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horrorfan [7]
3 years ago
5

"Death Valley Hike"

Mathematics
2 answers:
Amanda [17]3 years ago
7 0
I would go with Either E or A
Goryan [66]3 years ago
7 0

Answer:

I'm pretty sure E

Step-by-step explanation:

think about it he traveled 70 feet in 5 minutes so he would still be below sea level

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CAN ANYONE HELP ME ANSWER THIS PLEASE​
HACTEHA [7]

Answer: 14/45

Step-by-step explanation: The probability that they first event will occur is 2/10. The probability that the second event will occur is 1/9. If you multiply those like it says in the hint you will get 14/45. (I think this is right)

5 0
3 years ago
What are the solutions of the equation?
Olegator [25]
The correct answer would be 41
7 0
3 years ago
Help me with this thanks for anyone who helps! :D
Kobotan [32]

Answer:

x= 90-60

x=30

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
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
WHAT IS THE RATE OF CHANGE FOR THIS
stealth61 [152]
Well first find 2 points, I am going to choose (0,0.5) and (1,0.3). So to find rate of change or Slope it is y2-y1/x2-x1 so we are doing that, .3-.5/1-0. which will be -.2/1

Your answer: -.2/1
4 0
3 years ago
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