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Shkiper50 [21]
3 years ago
14

Solve −5+1/4x=3x+6 by graphing.

Mathematics
1 answer:
erma4kov [3.2K]3 years ago
5 0

Answer:

hope its right lol

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Help me please anybody. And I am in my school iPad
Debora [2.8K]

Answer:

(0,5)

Step-by-step explanation:

I'm really not sure but I'm pretty positive that this is the answer if not than my bad

8 0
2 years ago
What is the area of this regular pentagon that has been divided into five congruent triangles?
Len [333]

Answer:

120 cm^2.

Step-by-step explanation:

The area of one triangle = 1/2 * base * height

= 1/2 * 8 * 6

= 1/2 * 48

= 24 cm^2

So area of the pentagon = 5 * 24

= 120 cm^2.

7 0
3 years ago
How do you do solve this?
solniwko [45]
The relative frequency is the frequency of the item divided by the total number of all items. The total number of all items here is: 50 + 40 + 90 +20 = 200.

Burgers =  \frac{50}{200} = 25%
Pasta =  \frac{40}{200} = 20%
Pizza =  \frac{90}{200} = 45%
Salad =  \frac{20}{200} = 10%

Hope this helps!
3 0
3 years ago
PLZ HELP IM IN A HURRY!
AysviL [449]

Answer:

0.078 rounded to the nearest hundreth is 0.08 and 0.078 rounded to the nearest thousandth is 0.078

Step-by-step explanation:

if you need help i will give you my email

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
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