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Luda [366]
3 years ago
8

Miss has a sheet of wrapping paper 85cm in length and 60 cm in width. She cuts the paper into rectangular pieces of length 20 cm

and width 10 cm to wrap a gift.
(a) what is the greatest number of pieces that can be cut from the sheet of paper?
(b) Miss wants to wrap 100 gifts. How many sheets of wrapping paper will she need?

Mathematics
1 answer:
dalvyx [7]3 years ago
7 0

Answer:

Part a - 25 pieces

Part b - 4 sheets

Step-by-step explanation:

<u>Part a</u>

Think of this in terms of how many "units" can the total paper be cut into. Easiest way to keep track of units is by their area. In other words, take the total area and see how many unit areas fit into it:

Number of units = total area / unit area

N = (85 x 60) / (20 x 10)

N = 5100 / 200

N = 25.5

Therefore the greatest number of pieces the paper can be cut into is 25 units (the 0.5 doesn't count since it's not a full pieces).

<u>Part b</u>

Again, back to units. This time the "total" is 100, and our unit size is 25.5. So:

Number of units = total gifts / gifts per unit

N = 100 / 25.5

N = 3.9

Therefore 4 sheets of wrapping paper will be needed (you need to round the 3.9 up because wrapping paper sheets only come in whole numbers for the purposes of this question).

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<u></u>RT corresponds to TR. correct option b.

<u>Step-by-step explanation:</u>

In the given parallelogram or rectangle , we have a diagonal RT . We need to find which side is in correspondence with side/Diagonal RT of parallelogram URST .

<u>Side TU:</u>

In triangle UTR , we see that TR is hypotenuse and is the longest side among UR & TU . So , TR can never be equal in length to UR & TU . So there's no correspondence of Side TU with RT.

<u>Side TR:</u>

Since, direction of sides are not mentioned here , we can say that TR & RT is parallel & equal to each other . So , TR is in correspondence with side/Diagonal RT of parallelogram URST .

<u>Side UR:</u>

In triangle UTR , we see that TR is hypotenuse and is the longest side among UR & TU . So , TR can never be equal in length to UR & TU . So there's no correspondence of Side UR with RT.

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What is a equations ?
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Find the coefficient of variation for each of the two sets of data, then compare the variation. Round results to one decimal pla
svp [43]

Here is  the correct computation of the question given.

Find the coefficient of variation for each of the two sets of data, then compare the variation. Round results to one decimal place. Listed below are the systolic blood pressures (in mm Hg) for a sample of men aged 20-29 and for a sample of men aged 60-69.

Men aged 20-29:      117      122     129      118     131      123

Men aged 60-69:      130     153      141      125    164     139

Group of answer choices

a)

Men aged 20-29: 4.8%

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There is substantially more variation in blood pressures of the men aged 60-69.

b)

Men aged 20-29: 4.4%

Men aged 60-69: 8.3%

There is substantially more variation in blood pressures of the men aged 60-69.

c)

Men aged 20-29: 4.6%

Men aged 60-69: 10.2 %

There is substantially more variation in blood pressures of the men aged 60-69.

d)

Men aged 20-29: 7.6%

Men aged 60-69: 4.7%

There is more variation in blood pressures of the men aged 20-29.

Answer:

(c)

Men aged 20-29: 4.6%

Men aged 60-69: 10.2 %

There is substantially more variation in blood pressures of the men aged 60-69.

Step-by-step explanation:

From the given question:

The coefficient of variation can be determined by the relation:

coefficient \ of  \ variation = \dfrac{standard \ deviation}{mean}*100

We will need to determine the coefficient of variation both men age 20 - 29 and men age 60 -69

To start with;

The coefficient of men age 20 -29

Let's first find the mean and standard deviation before we can do that ;

SO .

Mean = \dfrac{\sum \limits^{n}_{i-1}x_i}{n}

Mean = \frac{117+122+129+118+131+123}{6}

Mean = \dfrac{740}{6}

Mean = 123.33

Standard deviation  = \sqrt{\dfrac{\sum (x_i- \bar x)^2}{(n-1)} }

Standard deviation =\sqrt{\dfrac{(117-123.33)^2+(122-123.33)^2+...+(123-123.33)^2}{(6-1)} }

Standard deviation  = \sqrt{\dfrac{161.3334}{5}}

Standard deviation = \sqrt{32.2667}

Standard deviation = 5.68

The coefficient \ of  \ variation = \dfrac{standard \ deviation}{mean}*100

coefficient \ of  \ variation = \dfrac{5.68}{123.33}*100

Coefficient of variation = 4.6% for men age 20 -29

For men age 60-69 now;

Mean = \dfrac{\sum \limits^{n}_{i-1}x_i}{n}

Mean = \frac{   130 +    153    +  141  +    125 +   164  +   139}{6}

Mean = \dfrac{852}{6}

Mean = 142

Standard deviation  = \sqrt{\dfrac{\sum (x_i- \bar x)^2}{(n-1)} }

Standard deviation =\sqrt{\dfrac{(130-142)^2+(153-142)^2+...+(139-142)^2}{(6-1)} }

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Thus; Option C is correct.

Men aged 20-29: 4.6%

Men aged 60-69: 10.2 %

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