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Lena [83]
4 years ago
8

In the diagram, the areas of and are in a ratio of 3 : 4. What are the coordinates of point C?

Mathematics
2 answers:
Naddika [18.5K]4 years ago
8 0
Sorry but cant answer without a grid or something
ehidna [41]4 years ago
8 0
4,-6 i think hope this helped
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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
What is the answer to this question because i do not know 5/12 x =1
Otrada [13]

Answer:

x=12/5

Step-by-step explanation:

x*5/12=1

x=1*12/5

x=12/5

3 0
2 years ago
Read 2 more answers
True or false ? If a parallelogram is inscribed in a circle , it must be a square
TEA [102]

The answer is TRUE. The parallelogram must be a rectangle

5 0
3 years ago
Read 2 more answers
The overhead reach distances of adult females are normally distributed with a mean of 205 cm and a standard deviation of 7.8 cm.
devlian [24]

Answer:

Given the mean = 205 cm and standard deviation as 7.8cm

a. To calculate the probability that an individual distance is greater than 218.4 cm, we subtract the probability of the distance given (i.e 218.4 cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) from 1. Therefore, we have 1- P(Z\leq 1.72). Using the Z distribution table we have 1-0.9573. Therefore P(X >218.4)= 0.0427.

b. To calculate the probability that mean of 15 (i.e n=15) randomly selected distances is greater than 202.8, we subtract the probability of the distance given (i.e 202.8cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) divided by the square root of mean (i.e n= 15)  from 1. Therefore, we have 1- P(Z\leq -1.09). Using the Z distribution table we have 1-0.1378. Therefore P(X >202.8)= 0.8622.

c. This will also apply to a normally distributed data even if it is not up to the sample size of 30 since the sample distribution is not a skewed one.

Step-by-step explanation:

Given the mean = 205 cm and standard deviation as 7.8cm

a. To calculate the probability that an individual distance is greater than 218.4 cm, we subtract the probability of the distance given (i.e 218.4 cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) from 1. Therefore, we have 1- P(Z\leq 1.72). Using the Z distribution table we have 1-0.9573. Therefore P(X >218.4)= 0.0427.

b. To calculate the probability that mean of 15 (i.e n=15) randomly selected distances is greater than 202.8, we subtract the probability of the distance given (i.e 202.8cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) divided by the square root of mean (i.e n= 15)  from 1. Therefore, we have 1- P(Z\leq -1.09). Using the Z distribution table we have 1-0.1378. Therefore P(X >202.8)= 0.8622.

c. This will also apply to a normally distributed data even if it is not up to the sample size of 30 since the sample distribution is not a skewed one.

4 0
3 years ago
2(2x - 8) = x - 46 ?
Anna [14]

Answer: X = -10

Step-by-step explanation:

Times two to the parentheses

Minus X from both sides, left with 3X-16=-46

Add 16 to both sides 3X=-30

Divide both sides by 3

Answer X=-10

4 0
3 years ago
Read 2 more answers
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