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Bezzdna [24]
3 years ago
7

Find the domain and range (0,1), (3,1), (3,9)

Mathematics
1 answer:
Elza [17]3 years ago
7 0

Answer:

domain={0,3}

range={1,9}

Step-by-step explanation:

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A padlock has a four-digit code that includes digits from 0 to 9, inclusive. What is the probability that the code does not cons
Eduardwww [97]
Since there is no repetition allowed, there are 10 possibilities for the 1st digit, 9 for the 2nd, 8 for the 3rd, and 7 for the 4th. This gives a total of (10)(9)(8)(7) = 5040 four-digit codes.
For all odd digits to be used, there are 5 possibilities for the 1st digit (1,3,5,7,9), 4 for the 2nd, 3 for the 3rd, 2 for the 4th. This gives a total of (5)(4)(3)(2) = 120 codes that only use odd digits.
Therefore there are 5040 - 120 = 4920 codes that do not consist of all odd digits. The probability is 4920/5040 = 41/42.

4 0
2 years ago
Read 2 more answers
find the probability of being delt 5 clubs and 3 cards with one of each remaining suit in 8 card poker
kumpel [21]

Answer: 0.003757(approx).

Step-by-step explanation:

Total number of combinations of selecting r things out of n things is given by:-

^nC_r=\dfrac{n!}{r!(n-r)!}

Total cards in a deck = 52

Total number of ways of choosing 8 cards out of 52 = ^{52}C_8

Total number of ways to choose 5 clubs and 3 cards with one of each remaining suit = ^{13}C_5\times^{13}C_1\times^{13}C_1\times^{13}C_1  [since 1 suit has 13 cards]

The required probability = =\dfrac{^{13}C_5\times^{13}C_1\times^{13}C_1\times^{13}C_1}{^{52}C_8}

=\dfrac{\dfrac{13!}{5!8!}\times13\times13\times13}{\dfrac{52!}{8!44!}}\\\\=\dfrac{24167}{6431950}\\\\\approx0.003757

Hence, the required probability is 0.003757 (approx).

5 0
2 years ago
A chef buys 6 pounds of salmon for $51. How much will he pay for 9 pounds of salmon?
kotegsom [21]
$76.5

51/6 = 8.5
8.5 x 9 = 76.5
7 0
2 years ago
Tell whether the lines through the given points are parallel, perpendicular, or neither. (-3,1), (1,9), (-2-9), (-1,-7)​
Brrunno [24]

Answer:

parallel

Step-by-step explanation:

all the details are in the attached picture.

4 0
3 years ago
Can someone help me with this and show their work and how they found the answer?
Semenov [28]

Answer:

2kg

Step-by-step explanation:

Let the sum of the other 2 bricks be R. Given that the heaviest brick weighs 2/3 times as much as the other 2 bricks in total, then the weight of the heaviest brick H in terms of the weight of the other 2 will be

H = 2/3 * R

= 2R/3

Given that the 3 bricks weigh 5kg in total then,

2R/3 + R = 5

Multiply through the equation by 3

2R + 3R = 15

5R = 15

Divide both sides by 5

R = 15/5

= 3

The weight of the heaviest brick is 2R/3. Since R = 3, the heaviest block weighs

= 2 * 3/3

= 2kg

4 0
3 years ago
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