Question

question 1
the volume of a fish tank is 20 cubic feet. if the density is 0.2 fish/feet cubed, how many fish are in the tank?

a 2
b 4
c 50
d 100
e i have not learned it as yet.

question 2
a pair of dice is rolled. what is the probability that the sum of the two dice will be greater than 8 given that the first die rolled is a 5?

a 1/6
b 2
c 1/2
d 3/4
e i have not learned this yet.

Explanation:

Question 1

Step1: Recall the formula for density

Density ($
ho$) is defined as the number of fish ($N$) per unit volume ($V$), so the formula is $
ho=\frac{N}{V}$, which can be rearranged to $N =
ho\times V$.

Step2: Substitute the given values

We know that $
ho = 0.2$ fish/ cubic foot and $V = 20$ cubic feet. Substituting these values into the formula, we get $N=0.2\times20$.

Step3: Calculate the result

$0.2\times20 = 4$.

Step1: Determine the sample space for the second die

Given that the first die is 5, the second die can take values from 1 to 6 (since a die has 6 faces). So the sample space $S=\{1,2,3,4,5,6\}$, and $n(S) = 6$.

Step2: Find the favorable outcomes

We need the sum of the two dice to be greater than 8. Let the second die be $x$. Then $5 + x>8$, which simplifies to $x > 3$. So the favorable values of $x$ are 4, 5, 6. Thus, the set of favorable outcomes $E=\{4,5,6\}$, and $n(E)=3$.

Step3: Calculate the probability

The probability $P$ is given by the formula $P=\frac{n(E)}{n(S)}$. Substituting $n(E) = 3$ and $n(S)=6$, we get $P=\frac{3}{6}=\frac{1}{2}$.

Answer:

b. 4

Question 2