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Yuliya22 [10]
3 years ago
6

there are 5 red counters and 3 blue counters in bag A. there are 4 red counters in bag B. Tina takes at random a counter from ea

ch bag. what is the probability that Tina takes two blue counters
Mathematics
1 answer:
mars1129 [50]3 years ago
7 0

Answer:

The probability is 0

Step-by-step explanation:

From the first bag, the total number of counters is 5 + 3 = 8

The probability of taking a blue counter from the first bag is the number of blue counters divided by the total number of counters

Mathematically, we have this as;

3/8

For the second bag, all the counters are red with no blue counter

So the probability of taking a blue counter here is 0

Now, the probability that he takes a blue counter in A and a blue counter in B will

be;

3/8 * 0 = 0

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Evgesh-ka [11]
Change 212,514 into 215,000.

Change 396,705 into 400,000.

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Estimate:

215,000 + 400,000 = 615,000

Therefore:

212,514 + 396,705 ≈ 615,000
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3 years ago
May someone please help with the way please
Oksana_A [137]

Answer:

17.6 m²

Step-by-step explanation:

Given the ratio of similar shapes = a : b, then

area of shapes = a ² : b²

Δ PTQ and Δ PRS are similar and so the ratio of corresponding sides are equal, that is

PT : PR = 6 : 9 = 2 : 3, thus

ratio of areas = 2² : 3² = 4 : 9

let the area of Δ PQT be x, then using proportion

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9x = 4(x + 22) ← distribute

9x = 4x + 88 ( subtract 4x from both sides )

5x = 88 ( divide both sides by 5 )

x = 17.6

Thus area of Δ PQT = 17.6 m²

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3 years ago
Find the slope between the points ( − 2 , 5 ) (−2,5) and ( 4 , − 7 ) (4,−7)
emmainna [20.7K]

Answer:

1) undefined

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Step-by-step explanation:

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4 years ago
Which statement is an accurate comparison of Mrs. Liu’s 3rd and 5th period classes?
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Read 2 more answers
According to DeMorgan 's theorem, the complement of W · X + Y · Z is W' + X' · Y' + Z'. Yet both functions are 1 for WXYZ = 1110
sergeinik [125]

Answer:

The parenthesis need to be kept intact while applying the DeMorgan's theorem on the original equation to find the compliment because otherwise it will introduce an error in the answer.

Step-by-step explanation:

According to DeMorgan's Theorem:

(W.X + Y.Z)'

(W.X)' . (Y.Z)'

(W'+X') . (Y' + Z')

Note that it is important to keep the parenthesis intact while applying the DeMorgan's theorem.

For the original function:

(W . X + Y . Z)'

= (1 . 1 + 1 . 0)

= (1 + 0) = 1

For the compliment:

(W' + X') . (Y' + Z')

=(1' + 1') . (1' + 0')

=(0 + 0) . (0 + 1)

=0 . 1 = 0

Both functions are not 1 for the same input if we solve while keeping the parenthesis intact because that allows us to solve the operation inside the parenthesis first and then move on to the operator outside it.

Without the parenthesis the compliment equation looks like this:

W' + X' . Y' + Z'

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Here, the 'AND' operation will be considered first before the 'OR', resulting in 1 as the final answer.

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