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dimulka [17.4K]
3 years ago
15

Enter an equation of the form y= mx for the line that passes through the

Mathematics
1 answer:
hammer [34]3 years ago
5 0

Answer:

y = 13x

Step-by-step explanation:

gradient = 13

x = 0

y = 0

\frac{Δy}{Δx}

so  \frac{y - 0}{x - 0} = 13

cross multiply and you get

y = 13x

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Which situation CANNOT be represented by this equation? 3x+4=19
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For a standard normal random variable, what z-score has (a) probability 0.175 to the right?
Mumz [18]
Answer: 0.935

Explanation: 

Let S = z-score that has a probability of 0.175 to the right. 

In terms of normal distribution, the expression "probability to the right" means the probability of having a z-score of more than a particular z-score, which is Z in our definition of variable Z. In terms of equation:

P(z ≥ S) = 0.175   (1)

Equation (1) is solvable using a normal distribution calculator (like the online calculator in this link: http://stattrek.com/online-calculator/normal.aspx). However, the calculator of this type most likely provides the value of P(z ≤ Z), the probability to the left of S.

Nevertheless, we can use the following equation:

P(z ≤ S) + P(z ≥ S) = 1 
⇔ P(z ≤ S) = 1 - P(z ≥ S)  (2)

Now using equations (1) and (2):

P(z ≤ S) = 1 - P(z ≥ S)
P(z ≤ S) = 1 - 0.175 
P(z ≤ S) = 0.825

Using a normal distribution calculator (like in this link: http://stattrek.com/online-calculator/normal.aspx),

P(z ≤ S) = 0.825
⇔ S = 0.935 

Hence, the z-score of 0.935 has a probability 0.175 to the right.
7 0
3 years ago
Find the probability of this event. Enter the answer as a fraction in simplest form, as a decimal, and as a percent.
olasank [31]

Answer:

12/60, 0.2, 20%

hope i was helpful :)

5 0
3 years ago
The computers of six faculty members in a certain department are to be replaced. Two of the faculty members have selected laptop
Anettt [7]

Answer:

a. \frac{1}{15}

b. \frac{2}{5}

c. \frac{14}{15}

d. \frac{8}{15}

Step-by-step explanation:

Given that there are two laptop machines and four desktop machines.

On a day, 2 computers to be set up.

To find:

a. probability that both selected setups are for laptop computers?

b. probability that both selected setups are desktop machines?

c. probability that at least one selected setup is for a desktop computer?

d. probability that at least one computer of each type is chosen for setup?

Solution:

Formula for probability of an event E can be observed as:

P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}

a. Favorable cases for Both the laptops to be selected = _2C_2 = 1

Total number of cases = 15

Required probability is \frac{1}{15}.

b. Favorable cases for both the desktop machines selected = _4C_2=6

Total number of cases = 15

Required probability is \frac{6}{15} = \frac{2}{5}.

c. At least one desktop:

Two cases:

1. 1 desktop and 1 laptop:

Favorable cases = _2C_1\times _4C_1 = 8

2. Both desktop:

Favorable cases = _4C_2=6

Total number of favorable cases = 8 + 6 = 14

Required probability is \frac{14}{15}.

d. 1 desktop and 1 laptop:

Favorable cases = _2C_1\times _4C_1 = 8

Total number of cases = 15

Required probability is \frac{8}{15}.

8 0
3 years ago
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