a pet store keeps 4 small fish in every 10 gallons of water how many gallons of water is needed for 36 fish

Mathematics

2 answers:

Lynna [10]3 years ago

8 0

90 gallons of water to keep 36 fish

faltersainse [42]3 years ago

7 0

90 gallons of water

You might be interested in

How do I prove that 2 cot 2x = cot x-tan x?

BlackZzzverrR [31]

We will start off working on the right hand side.
cot x - tan x 
= [cos x / sin x] - [sin x / cos x] 
= [(cos x)^ 2 - (sin x)^2] / [sin x cos x] 

This is where it gets a bit tougher if you do not have your formula list with you. 
(cos x)^ 2 - (sin x)^2 = cos(2x) 
sin 2x = 2 sin x cos x 

Note that by arranging the second formula, we will have sin x cos x = (1/2) sin 2x 

Hence, we will get: 
[(cos x)^ 2 - (sin x)^2] / [sin x cos x] 
= [cos 2x] / (1/2)[sin 2x] 
= 2[cos 2x] / [sin 2x] 
= 2cot 2x

7 0

3 years ago

Suppose that the probability that any particle emitted by a radioactive material will penetrate a certain shield is 0.01. If 10

AnnyKZ [126]

Answer:

a)P=0.42

b) $n\geq 297$

Step-by-step explanation:

We have a binomial distribution, since the result of each experiment admits only two categories (success and failure) and the value of both possibilities is constant in all experiments. The probability of getting k successes in n trials is given by:

$P=\begin{pmatrix}n\\ k\end{pmatrix} p^k(1-p)^{n-k}=\frac{n!}{k!(n!-k!)}p^k(1-p)^{n-k}$

a) we have k=2, n=10 and p=0.01:

$P=\frac{10!}{2!(10!-2!)}0.01^2(1-0.01)^{10-2}\\P=\frac{10!}{2!*8!}0.01^2(0.99)^{8}\\P=45*0.01^2(0.99)^8=0.42$

b) We have, $1-(1-p)^n=P$ , Here P is the probability that at least one particle will penetrate the shield, this probabity has to be equal or greater than 0.95. Therefore, this will be equal to subtract from the total probability, the probability that the particles do not penetrate raised to the total number of particles.