Someone plz help me.............

Dominik [7]

Step-by-step answer:

This is a regular heptagon, means it has 7 congruent sides and 7 congruent vertex angles.

To work with polygons, there is a very important piece of information that you must know to solve the majority of related problems.

This is:

sum of exterior angles of polygons = 360 degrees.

If you don't remember the 360 degrees, think of the sum of exterior angles of an equilateral triangle, which is 3*(180-60)=3*120=360! It works!

For a regular heptagon, c = 360/7=51.43 degrees approx.

This means that each vertex angle measures

vertex angle = 180-c

So since 2d+the vertex angle = 360, we have

2d+(180-c)=360

solve for d:

2d=360-(180-c)=180+c

d=(180+c)/2=90+c/2=115.71 degrees. (approx.)

8 0

3 years ago

A:1/360 B:1/90 C:1/180 D:1/100

valkas [14]

the answer is 1/360.

$\frac{1}{1} = 360degrees \\ x= 1degree$

cross multiplication and u get the answer

5 0

3 years ago

The rates of on-time flights for commercial jets are continuously tracked by the U.S. Department of Transportation. Recently, So

tatyana61 [14]

Answer:

The probability that at least 13 flights arrive late is 2.5196 $\times 10^{-6}$ .

Step-by-step explanation:

We are given that Southwest Air had the best rate with 80 % of its flights arriving on time.

A test is conducted by randomly selecting 18 Southwest flights and observing whether they arrive on time.

The above situation can be represented through binomial distribution;

$P(X = x) = \binom{n}{r}\times p^{r} \times (1-p)^{n-r} ; x = 0,1,2,3,.........$

where, n = number of trials (samples) taken = 18 Southwest flights

r = number of success = at least 13 flights arrive late

p = probability of success which in our question is probability that

flights arrive late, i.e. p = 1 - 0.80 = 20%

Let X = Number of flights that arrive late.

So, X ~ Binom(n = 18, p = 0.20)

Now, the probability that at least 13 flights arrive late is given by = P(X $\geq$ 13)

P(X $\geq$ 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18)

= $\binom{18}{13}\times 0.20^{13} \times (1-0.20)^{18-13}+ \binom{18}{14}\times 0.20^{14} \times (1-0.20)^{18-14}+ \binom{18}{15}\times 0.20^{15} \times (1-0.20)^{18-15}+ \binom{18}{16}\times 0.20^{16} \times (1-0.20)^{18-16}+ \binom{18}{17}\times 0.20^{17} \times (1-0.20)^{18-17}+ \binom{18}{18}\times 0.20^{18} \times (1-0.20)^{18-18}$

= $\binom{18}{13}\times 0.20^{13} \times 0.80^{5}+ \binom{18}{14}\times 0.20^{14} \times 0.80^{4}+ \binom{18}{15}\times 0.20^{15} \times 0.80^{3}+ \binom{18}{16}\times 0.20^{16} \times 0.80^{2}+ \binom{18}{17}\times 0.20^{17} \times 0.80^{1}+ \binom{18}{18}\times 0.20^{18} \times 0.80^{0}$