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Sati [7]
3 years ago
15

Andy, Beth, Charlie, and Daniel take a test with thirty questions. Andy and Beth together get the same number of questions wrong

as Charlie and Daniel together. Andy and Daniel together get four more questions wrong than Beth and Charlie do together. If Charlie gets five questions wrong, how many questions does Andy get wrong?
Mathematics
1 answer:
olga_2 [115]3 years ago
5 0

Answer:

Andy gets 7 questions wrong.

Step-by-step explanation:

Let A, B, C and D be the number of questions answered wrong by Andy, Beth, Charlie, and Daniel, respectively.

Andy and Beth together get the same number of questions wrong as Charlie and Daniel together:

A+B =C+D

Andy and Daniel together get four more questions wrong than Beth and Charlie do together:

A+D =B+C+4

Charlie gets five questions wrong:

C=5

Solving the linear system:

C=5\\A+B =5+D\\A+D =B+9\\\\A+B =5+D\\A-B =9-D\\\\2A = 14\\A=7

Andy gets 7 questions wrong.

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Point K is rotated 90°. The coordinate of the pre-image point K was (2, –6) and its image K’ is at the coordinate (−6, −2). Find
aalyn [17]

Answer:

A clockwise rotation.

Step-by-step explanation:

The rule for 90 degree clockwise rotation is (x,y) turns into (y,-x) therefore if you use your numbers (2,-6) and then use the 90 degree clockwise rotation rule it becomes (-6,-2)

5 0
3 years ago
The probability of tossing heads with a standard coin is 1/2, because it is one of two possible outcomes. The probability of tos
spayn [35]

Answer:

  a) 1/64

  b) 1/4096

Step-by-step explanation:

As you can tell from the example, the exponent of 1/2 is the number of heads in a row.

a) p(6 heads in a row) = (1/2)^6 = 1/(2^6) = 1/64

b) p(12 heads in a row) = (1/2)^12 = 1/(2^12) = 1/4096

_____

<em>Additional comment</em>

The probability of a head is 1/2 because we generally are concerned with a "fair coin." That is defined as a coin in which each of the 2 possible outcomes has the same probability, 1/2. Similarly, a "fair number cube" has 6 faces, and the probability of each is defined to be the same as any other, 1/6. Loaded dice and unfair coins do sometimes show up in probability problems.

3 0
3 years ago
Find the value of w (u(2))
katrin [286]
<h3>Answer:  17</h3>

==================================================

Explanation:

We'll start things off by computing the inner function u(2)

Plug x = 2 into the u(x) function

u(x) = -x-1

u(2) = -2-1

u(2) = -3

This tells us that w(u(2)) is the same as w(-3). I replaced u(2) with -3.

We'll plug x = -3 into the w(x) function

w(x) = 2x^2-1

w(-3) = 2(-3)^2 - 1

w(-3) = 2(9) - 1

w(-3) = 18-1

w(-3) = 17

Therefore, w(u(2)) = 17

------------------------

Here's a slightly different approach:

Let's find what w(u(x)) is in general

w(x) = 2x^2 - 1

w(u(x)) = 2(u(x))^2 - 1

w(u(x)) = 2(-x-1)^2 - 1

Then we can plug in x = 2

w(u(x)) = 2(-x-1)^2 - 1

w(u(2)) = 2(-2-1)^2 - 1

w(u(2)) = 2(-3)^2 - 1

w(u(2)) = 2(9) - 1

w(u(2)) = 18 - 1

w(u(2)) = 17

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Answer:

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Step-by-step explanation:

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