Answer:
0
There is no x or number with an x (example: y= 2x+3, the 2 would be the slope)
Answer:
0
Step-by-step explanation:
B(−45, 42), (−17, 42), (−17, 21), (−45, 21)
C(−105, 42), (−77, 42), (−77, 21), (−105, 21)
D(−35,−28), (−7,−28), (−7,−49), (−35,−49)
By first dilating and then translating the given coordinates of the rectangle, we obtain the final coordinates as (−35, 32), (−7, 32), (−7, 11), (−35, 11). The correct answer is A.
The process of dilation involves multiplying the x and y coordinates of each vertex of the rectangle by a certain factor. In this case the factor is 7, which gives us the following set of coordinates after dilation: (−35, 42), (−7, 42), (−7, 21), (−35, 21). After the dilation, the rectangle is then translated down 10 units, which involves subtracting 10 from each of the y-coordinates. Following the translation, the final set of coordinates of the rectangle is (−35, 32), (−7, 32), (−7, 11), (−35, 11). Therefore, the correct answer is A.
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Answer:
d = 4√5
Step-by-step explanation:
d² = 8² + 4²
d² = 64 + 16
d² = 80
d =√80
d = √(16 * 5)
d = 4√5
Answer:
Step-by-step explanation:
hard to see graph ....
we are given
A coin is flipped until 3 heads in succession occur
so, firstly, we will find sample space
S={HHH , THHH , HTHHH, TTHHH , TTTHHH , HTTHHH , THTHHH , HHTHHHH, ......}
now, we are given that
list only those elements of the sample space that require 6 or less tosses
so, we can see that sample space
S={HHH , THHH , HTHHH, TTHHH , TTTHHH , HTTHHH , THTHHH , HHTHHHH, ......}
There are infinite such possibilities
so, there are infinite number of elements in space
and we know that
discrete sample space will always have finite elements
but we have infinite number of sample space elements here
so, this is not discrete sample space...........Answer
The sample space of flipping a coin until getting 3 heads in a row, and listing the elements that require 6 or fewer tosses.
In this case, we are looking at the sample space of flipping a coin until we get 3 heads in a row. To list the elements of the sample space that require 6 or fewer tosses, we need to consider the possible outcomes:
Yes, this is a discrete sample space because each toss of the coin has a finite number of possible outcomes (H or T).
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Answer:
Step-by-step explanation:
Whenthereis a(+)infrontofanexpressioninparentheses,theexpressionremainsthesame:
Calculatethesumof difference
Hopethishelps..
Goodluck on your assignment...
The probability that a randomly selected traveller who checks work email also uses a cell phone to stay connected is 57.5%.
The probability that someone who brings a laptop on vacation also uses a cell phone to stay connected is 70%.
If the randomly selected traveller checked their work email and brought a laptop, the probability that he/she uses a cell phone to stay connected is 58.8%.
We have,
Let:
C = Check work email
P = Use a cell phone to stay connected
L = Bring a laptop
Given information:
P(C) = 0.40 (Probability of checking work email)
P(P) = 0.30 (Probability of using a cell phone to stay connected)
P(L) = 0.25 (Probability of bringing a laptop)
P(C ∩ P) = 0.23 (Probability of both checking work email and using a cell phone to stay connected)
P(Neither) = 0.50 (Probability of neither checking work email, using a cell phone to stay connected, nor bringing a laptop)
Additional information:
P(C | L) = 0.84 (Probability of checking work email given that a laptop is brought)
P(P | L) = 0.70 (Probability of using a cell phone to stay connected given that a laptop is brought)
a. For the value of P(P | C), use the conditional probability formula:
P(P | C) = P(C ∩ P) / P(C)
P(P | C) = 0.23 / 0.40
P(P | C) = 0.575
b. For the value of P(P | L), use the conditional probability formula:
P(P | L) = P(P ∩ L) / P(L)
P (P | L) = 0.70
c. For the value of P(P | C ∩ L), use the conditional probability formula:
P(P | C ∩ L) = P(C ∩ P ∩ L) / P(C ∩ L)
Since we don't have the direct probability of P(C ∩ P ∩ L), we can use the information provided:
P(C | L) = 0.84
P(P | C ∩ L) = P(C | L) × P(P | L)
P(P | C ∩ L) = 0.84 × 0.70
P(P | C ∩ L) = 0.588
Thus, The probability that a randomly selected traveller who checks work email also uses a cell phone to stay connected is 57.5%.
The probability that someone who brings a laptop on vacation also uses a cell phone to stay connected is 70%.
If the randomly selected traveller checked their work email and brought a laptop, the probability that he/she uses a cell phone to stay connected is 58.8%.
Learn more about probability here:
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