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ANTONII [103]
2 years ago
15

Ryan’s mom randomly chooses two days each week for Ryan to do his chores. What is the probability that she picks Saturday and Su

nday?
Mathematics
2 answers:
torisob [31]2 years ago
4 0
There are a total of 7C2 of options for Ryan's mom to choose.

If Ryan's mom picks Saturday and Sunday, this is the only 1 option in those 7C2 of options. Thus, the propability is (1 / 7C2) × 100% = 4.762%.
Alona [7]2 years ago
4 0

Answer:

Probability that she picks Saturday and Sunday = 1/21 = 0.0476

Step-by-step explanation:

Total number of days in a week = 7

Number of days selected = 2

Total number of combinations = 7C₂ =\frac{7*6}{1*2} =21

Total number of favorable ways = 1 that is  Sunday and  Saturday.

Probability that she picks Saturday and Sunday = 1/21 = 0.0476

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The probability p of an orangecandy is 0.2. The sample size = 100.
The mean is given by:
n\times p=100\times0.2=20
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The answers are: Mean = 20. Standard deviation = 4.

4 0
2 years ago
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Find the value of x in each case:
Zanzabum

Answer:

x=30\°

Step-by-step explanation:

<u>Find the measures of interior angles in each triangle</u>

Triangle BGC

m

The measures of triangle BGC are 90\°-60\°-30\°

Triangle CGH

we know that

m -----> by consecutive interior angles

we have that

m

so

m

m

substitute

m

we have

m

m

m

remember that

m

60\°+2x+180\°-4x=180\°

60\°=2x

x=30\°

The measures of triangle CGH are 60\°-60\°-60\°

Triangle GHE

m< EGH=90\°-2x=90-2(30\°)=30\°

m< GHE=4x=4(30\°)=120\°

remember that

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substitute and solve for m<GEH

30\°+120\°+m

150\°+m

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The measures of triangle GHE are 30\°-120\°-30\°

6 0
2 years ago
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Use the formula A=bh , where A is the area, b is the base length, and h is the height of the parallelogram, to solve this proble
Jlenok [28]
33 in because 1056 divided by 32 is 33
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1 year ago
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Find the smallest relation containing the relation {(1, 2), (1, 4), (3, 3), (4, 1)} that is:
professor190 [17]

Answer:

Remember, if B is a set, R is a relation in B and a is related with b (aRb or (a,b))

1. R is reflexive if for each element a∈B, aRa.

2. R is symmetric if satisfies that if aRb then bRa.

3. R is transitive if satisfies that if aRb and bRc then aRc.

Then, our set B is \{1,2,3,4\}.

a) We need to find a relation R reflexive and transitive that contain the relation R1=\{(1, 2), (1, 4), (3, 3), (4, 1)\}

Then, we need:

1. That 1R1, 2R2, 3R3, 4R4 to the relation be reflexive and,

2. Observe that

  • 1R4 and 4R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 4R1 and 1R2, then 4 must be related with 2.

Therefore \{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(4,1),(4,2)\} is the smallest relation containing the relation R1.

b) We need a new relation symmetric and transitive, then

  • since 1R2, then 2 must be related with 1.
  • since 1R4, 4 must be related with 1.

and the analysis for be transitive is the same that we did in a).

Observe that

  • 1R2 and 2R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 2R1 and 1R4, then 2 must be related with 4.
  • 4R1 and 1R2, then 4 must be related with 2.
  • 2R4 and 4R2, then 2 must be related with itself

Therefore, the smallest relation containing R1 that is symmetric and transitive is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

c) We need a new relation reflexive, symmetric and transitive containing R1.

For be reflexive

  • 1 must be related with 1,
  • 2 must be related with 2,
  • 3 must be related with 3,
  • 4 must be related with 4

For be symmetric

  • since 1R2, 2 must be related with 1,
  • since 1R4, 4 must be related with 1.

For be transitive

  • Since 4R1 and 1R2, 4 must be related with 2,
  • since 2R1 and 1R4, 2 must be related with 4.

Then, the smallest relation reflexive, symmetric and transitive containing R1 is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

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1 year ago
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X^2+4x-1+6=0
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2 years ago
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