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3 years ago

What does “E” and “D” equal? I forgot how to do this lol

Mathematics

2 answers:

Akimi4 [234]3 years ago

6 0

Answer:

<h2>d = 49 degrees</h2><h2>e = 41 degrees</h2>

Step-by-step explanation:

d = 180 - 131 = 49

e = 180 - (90 + 49) =

180 - 139 =

Hope that helps!

Elan Coil [88]3 years ago

3 0

Answer:

d: 49

e:41

Step-by-step explanation:

180-131=d which is 49

90+49=139

180-139= e which is 41

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4) You randomly select one card from a 52-card deck. Find the probability of selecting the 6 of hearts or the ace of diamonds.

MakcuM [25]

Answer:

1/26

Step-by-step explanation:

There is only one 6 of hearts and only one ace of diamonds.

The probability of selecting the 6 of hearts is thus 1/52, and that of selecting the ace of diamonds is also 1/52.

The probability of selecting the 6 of hearts or the ace of diamonds is the SUM of these two results: 1/52 + 1/52 = 2/52 = 1/26.

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3 years ago

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Eric's class consists of 12 males and 16 females. If 3 students are selected at random, find the probability that they

Reptile [31]

Answer:

The probability that all are male of choosing '3' students

P(E) = 0.067 = 6.71%

Step-by-step explanation:

Let 'M' be the event of selecting males n(M) = 12

Number of ways of choosing 3 students From all males and females

$n(M) = 28C_{3} = \frac{28!}{(28-3)!3!} =\frac{28 X 27 X 26}{3 X 2 X 1 } = 3,276$

Number of ways of choosing 3 students From all males

$n(M) = 12C_{3} = \frac{12!}{(12-3)!3!} =\frac{12 X 11 X 10}{3 X 2 X 1 } =220$

The probability that all are male of choosing '3' students

$P(E) = \frac{n(M)}{n(S)} = \frac{12 C_{3} }{28 C_{3} }$

$P(E) = \frac{12 C_{3} }{28 C_{3} } = \frac{220}{3276}$

P(E) = 0.067 = 6.71%

<u><em>Final answer</em></u>:-

The probability that all are male of choosing '3' students

P(E) = 0.067 = 6.71%

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3 years ago

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Answer:

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Step-by-step explana

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3 years ago

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Masja [62]

x²-6x=0

x(x-6)=0

Two solutions are possible:

x=0

x=6

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3 years ago