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loris [4]
3 years ago
6

Ggl meet code jgz-inwt-sdp

Mathematics
2 answers:
PolarNik [594]3 years ago
7 0

Answer:

y

Step-by-step explanation:

Oxana [17]3 years ago
3 0

Answer:

Thank you for the points

Step-by-step explanation:

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Which of the following equations describes the line shown
wariber [46]
(-4,4) (2,1)
gradient = (1-4)/(2--4) = -1/2
y = mx + c
y = -1/2x + c
Replace point (2,1) in the equation
1=-1/2(2) +c
c = 2
Equation : y = -1/2x + 2
y-2 = -1/2x
Answer is C.
Hope it helped!
6 0
3 years ago
What is <br> equivalent to 15x + 3?
scZoUnD [109]
The answer is
3(5x+1)
8 0
3 years ago
Read 2 more answers
14.4% of a voters in middleville voted for rocky rufus for mayor. if 2500 people voted, how many votes did rocky get?
il63 [147K]

Answer:

360 votes

Step-by-step explanation:

Move the decimal to the left 2 times and you'll get .144. Multiply that by 2500, and you have your answer.

3 0
3 years ago
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!}&#10;

                                                           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!}&#10;

                      \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!}&#10;

                      \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
On a map the scale is 1 inch = 125 miles. What is the actual distance between two cities of the map distance is 4 inches. ( PLS
kogti [31]

Answer:

500 miles

Step-by-step explanation:

Please mark as brainliest.

8 0
3 years ago
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