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lina2011 [118]
3 years ago
13

PLEASE HELP ASAP!!

Mathematics
1 answer:
alex41 [277]3 years ago
7 0

Step-by-step explanation:

e = 20sin(πt/4 - π/2)

= 20[(sinπt/4)(cosπ/2) - (cosπt/4)(sinπ/2)]

Used sin(x - y) = sinxcosy - cosxsiny.

= 20[0 - cosπt/4]

Since cosπ/2 = 0 and sinπ/2 = 1.

= -20cosπt/4.

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bezimeni [28]

Answer:

the answer would be y=2x+4

Step-by-step explanation:

4 0
3 years ago
A survey of 500 high school students was taken to determine their favorite chocolate candy. Of the 500 students surveyed, 42 lik
Step2247 [10]

Answer:

Step-by-step explanation:

Universal set

U = 500

The number that likes snickers

n(S) = 42

The number that like Twix.

n(T) = 110

The number that like Reeses

n(R) = 125

n(S n T) = 33

n(T n R) = 62

n(S n R) = 26

n( S n R n T) = 22

Then,

n(S n T) only = n(S n T) - n(S n R n T)

n(S n T) only =33 - 22 = 11

n(T n R) only = n(T n R) - n(S n R n T)

n(T n R) only =62 - 22 = 40

n(S n R) only = n(S n R) - n(S n R n T)

n(S n R) only =26 - 22 = 4.

Also,

n(S) only = n(S) - n(S n R) - n(S n T) only

n(S) only = 42 - 26 - 11 = 5

n(T) only = n(T) - n(T n R) - n(T n S) only

n(T) only = 110 - 62 - 11 = 37

n(R) only = n(R) - n(S n R) - n(R n T) only

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Then, to know if some student don't like any of the of chocolate,

Let know the number of students that like the chocolate candy

n(S) + n(T)only + n(T n R)only + n(R) only

42 + 37 + 40 + 58 = 177 students.

Therefore, the total students that like chocolate candy is 177, so those  that does not like any of them are

n(S U R U T)' = U – n(S U R U T)

n(S U R U T) = 500 - 177 = 323.

So, the question is how many student likes at most 2 kinds of these chocolates, this means that they can like exactly 2 or less or even none.

So, this category are

n(2 most) = n(S n T)only + n(R n T)only + n(S n R)only + n(s)only + n(T)only + n(R)only + n(S U R U T)'

n(2 most) = 11 + 40 + 4 + 5 + 37 + 58 + 323

n(2 most) = 478

The correct answer is E.

Check attachment for Venn diagram

5 0
3 years ago
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Step-by-step explanation:

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3 years ago
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The value of n from the set {6, 7, 8, 9} that holds true for 4n − 12 < 2n + 2 is .
IrinaVladis [17]

Answer:

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so, only for n=6 is the expression true.

3 0
2 years ago
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