All categories

3 years ago

What is the radius of the circle x2+(y+6)2=25

Mathematics

1 answer:

GenaCL600 [577]3 years ago

6 0

Answer: y=7

Step-by-step explanation:

Use the vertex form to determine the values of

, and

and use these values to find the direction the parabola opens, vertex, focus, axis of symmetry, and directrix.

Direction: Opens Down

Vertex:

(

)

Focus:

(

)

You might be interested in

What is an expression that represents the situation $5.50 divided by $44

Dahasolnce [82]

Answer:

5.50/40

Step-by-step explanation:

5.50 divided by $44

this 5.50/40 is ths same as this

5.50 divided by $44

5.50/40 is your answer

6 0

4 years ago

A certain whale can eat up to 2,400 pounds of food in a day. Which of the following best represents an estimate of the pounds th

steposvetlana [31]

The answer is A because if you do 10*10*10 that would give you 1k and if u multiply that by 2 you get 2k

6 0

3 years ago

Read 2 more answers

The average of 6 number is 4. A seventh number is added and the new average is 5. Find the seventh number

KatRina [158]

The seventh number would have to be 11

6 0

3 years ago

Can someone help me with this PLZ

nydimaria [60]

What should we be helping you with?

I don't see the question?

Please state it clearly so that I can help you. I would love to help you!

7 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!}
$

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