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o-na [289]
3 years ago
7

What are expressions always equivalent to 4(6+2x)

Mathematics
1 answer:
timurjin [86]3 years ago
4 0
24+12x is the answer
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seropon [69]
No. because he just isnt
5 0
3 years ago
What is x for Unknown angle problems with algebra?
yulyashka [42]

The value of x=20

Step-by-step explanation:

We need to find the value of x

We are given a right angle which is equal to 90°

So, all three angles must sum to 90°

Our equation will be:

x+2x+(x+10)=90

Solving and finding value of x

x+2x+x+10=90\\4x=90-10\\4x=80\\x=80/4\\x=20

So, the value of x=20

Keywords: Angles

Learn more about angles at:

  • brainly.com/question/3202836
  • brainly.com/question/7151553

#learnwithBrainly

5 0
3 years ago
Jill drives 75 1/3 miles every hour how many miles can Jill Drive in 8 and 3/4 hours
11Alexandr11 [23.1K]
First, multiply 75 by 8 to get 600. 
Then multiply 1/3 by 8 to get 8/3 or 2 2/3. 
Add those together to get 602 2/3. 

Next multiply 1/3 by 3/4 by multiplying the numerator by 3 and the denominator by 4. That is 3/12. Also known as 1/4. 
Multiply 75 by 3/4 by dividing it by 4 then multiplying it by 3. That is 56 1/4. 
Add 56 1/4 and 1/4 together to get 56 1/2. 

Lastly, add together 56 1/2 and 602 2/3.
Convert the fractions to sixths.
56 3/6 + 602 4/6 = 659 1/6 miles
Hope this helps! 
6 0
3 years ago
Plz help asap almost done ​
Makovka662 [10]

Answer:

5(-2.1)-4(-8.9)= 25.1

Step-by-step explanation:

lol

7 0
3 years ago
A box contains 20 light box of which five or defective it for lightbulbs or pick from the box randomly what's the probability th
Snowcat [4.5K]

Answer:

1

Step-by-step explanation:

Given:-

- The box has n = 20 light-bulbs

- The number of defective bulbs, d = 5

Find:-

what's the probability that at most two of them are defective

Solution:-

- We will pick 2 bulbs randomly from the box. We need to find the probability that at-most 2 bulbs are defective.

- We will define random variable X : The number of defective bulbs picked.

Such that,               P ( X ≤ 2 ) is required!

- We are to make a choice " selection " of no defective light bulb is picked from the 2 bulbs pulled out of the box.

- The number of ways we choose 2 bulbs such that none of them is defective, out of 20 available choose the one that are not defective i.e n = 20 - 5 = 15 and from these pick r = 2:

        X = 0 ,       Number of choices = 15 C r = 15C2 = 105 ways

- The probability of selecting 2 non-defective bulbs:

      P ( X = 0 ) = number of choices with no defective / Total choices

                       = 105 / 20C2 = 105 / 190

                       = 0.5526

- The number of ways we choose 2 bulbs such that one of them is defective, out of 20 available choose the one that are not defective i.e n = 20 - 5 = 15 and from these pick r = 1 and out of defective n = 5 choose r = 1 defective bulb:

        X = 1 ,       Number of choices = 15 C 1 * 5 C 1 = 15*5 = 75 ways

- The probability of selecting 1 defective bulbs:

      P ( X = 1 ) = number of choices with 1 defective / Total choices

                       = 75 / 20C2 = 75 / 190

                       = 0.3947

- The number of ways we choose 2 bulbs such that both of them are defective, out of 5 available defective bulbs choose r = 2 defective.

        X = 2 ,       Number of choices = 5 C 2 = 10 ways

- The probability of selecting 2 defective bulbs:

      P ( X = 2 ) = number of choices with 2 defective / Total choices

                       = 10 / 20C2 = 10 / 190

                       = 0.05263

- Hence,

    P ( X ≤ 2 ) = P ( X =0 ) + P ( X = 1 ) + P (X =2)

                     = 0.5526 + 0.3947 + 0.05263

                     = 1

7 0
3 years ago
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