]

Answer:

The quotient is 48.

Step-by-step explanation:

Estimate the quotient using compatible numbers:

45 (or I guess any number close the answer above, not sure tbh)

Multiply the estimate by 21:

945

45*21

Is the estimate to high or too low?

too low

Adjust and continue until the product is 1008.

1008/21=48

Have a good day/evening! I hope my answer is correct!

8 0
1 year ago
Students in a zoology class took a final exam. They took equivalent forms of the exam at monthly intervals thereafter. After t​
juin [17]

Answer:

a. 73; b. 48.9; c. 2; d. 33.8; e. 73

Step-by-step explanation:

Assume the function was

S(t)= 73 - 15 ln(t + 1), t  ≥ 0

a. Average score at t = 0

S(0) = 73 - 15 ln(0 + 1) = 73 - 15 ln(1) = 73 - 15(0) =73 - 0 = 73

b. Average score at t = 4

S(4) = 73 - 15 ln(4 + 1) = 73 - 15 ln(5) = 73 - 15(1.61) =73 - 24.14 = 48.9

c. Average score at t =24

S(24) = 73 - 15 ln(24 + 1) = 73 - 15 ln(25) = 73 - 15(3.22) =73 - 48.28 = 24.7

d. Percent of answers retained

At t = 0. the students retained 73 % of the answers.

At t = 24, they retained 24.7 % of the answers.

\text{Percent retention} = \dfrac{\text{24.7}}{\text{73}} \times 100 \, \% = \text{33.8 \%}\\\\\text{The students retained $\large \boxed{\mathbf{33.8 \, \%}}$ of their original knowledge after two years.}

e. Maximum of the function

The maximum of the function is at t= 0.

Max = 73 %

The graph below shows your knowledge decay curve. Knowledge decays rapidly at first but slows as time goes on.

 

6 0
3 years ago
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