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s344n2d4d5 [400]
2 years ago
7

PLEASE HELP PLEASE PLEASE PLEASE

Mathematics
1 answer:
sleet_krkn [62]2 years ago
8 0

Answer:

oh no that's tough bro

hope you get through

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Be sure to show your work and solve for g:<br> g<br> +<br> 6<br> −<br> 3<br> =<br> 31
Iteru [2.4K]

28 + 6 = 34 and 34 - 3 = 31!!!!!!!

4 0
3 years ago
1. If the length of a rectangle is 6 more than its width, and its perimeter is 48
Salsk061 [2.6K]

For this case we have that by definition, the perimeter of the rectangle is given by:

P = 2w + 2L

Where:

W: Is the width of the rectangle

L: is the length of the rectangle

According to the data we have:

L = 6 + w\\P = 48

Substituting:

48 = 2w + 2 (6 + w)\\48 = 2w + 12 + 2w\\48 = 4w + 12\\48-12 = 4w\\36 = 4w\\w = \frac {36} {4}\\w = 9

So, the width of the rectangle is 9 inches

L = 6 + 9 = 15

So, the length of the rectangle is 15 inches

Answer:

the width of the rectangle is 9 inches

the length of the rectangle is 15 inches

5 0
3 years ago
The medians $AD$, $BE$, and $CF$ of triangle $ABC$ intersect at the centroid $G$. The line through $G$ that is parallel to $BC$
Flauer [41]
G divides each median in the ratio of 2 to 1 (the longer side goes from G to the angle) triangle AMN is similar to triangle ABC (why?) AMN is a scaled version of ABC (by a factor of ⅔) its area should be scaled by (⅔)^2
6 0
3 years ago
Find the area of the parallelogram
iogann1982 [59]

Answer:

The answer is 32.

Step-by-step explanation:

Just multiply the height by the base length. 4*8=32

8 0
3 years ago
Suppose that you are given a bag containing n unbiased coins. You are told that n-1 of these coins are normal, with heads on one
gladu [14]

Answer:

The (conditional) probability that the coin you chose is the fake coin is 2/(1 + n)

Step-by-step explanation:

Given

Total unbiased coin = n

Normal coins =n - 1

Fake = 1

The (conditional) probability that the coin you chose is the fake coin is represented by

P(Fake | Head)

And it's calculated as follows;

P(Fake | Head) = P(Fake, Head) ÷ P(Head) ----- (1)

Where P(Fake, Head) = P(Fake) * P(Head | Fake)

P(Fake) = 1/n --- because only one is fake

P(Head | Fake) = n/n because all coins (including the fake) have head

So, P(Fake, Head) = P(Fake) * P(Head | Fake) becomes

P(Fake, Head) = 1/n * n/n

P(Fake, Head) = 1/n

P(Head) is calculated by

P(Fake) * P(Head | Fake) + P(Normal) * P(Head | Normal)

P(Fake) * P(Head | Fake) = P(Fake, Head) = 1/n (as calculated above)

P(Normal) * P(Head | Normal) = ½ * (n - 1)/n ----- considering that the coin also has a tail with equal probability as that of the head.

Going back to (1)

P(Fake | Head) = P(Fake, Head) ÷ P(Head) becomes

P(Fake | Head) = (1/n) ÷ ((1/n) + (½(n-1)/n))

= (1/n) ÷ ((1/n) + (½(n-1)/n))

= (1/n) ÷ (1/n + (n - 1)/2n)

= (1/n) ÷ (2 + n - 1)/(2n)

= (1/n) ÷ (1 + n)/(2n)

= (1/n) * (2n)/(1 + n)

= 2/(1 + n)

Hence, the (conditional) probability that the coin you chose is the fake coin is 2/(1 + n)

5 0
3 years ago
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