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Murrr4er [49]
2 years ago
12

What is the value for x?

Mathematics
1 answer:
Stella [2.4K]2 years ago
5 0
Can we get some answer choices? 
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Use the function below to find F(4)
Ne4ueva [31]

Answer:

A 256/3

Step-by-step explanation:

1/3 * 4^4 =

1/3 * (4*4*4*4) =

1/3 * 256 = 256/3

4 0
1 year ago
Read 2 more answers
What fraction is equivalent to 5/9
Mama L [17]
1. Any fraction that you multiply the numerator by the same number as the denominator. 
Example:
1 x 2 = 2
-------------
2 x 2 = 4
2. The simplest form of 24/60 is 2/5 I believe.
3. 4/9
4. 23/8





Please vote me brainliest on this one too
thx
3 0
3 years ago
What fraction is equivalent to 123.5℅
iVinArrow [24]
123.5%
= 123.5/100
= 1235/1000
= (1235/5) / (1000/5)
= 247/200
= 1 47/200

The final answer is 247/200 or 1 47/200~
5 0
2 years ago
Please help this is on one of my finals <br> Find the sum <br> 8/14 + 7/14 + 2/14
mylen [45]
The answer would be 1 3/14
6 0
2 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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