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Lelechka [254]
3 years ago
9

Multiply.

Mathematics
1 answer:
wolverine [178]3 years ago
7 0
So you're initial equation is (2/6)*24=m. Well since 2/6=1/3, plug in 24 for the 1 in 1/3. So 24/3=m, and 24/3=8 so m=8.
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Marcus daims that in the expression 3x + 3y, the terms 3x and 3y are like terms because both are variable terms and both have th
kumpel [21]

Answer:

Step-by-step explanation:

3x and 3y are not "like terms," despite having the same coefficient.

Like terms have the same variable. For example, 3x and 2x are like terms, so

3x+2x = 5x

7 0
3 years ago
Calculate the perimeter of the figure below.
mote1985 [20]

Well, we know perimeter is just adding all the lengths together.

And we already know 2 of the lengths, so we know the rest. 11 1/2 is the same as the opposite side, and 6 1/3 is the same as the opposite side.

So the sides are 6 1/3, 6 1/3, 11 1/2, 11 1/2.

Add them up:

6 1/3 + 6 1/3 + 11 1/2 + 11 1/2.

Improper fractions:

19/3 + 19/3 + 23/2 + 23/2.

Common denominator, (6):

38/6 + 38/6 + 69/6 + 69/6 = 214/6.

Simplify:

214/6 =  35 2/3

7 0
3 years ago
Give an example of an equation with an infinite number of solutions. Then make one change to the equation so that it has no solu
mrs_skeptik [129]
Two equations with infinite solutions would look the exact same. Example:
y=mx+b
y=mx+b
Example 2
y=2x+5
y=2x+5
For an equation with no solution they would have the same slope but different y intercepts. An equation with same slope and same y intercepts would have infinite solutions. 
5 0
3 years ago
The Resource Conservation and Recovery Act mandates the tracking and disposal of hazardous waste produced at U.S. facilities. Pr
aniked [119]

Answer:

(a) E(X) =  0.383

The expected number in the sample that treats hazardous waste on-site is 0.383.

(b) P(x = 4) = 0.000169

There is a 0.000169 probability that 4 of the 10 selected facilities treat hazardous waste on-site.

Step-by-step explanation:

Professional Geographer (Feb. 2000) reported the hazardous waste generation and disposal characteristics of 209 facilities.

N = 209

Only eight of these facilities treated hazardous waste on-site.

r = 8

a. In a random sample of 10 of the 209 facilities, what is the expected number in the sample that treats hazardous waste on-site?

n = 10

The expected number in the sample that treats hazardous waste on-site is given by

$ E(X) =  \frac{n \times r}{N} $

$ E(X) =  \frac{10 \times 8}{209} $

$ E(X) =  \frac{80}{209} $

E(X) =  0.383

Therefore, the expected number in the sample that treats hazardous waste on-site is 0.383.

b. Find the probability that 4 of the 10 selected facilities treat hazardous waste on-site

The probability is given by

$ P(x = 4) =  \frac{\binom{r}{x} \binom{N - r}{n - x}}{\binom{N}{n}} $

For the given case, we have

N = 209

n = 10

r = 8

x = 4

$ P(x = 4) =  \frac{\binom{8}{4} \binom{209 - 8}{10 - 4}}{\binom{209}{10}} $

$ P(x = 4) =  \frac{\binom{8}{4} \binom{201}{6}}{\binom{209}{10}} $

$ P(x = 4) =  \frac{70 \times 84944276340}{35216131179263320}

P(x = 4) = 0.000169

Therefore, there is a 0.000169 probability that 4 of the 10 selected facilities treat hazardous waste on-site.

8 0
3 years ago
What is x?? I'm so confused!
LiRa [457]

Answer:

x = 468%

Step-by-step explanation:

I see a proportion here.

we want to solve for the variable x in the proportion:

18% / (2 weeks) =   ( x %) / (52 weeks)

here we should multiply 52 to boths sides

(52 weeks) * 18%  / (2 weeks) =   x %

26  * 18% =  x %

x = 468 %

3 0
3 years ago
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