Find M<1, M<2, M<3, M<4, M<5

Mathematics

1 answer:

Leokris [45]3 years ago

6 0

Answer:

Step-by-step explanation:

m∠2 = 52° [Alternate interior angles]

m∠1 + m∠2 = 180° [Linear pair of angles are supplementary]

m∠1 = 180° - 52°

m∠1 = 128°

m∠3 = 47° [Alternate interior angles]

m∠5 + 52° + 47° = 180° [Sum of linear angles is 180°]

m∠5 = 180° - 99°

= 81°

m∠4 + 47° = 180° [Sum of consecutive interior angles is 180°]

m∠4 = 180° - 47°

m∠4 = 137°

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A candy dish contains 30 jelly beans and 20 gumdrops. Ten candies are picked at random. What is the probability that 5 of the 10

AveGali [126]

Answer:

The probability of drawing 5 gumdrops in 10 picks from the dish is 0.215

Step-by-step explanation:

A hypergeometric distribution is a discrete probability distribution that describes the probability of successes in draws with each draw being a success or failure, without replacement from a finite population size that contains exactly the same objects. The general formula is given as:

$h(x) =\frac{({A}C_{x})({N-A}C_{n-x})}{^{N}C_{n} }$

where:

$^{n}C_{r}=\frac{n!}{(n-r)!r!}$

h(x) is the probability of x successes,

n is the number of attempts,

A is the number of successes

N is the number of elements

For this problem:

A = 30, x = 5, N = 50, n = 10

The probability of drawing 5 gumdrops in 10 picks from the dish P(x=5) is

$P(x=5)=\frac{({30}C_{5})({50-30}C_{10-5})}{^{50}C_{10} }\\P(x=5)=\frac{({30}C_{5})({20}C_{5})}{^{50}C_{10} }$

$P(x=5)=\frac{\frac{30!}{(30-5)!5!}*\frac{20!}{(20-5)!5!} }{\frac{50!}{(50-10)!10!} }\\P(x=5)=\frac{\frac{30!}{25!5!}*\frac{20!}{15!5!} }{\frac{50!}{40!10!} }$