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Alinara [238K]
2 years ago
15

If it requires 6 pounds of wild caught animals to create 1 pound of salmon, what percent of the wild caught animals are wasted?

Round your answer to the nearest 10th.
Mathematics
1 answer:
Oduvanchick [21]2 years ago
4 0

The percentage of animal wasted is the number of animal wasted over 100

85.7%  of the wild caught animals are wasted

<h3>How to calculate the animal wasted</h3>

From the question, we have:

Animals caught = 6 pounds

Salmon = 1 pound

The total weight in the system is:

Total = 6 pounds + 1 pound

So, we have:

Total = 7 pounds

The percent of wild caught animals wasted is then calculated as:

\%Wasted = \frac{6}{7} * 100\%

Simplify

\%Wasted = 85.7\%

Hence, 85.7%  of the wild caught animals are wasted

Read more about percentage at:

brainly.com/question/843074

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There are 8 members on a board of directors. If they must form a subcommittee of 3 members, how many different subcommittees are
Katyanochek1 [597]

Answer:

56 possible different subcommittees

Step-by-step explanation:

This is a combination problem.

In combination probability, the formula is;

nCr = n!/((n - r)! × r!)

Now, out of the 8 members on the board of directors, they now want to form a subcommittee of 3 members.

Thus, the different subcommittees possible is given by;

8C3 = 8!/((8 - 3)! × (3!)) = 56

Thus, there are 56 possible different subcommittees

3 0
2 years ago
A polygon has vertices whose coordinates are A(1, 4), B(4, -1), C(-1, -4), and D(-4, 1). Use the midpoint formula to determine w
Alenkinab [10]

Answer:

The diagonals of the polygon ABCD bisect each other

Step-by-step explanation:

The given vertices of the polygon are;

A(1, 4), B(4, -1), C(-1, -4), and D(-4, 1)

Given that the vertices of the polygon in clockwise order from the top left corner are ABCD, we have the diagonals as AC and DB

By the midpoint formula, we have;

(x_m , \ y_m) = \left (\dfrac{x_1 + x_2}{2} , \ \dfrac{y_1 + y_2}{2} \right )

Where;

(x_m , \ y_m) = The midpoint coordinates

Therefore, the midpoint of diagonal AC = \left (\dfrac{1 + (-1)}{2} , \ \dfrac{4 + (-4)}{2} \right ) = (0, 0)

The midpoint  of diagonal DB = \left (\dfrac{(-4) +4}{2} , \ \dfrac{1 + (-1)}{2} \right ) = (0, 0)

Therefore, given that the coordinates of the location of the midpoint of the diagonals AC and DB is the same, the diagonals bisect each other

Each diagonal passes through the midpoint of the other diagonal.

3 0
3 years ago
Daniel is starting his own sewing business. Daniel has calculated that he needs to earn $360 per week to support his family. If
Anna007 [38]

Answer:

$12 the hour

Step-by-step explanation: $360 divided by 30 is 12, meaning he will need to make a minimum of 12 an hour to support his family.

4 0
3 years ago
Two friends shop for fresh fruit. Elena buys a watermelon for $7.45 and 3 pounds ofcherries. Taniy buys a pineapple for $3.35 an
Scrat [10]

Let's use the variable E to represent the total value spent by Elena and T to represent the total value spent by Taniy.

If Elena spent $7.45 plus the cost of 3 pounds of cherries, the expression for the amount she spent is:

E=7.45+3p

Taniy spent $3.35 plus the cost of 2 pounds of cherries, so the expression for the amount the spent is:

T=3.35+2p

To find how much more Elena spent, let's subtract both expressions:

\begin{gathered} E-T\\ \\ =7.45+3p-(3.35+2p)\\ \\ =7.45+3p-3.35-2p\\ \\ =4.1+p \end{gathered}

Therefore the answer is 4.1 + p.

3 0
1 year ago
In a recent survey of college professors, it was found that the average amount of money spent on entertainment each week was nor
Alenkinab [10]

Answer:

0.0918

Step-by-step explanation:

We know that the average amount of money spent on entertainment is normally distributed with mean=μ=95.25 and standard deviation=σ=27.32.

The mean and standard deviation of average spending of sample size 25 are

μxbar=μ=95.25

σxbar=σ/√n=27.32/√25=27.32/5=5.464.

So, the average spending of a sample of 25 randomly-selected professors is normally distributed with mean=μ=95.25 and standard deviation=σ=27.32.

The z-score associated with average spending $102.5

Z=[Xbar-μxbar]/σxbar

Z=[102.5-95.25]/5.464

Z=7.25/5.464

Z=1.3269=1.33

We have to find P(Xbar>102.5).

P(Xbar>102.5)=P(Z>1.33)

P(Xbar>102.5)=P(0<Z<∞)-P(0<Z<1.33)

P(Xbar>102.5)=0.5-0.4082

P(Xbar>102.5)=0.0918.

Thus, the probability that the average spending of a sample of 25 randomly-selected professors will exceed $102.5 is 0.0918.

5 0
3 years ago
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