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3 years ago

In the circle above, Line EF and CD intersect at G. Determine the measure of arc DE.

Mathematics

1 answer:

Aliun [14]3 years ago

6 0

Answer:

100

Step-by-step explanation:

It is a rule that for two opposite arcs, with intersecting lines, between them, the average of the arcs is equivalent to the inside angle. Given this, we can say that:

(CF + ED)/2 = angle G

plug values in

(12x+9x+10)/2 = 110

multiply both sides by 2

21x + 10 = 220

subtract 10 from both sides

21x = 210

divide both sides by 21

x=10

Plug this into DE

10*9+10 = 100

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Ship collisions in the Houston Ship Channel are rare. Suppose the number of collisions are Poisson distributed, with a mean of 1

alexandr1967 [171]

Answer:

$a) \simeq 0.3012$ $b) \simeq 0.0494$ c) $\simeq 0.2438$

Step-by-step explanation:

Rate of collision,

1.2 collisions every 4 months

or, $\frac{1.2}{4}$

= 0.3 collisions per month

So, the Poisson distribution for the random variable no. of collisions per month (X) is given by,

P(X =x) = $\frac{e^{-\lambda}\times {\lambda}^{x}}}{x!}
$

for x ∈ N ∪ {0}

= 0 otherwise --------------------------------------(1)

here, $\lambda = 0.3$ collision / month

No collision over a 4 month period means no collision per month or X =0

Putting X = 0 in (1) we get,

P(X = 0) = $\frac{e^{-0.3}\times {\0.3}^{0}}{0!}
$

$\simeq 0.7408182207$ ------------------------------------(2)

Now, since we are calculating this for 4 months,

so, P(No collision in 4 month period)

= $0.7408182207^{4}$

$\simeq 0.3012$ -----------------------------------------------------------(3)

2 collision in 2 month period means 1 collision per month or X =1

Putting X =1 in (1) we get,

P(X =1) = $\frac{e^{-0.3}\times {\0.3}^{1}}{1!}
$

$\simeq 0.2222454662$ ------------------------------------(4)

Now, since we are calculating this for 2 months, so ,

P(2 collisions in 2 month period)

= $0.2222454662^{2}$

$\simeq 0.0494$ -----------------------------------------(5)

1 collision in 6 months period means

$\frac{1}{6}$ collision per month

Now, P(1 collision in 6 months period)

= P( X = 1/6] (which is to be estimated)

= $\frac {P(X=0)\times 5 + P(X =1)\times 1}{6}$

= \frac {0.7408182207 \times 5 + 0.2222454662 \times 1}{6}[/tex]

$\simeq 0.6543894283$ -------------------------------------------(6)

So,

P(1 collision in 6 month period)

= $0.6543894283^{6}$

$\simeq 0.0785267444$ ------------------------------------------------(7)

So,

P(No collision in 6 months period)

= $(P(X =0)^{6}$

$\simeq 0.1652988882$ ---------------------------------(8)

so,

P(1 or fewer collision in 6 months period)

= (8) + (7 ) = 0.0785267444 +0.1652988882

$\simeq 0.2438$ ---------------------------------------------(9)

7 0

3 years ago

A certain ore is 31.8% nickel by mass. How many kilograms of this ore would you need to dig up to have 25.0 g of nickel?

natulia [17]

It will need 1.8 kilograms to have 25.0 g of nickel

6 0

3 years ago

Can someone explain how you find the answer!

ahrayia [7]

Answer:

21m

Step-by-step explanation:

So for the rectangle, do 5*6, and you will get the area of the whole backyard. Then to get the area of the grass, you do 3*3 to get 9.

Finally, to get the cement, you subtract the area of the grass from the area of the whole backyard so 30-9 to get 21.

5*6=30

3*3=9

30-9+21

Please mark brainliest if you can!

Funny, I think you put the question in 2 times, this is the second time I'm answering this same question by you. :D

7 0

3 years ago

The ratio of boys to girls at a movie is 3:4 . If there are 16 girls , how many boys are at the movie?

Annette [7]

Answer:

12 boys

Step-by-step explanation:

Set up a proportion: $\frac{3}{4} = \frac{x}{16}$
Cross multiply, then divide: 3 × 16 = 48, 48 ÷ 4 = 12

I hope this helps!

8 0

3 years ago

Read 2 more answers

What is the common factor of 49 and 50?

Zinaida [17]

The factors of 49 are: 1, 7 and 49.
The factors of 50 are: 1, 2, 5, 10, 25 and 50.

The common factor is 1.

6 0

3 years ago

Read 2 more answers