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Sliva [168]
3 years ago
5

Which of the quotient of 6 divided by 1/3​

Mathematics
1 answer:
dmitriy555 [2]3 years ago
5 0

Answer:

18

Step-by-step explanation:

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Toby thinks of a number less than 15. He divides it by 2 and then he adds 4. He then divides it by 3. His awnser is 3. 5
satela [25.4K]

Answer:

The number he thought of was 13

Step-by-step explanation:

Revers the problem

3.5 * 3 = 10.5

10.5 - 4 = 6.5

6.5 * 2 = 13

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2 years ago
Please help with this.
victus00 [196]

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neither, the slopes are the same

Step-by-step explanation:

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A function represented by the equation y=1/3x + 4 is a graph on a coordinate plane. What will happen to the line if the slope is
Stella [2.4K]

Answer:

Graph of the new function will be vertically stretched by 2 units.

Step-by-step explanation:

A linear function is represented by the equation, y = \frac{1}{3}x+ 4

This function has the slope = \frac{1}{3} and y-intercept = 4

If this function is stretched vertically by 'k' the the equation of the new function will be,

y = a(\frac{1}{3})x+ 4

y = \frac{a}{3}x+4 ---------(1)

If the slope of this equation is changed to \frac{2}{3},

New equation of the function will be,

y = \frac{2}{3}x+4 ---------(2)

Comparing equations (1) and (2),

a = 2

Therefore, graph of the new function will be stretched vertically by 2 units.

4 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
4 years ago
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