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LiRa
8 days ago
7

S=2(lw+lh+wh) solve for w

Mathematics
2 answers:
Leona [4.1K]8 days ago
7 0
S=2(lw+lh+wh)\\
S=2lw+2lh+2wh\\
S=w(2l+2h)+2lh\\
w(2l+2h)=S-2lh\\
w=\frac{S-2lh}{2l+2h}
tester [3.9K]8 days ago
7 0

We start with

s=2(lw+lh+wh)

To solve for w means isolating the variable w

s=2(lw+lh+wh)\\s=2lw+2lh+2wh

Combine terms that include the variable w, shifting the remaining terms to the other side of the equation

(s-2lh)=(2lw+2wh)

Factor out the variable w

(s-2lh)=w(2l+2h)

Then divide both sides by (2l+2h)

(s-2lh)/(2l+2h)=w(2l+2h)/(2l+2h)

w=(s-2lh)/(2l+2h)

Consequently,

the result is

w=(s-2lh)/(2l+2h)


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An urn contains 3 red and 7 black balls. Players and withdraw balls from the urn consecutively until a red ball is selected. Fin
Leona [4166]

Correct question:

An urn holds 3 red and 7 black balls. Players A and B take turns withdrawing balls until a red one is chosen. Calculate the probability that A picks the red ball. (A goes first, followed by B, with no replacement of drawn balls).

Answer:

The likelihood that A picks the red ball is 58.33 %

Step-by-step explanation:

A will select the red ball if it is drawn 1st, 3rd, 5th, or 7th.

1st draw: 9C2

3rd draw: 7C2

5th draw: 5C2

7th draw: 3C2

Calculating for all possible scenarios gives us:

9C2 = (9!) / (7!2!) = 36

7C2 = (7!) / (5!2!) = 21

5C2 = (5!) / (3!2!) = 10

3C2 = (3!) / (2!) = 3

Adding these possibilities results in 36 + 21 + 10 + 3 = 70.

The total outcomes for selecting a red ball = 10C3

10C3 = (10!) / (7!3!)

= 120.

The probability that A selects the red ball is determined by dividing the sum of possible events by the overall outcomes.

P( A selects the red ball) = 70 / 120

= 0.5833

= 58.33 %

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2 days ago
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Answer:

a) 0.00019923%

b) 47.28%

Step-by-step explanation:

a) To determine the likelihood that all sockets in the sample are defective, we can use the following approach:

The first socket is among a group that has 5 defective out of 38, leading to a probability of 5/38.

The second socket is then taken from a group of 4 defective out of 37, following the selection of the first defective socket, resulting in a probability of 4/37.

Extending this logic, the chance of having all 5 defective sockets is computed as: (5/38)*(4/37)*(3/36)*(2/35)*(1/34) = 0.0000019923 = 0.00019923%.

b) Using similar reasoning as in part a, the first socket has a probability of 33/38 of not being defective as it's chosen from a set where 33 sockets are functionally sound. The next socket has a proportion of 32/37, and this continues onward.

The overall probability calculates to (33/38)*(32/37)*(31/36)*(30/35)*(29/34) = 0.4728 = 47.28%.

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Let A represent the number of individuals without jobs.
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Answer and explanation:

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Algebra can also manifest itself in expressions, commonly referred to as algebraic expressions, which can be incorporated into equations, such as the previously mentioned 2a = $50. These expressions may take forms like 2a + 3b, where a and b designate the costs of different products that were acquired in quantities of 2 and 3, respectively.

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Answer:

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Step-by-step explanation:

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