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3 years ago

URGENT help me please

Mathematics

1 answer:

WARRIOR [948]3 years ago

6 0

Answer:

i have the same thing that im confused on

Step-by-step explanation:

confused help-

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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

write the slope-intercept form of the line with a slope of 2 and a y-intercept of -4. include your work in your final answer.

IgorLugansk [536]

I cant really show you how to graph this but i can try to explain it. you would make a 4 quadrant graph and do rise over run.
1)start by making the graph
2) on the line of the graph marked Y is where you would put -4
3) From -4, you would go up (rise) by 2 and go over to the right 1 where you end is the slop

5 0

3 years ago

Can someone help me find the answer?

raketka [301]

Answer:

AC=9 cm

ABC is not a right angled triangle

Step-by-step explanation:

AB=8cm

BC=5cm

22=AB+BC+AC

AC=22-8-5

AC=9

since AC is the longest,

using Pythagoras theorem

AC^2=BC^2+AB^2

9^2=8^2+5^2

81=64+25

81=89

therefore, it is not a right angled triangle

8 0

3 years ago

Read 2 more answers

Monique is going to draw a net of a triangular pyramid. How many triangles should there be in her drawing?

Salsk061 [2.6K]

Answer:

Step-by-step explanation:

8 0

3 years ago

Solve for g in the proportion.<br><br> 42/g = 14/10<br><br> g =

NeX [460]

Cross mutiply. 42*10= 420. 14*g= 14g

14g= 420

divide by 14 for both sides

14g/14= 420/14

g= 30

Check answer by using substitution method

42/30. Divide by 3 . 42/3= 14 , 30/3= 10

14/10 and 14/10 equal to each other by dividing 2.

14/2= 7 , 10/2= 5

7/5= 7/5

Answer: g= 30

6 0

2 years ago

Read 2 more answers