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2 years ago

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