Which decimals are between 10% and 20%? 0.11 0.09 0.13 0 0.19 0.011

Solve the proportion. 2/3/4 = 10 LO x = [?]

Setler79 [48]

Answer:

x = 6

Step-by-step explanation:

We want to find x,

$\frac{3}{5} = \frac{x}{10}$

to get x alone, we multiple both side by 10

$\frac{3}{5} * 10 = \frac{x}{10} *10$

$\frac{30}{5} = x$

30 divided by 5 is 6

Check:

We can see that 5 x ____ = 10 where ____ is 2

5 x 2 = 10

so 3 x 2 = 6

Learn more about Equivalent Fractions here: brainly.com/question/1612

8 0

2 years ago

How did the military contribute to the growth and longevity of the Roman Empire? Military leaders formed a balance of power with

Elenna [48]

Soldiers were organized into highly efficient legions

4 0

3 years ago

Read 2 more answers

Tavon has a $50 gift card that’s very loses $2 for each 30-day period it is not used. He has a $40 card that’s loses $1.50 for e

velikii [3]

What’s the question

3 0

3 years ago

The computers of six faculty members in a certain department are to be re- placed. two of the faculty members have selected lapt

FinnZ [79.3K]

Answer:

A) The probability that both selected setups are for laptop computers is 0.067

B)The probability that both selected setups are desktop machines is 0.4

C)The probability that at least one selected setup is for a desktop computer is 0.933

D)The probability that at least on computer of each type is chosen for setup is 0.533

Step-by-step explanation:

Number of laptops = 2

Number of desktops = 4

Total number of outcomes = 15

a) what is the probability that both selected setups are for laptop computers?

Total number of outcomes = 15

So, the probability that both selected setups are for laptop computers = $\frac{1}{15}=0.067$

b)what is the probability that both selected setups are desktop machines?

Number of desktops = 4

Number of desktops to be chosen = 4

We will use combination

No. of ways to select two desktops = $^4C_2=\frac{4!}{2!(4-2)!}=6$

So,the probability that both selected setups are desktop machines= $\frac{6}{15}=0.4$

(c) what is the probability that at least one selected setup is for a desktop computer?

P(at least 1 desktop)=1-P(No desktop)

P(at least 1 desktop)=1-P(both laptops)

P(at least 1 desktop)=1-0.067=0.933

So,the probability that at least one selected setup is for a desktop computer is 0.933

d) what is the probability that at least on computer of each type is chosen for setup?

No. of ways to select one desktop $=^4C_1=\frac{4!}{1!(4-1)!}=4$

No. of ways to select one laptop = $^2C_1=\frac{2!}{1!(2-1)!}=2$

So, No. of ways to select one laptop and one desktop= $4 \times 2 = 8$

So,the probability that at least on computer of each type is chosen for setup= $\frac{8}{15}=0.533$

6 0

3 years ago

Simplify 12ab/4a-8ab

iris [78.8K]

Answer:

12ab / 4a - 8ab

= 4a(3b)/ 4a(1 - 2b)

= 3b / 1 - 2b

Hope this helps

3 0

3 years ago