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4 years ago

5

a delivery man weighs 200 pounds. he is delivering cartons that each weigh 48 pounds. he wants to know how many cartons he can s

afely put on the elevator at one time. Let c represent the number of cartons. Write an inequality that represents this situation.

1 answer:

Anastasy [175]4 years ago

8 0

The inequality is:

48 c + 200 ≤ (the maximum load allowed on the elevator) .

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6. Calculate the area of the octagon in the<br> figure below.

Kryger [21]

Answer:

$41\text{ [units squared]}$

Step-by-step explanation:

The octagon is irregular, meaning not all sides have equal length. However, we can break it up into other shapes to find the area.

The octagon shown in the figure is a composite figure as it's composed of other shapes. In the octagon, let's break it up into:

4 triangles (corners)
3 rectangles (one in the middle, two on top after you remove triangles)

<u>Formulas</u>:

Area of rectangle with length $l$ and width $w$ : $A=lw$
Area of triangle with base $b$ and height $h$ : $A=\frac{1}{2}bh$

<u>Area of triangles</u>:

All four triangles we broke the octagon into are congruent. Each has a base of 2 and a height of 2.

Thus, the total area of one is $A=\frac{1}{2}\cdot 2\cdot 2=2\text{ square units}$

The area of all four is then $2\cdot 4=8$ units squared.

<u>Area of rectangles</u>:

The two smaller rectangles are also congruent. Each has a length of 3 and a width of 2. Therefore, each of them have an area of $3\cdot 2=6$ units squared, and the both of them have a total area of $6\cdot 2=12$ units squared.

The last rectangle has a width of 7 and a height of 3 for a total area of $7\cdot 3=21$ units squared.

Therefore, the area of the entire octagon is $8+12+21=\boxed{41\text{ [units squared]}}$

4 0

3 years ago

Calculate the sample mean and sample variance for the following frequency distribution of heart rates for a sample of American a

spin [16.1K]

Answer:

$Mean = 68.9$

$s^2 =18.1$ --- Variance

Step-by-step explanation:

Given

$\begin{array}{cccccc}{Class} & {51-58} & {59-66} & {67-74} & {75-82} & {83-90} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}$

Solving (a): Calculate the mean.

The given data is a grouped data. So, first we calculate the class midpoint (x)

For 51 - 58.

$x = \frac{1}{2}(51+58) = \frac{1}{2}(109) = 54.5$

For 59 - 66

$x = \frac{1}{2}(59+66) = \frac{1}{2}(125) = 62.5$

For 67 - 74

$x = \frac{1}{2}(67+74) = \frac{1}{2}(141) = 70.5$

For 75 - 82

$x = \frac{1}{2}(75+82) = \frac{1}{2}(157) = 78.5$

For 83 - 90

$x = \frac{1}{2}(83+90) = \frac{1}{2}(173) = 86.5$

So, the table becomes:

$\begin{array}{cccccc}{x} & {54.5} & {62.5} & {70.5} & {78.5} & {86.5} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}$

The mean is then calculated as:

$Mean = \frac{\sum fx}{\sum f}$

$Mean = \frac{54.5*4+62.5*3+70.5*11+78.5*13+86.5*4}{6+3+11+13+4}$

$Mean = \frac{2547.5}{37}$

$Mean = 68.9$ -- approximated

Solving (b): The sample variance:

This is calculated as:

$s^2 =\frac{\sum (x - \overline x)^2}{\sum f - 1}$

So, we have:

$s^2 =\frac{(54.5-68.9)^2+(62.5-68.9)^2+(70.5-68.9)^2+(78.5-68.9)^2+(86.5-68.9)^2}{37 - 1}$

$s^2 =\frac{652.8}{36}$

$s^2 =18.1$ -- approximated

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3 years ago

Proportions and Similarity

nekit [7.7K]

$\bf \cfrac{length}{width}\qquad \qquad \stackrel{A}{\cfrac{4}{5}}\qquad \stackrel{B}{\cfrac{20}{25}}\implies \cfrac{4}{5}\qquad \checkmark$

5 0

3 years ago

If u need points just answer this

sasho [114]

OK im answering ............................................................................................................

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3 years ago

Read 2 more answers

A salesman sold twice as much pears in the afternoon than in the morning. If he sold 360 kilograms of pears that day, how many k

serg [7]

Answer: 240 kg

Step-by-step explanation: Divide 360 by 3. That would equal 120. 120 times 2 equals 240

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3 years ago