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3 years ago

The environmental club has 8 sixth-graders, 11 seventh-graders, and 15 eighth

Mathematics

2 answers:

Tresset [83]3 years ago

8 0

Answer:

1320

Step-by-step explanation:

8 x 11 x 15 = 1320

Hope this helped :)

Bess [88]3 years ago

3 0

The correct answer is 1320. 8x11x15

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Which choice shows the numbers correctly ordered?

zvonat [6]

The correct answer is B

3 0

3 years ago

A rectangular piece of land borders a wall. The land is to be enclosed and to be into divided 3 equal plots with 200 feet of fen

baherus [9]

Answer:

Area = 2500 square feet is the largest area enclosed

Step-by-step explanation:

A rectangular piece of land borders a wall. The land is to be enclosed and to be into divided 3 equal plots with 200 feet of fencing

Let x be the length of each box and y be the width of the box

Perimeter of the box= 3(length ) + 4(width)

$200=3x+4y$

solve for y

$200=3x+4y$

$200-3x=4y$

divide both sides by 4

$y=50-\frac{3x}{4}$

Area of the rectangle = length times width

$Area = 3x \cdot y$

$Area = 3x \cdot (50-\frac{3x}{4})$

$A=150x-\frac{9x^2}{4}$

Now take derivative

$A'=150-\frac{9x}{2}$

Set it =0 and solve for x

$0=150-\frac{9x}{2}$

$150=\frac{9x}{2}$

multiply both sides by 2/9

$x=\frac{100}{3}$

$A''=-\frac{9}{2}$

For any value of x, second derivative is negative

So maximum at x= 100/3

$A=150x-\frac{9x^2}{4}$ , replace the value of x

$A=150(\frac{100}{3})-\frac{9(\frac{100}{3})^2}{4})$

Area = 2500 square feet is the largest area enclosed

3 0

3 years ago

The table shows the time Monique worked and the amount of money she earned during four different weeks.

spayn [35]

Answer:

330

Step-by-step explanation:

5 0

3 years ago

Watch help video

sertanlavr [38]

Answer:

awit wala akong maintindihan. srna agad

5 0

3 years ago

A company is studying the number of monthly absences among its 125 employees. The following probability distribution shows the l

Angelina_Jolie [31]

Answer:

(1)

$E(x) = 0.72$

$Var(x) = 1.1616$

$\sigma = 1.078$

(2)

$E(x) = 12$

$Var(x) = 1.3$

$\sigma = 1.14$

Step-by-step explanation:

Solving (1):

$\begin{array}{ccccccc}{Days} & {0} & {1} & {2} & {3} & {4}& {5} \ \\ {Probability} & {0.60} & {0.20} & {0.12} & {0.04} & {0.04} & {0.00} \ \end{array}$

(a): Mean

This is calculated as:

$E(x) = \sum x * P(x)$

So, we have:

$E(x) = 0 * 0.60 + 1 * 0.20 + 2 * 0.12 + 3 * 0.04 + 4 * 0.04 + 5 * 0.00$

$E(x) = 0.72$

Solving (b): The variance

This is calculated as:

$Var(x) = E(x^2) - (E(x))^2$

Where:

$E(x) = 0.72$

and $E(x^2) = \sum x^2 * P(x)$

So, we have:

$E(x^2) = 0^2 * 0.60 + 1^2 * 0.20 + 2^2 * 0.12 + 3^2 * 0.04 + 4^2 * 0.04 + 5^2 * 0.00$

$E(x^2) = 1.68$

So, we have:

$Var(x) = E(x^2) - (E(x))^2$

$Var(x) = 1.68 - 0.72^2$

$Var(x) = 1.1616$

Solving (c): The standard deviation.

This is calculated as:

$\sigma = \sqrt{Var(x)}$

$\sigma = \sqrt{1.1616}$

$\sigma = 1.078$

Solving (2):

$\begin{array}{ccccccc}{x} & {10} & {11} & {12} & {13} & {14}& { } \ \\ {P(x)} & {0.1} & {0.25} & {0.3} & {0.25} & {0.1} & { } \ \end{array}$

(a): Mean

This is calculated as:

$E(x) = \sum x * P(x)$

So, we have:

$E(x) = 10 * 0.10 + 11 * 0.25 + 12 * 0.3 + 13 * 0.25 + 14 * 0.1$

$E(x) = 12$

Solving (b): The variance

This is calculated as:

$Var(x) = E(x^2) - (E(x))^2$

Where:

$E(x) = 12$

and $E(x^2) = \sum x^2 * P(x)$

So, we have:

$E(x^2) = 10^2 * 0.10 + 11^2 * 0.25 + 12^2 * 0.3 + 13^2 * 0.25 + 14^2 * 0.1$

$E(x^2) = 145.3$

So, we have:

$Var(x) = E(x^2) - (E(x))^2$

$Var(x) = 145.3 - 12^2$

$Var(x) = 1.3$

Solving (c): The standard deviation.

This is calculated as:

$\sigma = \sqrt{Var(x)}$

$\sigma = \sqrt{1.3}$

$\sigma = 1.14$

8 0

3 years ago