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2 years ago

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:

You might be interested in

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)

HOPE THIS HELPS AND HAVE A NICE DAY <3

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 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!}
$

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