What is the solution for the system of equations that has a sum of 1 and<br> differece of 9?

Each of three coins has two sides, heads and tails. Represent the heads or tails status of each coin by a logical variable (A fo

jeka94

Answer:

(a) F(A, B, C) = AB'C' + A'BC' + A'B'C

(b) F(A, B, C) = (A' + B' + C')(A' + B + C)(A + B' + C)(A + B + C')(A + B + C)

Step-by-step explanation:

(a) If F(A, B, C) = 1 iff exactly one of the coins is heads, then either

A is heads and the others are tails (AB'C')

B is heads and the others are tails (A'BC')

C is heads and the others are tails (A'B'C)

Hence, as a minterm expansion,

F(A, B, C) = AB'C' + A'BC' + A'B'C

(b) To get the corresponding maxterm expansion, we convert to binary.

$F(A, B, C) = \sum (100, 010, 001) = \sum(4,2,1)$

The maxterm is the product of the complements.

$F(A, B, C) = \prod (0, 3, 5, 6, 7) = \prod(000, 011, 101, 110, 111)$

Expanding,

F(A, B, C) = (A' + B' + C')(A' + B + C)(A + B' + C)(A + B + C')(A + B + C)

3 0

2 years ago

What is the solution to the inequality 0.8x+20<1.2x−40?

LuckyWell [14K]

Answer:x=150

Step-by-step explanation:

Step 1: Subtract 1.2x from both sides.

0.8x+20−1.2x=1.2x−40−1.2x

−0.4x+20=−40

Step 2: Subtract 20 from both sides.

−0.4x+20−20=−40−20

−0.4x=−60

Step 3: Divide both sides by -0.4.

−0.4x

−0.4

=

−60

−0.4

3 0

2 years ago

Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive

LenaWriter [7]

Answer:

(a) The probability that all the next three vehicles inspected pass the inspection is 0.343.

(b) The probability that at least 1 of the next three vehicles inspected fail is 0.657.

(c) The probability that exactly 1 of the next three vehicles passes is 0.189.

(d) The probability that at most 1 of the next three vehicles passes is 0.216.

(e) The probability that all 3 vehicle passes given that at least 1 vehicle passes is 0.3525.

Step-by-step explanation:

Let <em>X</em> = number of vehicles that pass the inspection.

The probability of the random variable <em>X</em> is <em>P (X) = 0.70</em>.

(a)

Compute the probability that all the next three vehicles inspected pass the inspection as follows:

P (All 3 vehicles pass) = [P (X)]³

$=(0.70)^{3}\\=0.343$

Thus, the probability that all the next three vehicles inspected pass the inspection is 0.343.

(b)

Compute the probability that at least 1 of the next three vehicles inspected fail as follows:

P (At least 1 of 3 fails) = 1 - P (All 3 vehicles pass)

$=1-0.343\\=0.657$

Thus, the probability that at least 1 of the next three vehicles inspected fail is 0.657.

(c)

Compute the probability that exactly 1 of the next three vehicles passes as follows:

P (Exactly one) = P (1st vehicle or 2nd vehicle or 3 vehicle)

= P (Only 1st vehicle passes) + P (Only 2nd vehicle passes)

+ P (Only 3rd vehicle passes)

$=(0.70\times0.30\times0.30) + (0.30\times0.70\times0.30)+(0.30\times0.30\times0.70)\\=0.189$

Thus, the probability that exactly 1 of the next three vehicles passes is 0.189.

(d)

Compute the probability that at most 1 of the next three vehicles passes as follows:

P (At most 1 vehicle passes) = P (Exactly 1 vehicles passes)

+ P (0 vehicles passes)

$=0.189+(0.30\times0.30\times0.30)\\=0.216$

Thus, the probability that at most 1 of the next three vehicles passes is 0.216.

(e)

Let <em>X</em> = all 3 vehicle passes and <em>Y</em> = at least 1 vehicle passes.

Compute the conditional probability that all 3 vehicle passes given that at least 1 vehicle passes as follows:

$P(X|Y)=\frac{P(X\cap Y)}{P(Y)} =\frac{P(X)}{P(Y)} =\frac{(0.70)^{3}}{[1-(0.30)^{3}]} =0.3525$

Thus, the probability that all 3 vehicle passes given that at least 1 vehicle passes is 0.3525.

7 0

2 years ago

Simplify.

ValentinkaMS [17]

first, you would rearrange the equation so that like terms were next to each other.

7y - 5y - x2 + 2x2 +3x - 17x

then you would reduce the equation based on like terms

2y + x2 - 14x

then simply rearrange the equation to get the answer, which is <u>B</u>

5 0

3 years ago

Given log Subscript 3 Baseline 2 almost-equals 0. 631 and log Subscript 3 Baseline 7 almost-equals 1. 771, what is log Subscript

kari74 [83]

You can use the properties of logarithm to get to the solution.

The approximate value for given term is given by

$log_3(14) \approx 2.402$

<h3>What is logarithm and some of its useful properties?</h3>

When you raise a number with an exponent, there comes a result.

Lets say you get

$a^b = c$

Then, you can write 'b' in terms of 'a' and 'c' using logarithm as follows

$b = log_a(c)$

Some properties of logarithm are:

$log_a(b) = log_a(c) \implies b = c\\\\\log_a(b) + log_a(c) = log_a(b \times c)\\\\log_a(b) - log_a(c) = log_a(\frac{b}{c})$

<h3>Using the above properties</h3>

$log_3(2) + log_3(7) = log_3(2 \times 7) = log_3(14)\\\\0.631 + 1.771 = log_3(14)\\\\log_3(14) = 2.402$

Thus,

The approximate value for given term is given by

$log_3(14) \approx 2.402$

Learn more about logarithm here:

brainly.com/question/20835449

5 0

2 years ago

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What is the solution for the system of equations that has a sum of 1 and differece of 9?