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BartSMP [9]
2 years ago
5

I have a few questions!

Mathematics
2 answers:
UkoKoshka [18]2 years ago
6 0
First one is 6 units down
Sedaia [141]2 years ago
3 0

Answer:

okay first one is ,3

Step-by-step explanation:

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I don’t understand #6 question
romanna [79]

Answer:

B. (10, 2000)

Step-by-step explanation:

5000 - 300t = 1400ft - 1200f

=> t = 10

10 => (5000 - 300t) = 2000

So the answer is B. (10, 2000)

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6 0
2 years ago
The diameter of a circle has endpoints at (0,11) and (-6,-1). Write the equation of the circle in standard form.
Nata [24]

Answer:

(x - 3)^2 + (y - 6)^2 = 17

Step-by-step explanation:

The standard form of a circle's equation is (x-h)² + (y-k)² = r² where (h,k) is the center and r is the radius.

(x - 3)^2 + (y - 6)^2 = 17

8 0
3 years ago
Last year, there were 12,000 students at a local college. There were 2000 more female students than male students. How many male
krek1111 [17]
There is 7,000 female students, and 5,000 male students. Because, if there was 2,000 more female students then male, you have to find out what two numbers add up to 12,000. (Tip, to make it easier, take away the thousands, like 7+5)
5 0
3 years ago
Simplify: -8c + w + 6c - 5w<br><br> I will give Brainliest and Thanks
Gekata [30.6K]

Answer:

The answer is -2c - 4w

Step-by-step explanation:

You have to combine the c's and the w's and you get -2c and -4w

5 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!}&#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
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