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3 years ago

8

Correct answer please

1 answer:

ryzh [129]3 years ago

5 0

Answer:

Angle 3 and Angle 4

Step-by-step explanation:

Congruent angles are equal to each other, these can't be congruent.

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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

Please help with these! or at least some of them.

Fiesta28 [93]

Answer:

<h2>1) 9/2 = 5/4</h2><h2>5) False</h2><h2>6) $66</h2><h2>7) x=7</h2><h2>8) x=69</h2>

Step-by-step explanation:

5 0

3 years ago

The speed of 10 Ferris wheels is sampled. The speed, measured in miles per hour, are 10, 10, 12, 14, 10, 9, 8, 14, 12, 15 What i

liq [111]

Are u in k12? i didnt get it too

5 0

3 years ago

Read 2 more answers

Help please how many times longer is 3/4 than 2/15

liq [111]

3/4 is longer than 2/15 by 37/60

3 0

3 years ago

Read 2 more answers

Simplify the equation.

Phoenix [80]

Answer:

C. $3j^{3}l$

Step-by-step explanation:

1) First you cancel $k^{2}$ because they are completely the same. then you take out the constants and simply.

2)You then have to use the Quotient Rule:

$\frac{x^{a} }{x^{b} } =x^{a - b}$

You should have ...

$3j^{6-3} l^{5-4}$

3) Simplify 6 − 3 to 3.

$3j^{3} l^{1}$

4) Simplify 5 − 4 to 1.

5) Use Rule of One: $x^{1} =x$

Answer is !!!

$3j^{3} l$

5 0

3 years ago