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dimaraw [331]
3 years ago
5

A student measured the dimensions of a table and recorded the length as 103.50cm and the width as 73.75cm. according to the stud

ent’s calculator, the area is 7633.125cm2. what should the student report? explain your answer.
Mathematics
1 answer:
PolarNik [594]3 years ago
8 0
The table is usually the shape of a rectangle. The area is equal to the length times the width. To make sure, let's recalculate the solution.

Area = length × width
Area = 103.50 cm × 73.75 cm
Area = <span>7,633.125 cm</span>²

So, we established that the answer is correct. Now, I'll introduce the concept of significant figures. There are rules regarding significant figures to be applied conventionally in measurements and calculations. This is to increase precision of the data. Significant figures are digits that carry meaning. When you are given data with a specific number of significant figures, the final answer should also contain the same number of significant figures. In convention, the rule says that the zero's after non-zero digits after the decimal point are significant. Also, all non-zero digits are significant. The value 103.50 has 5 significant figures, and 73.75 has 4. When it comes to multiplication, the answer should contain the least significant figures. So, we should have 4 significant figures. Therefore, the final answer to be reported should be 7,633 cm².
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Answer:

T = 75 + 116.9*N

Step-by-step explanation:

We have that the equation for Michael's savings is:

A = 75 + 50.85*N, where N is the number of weeks

And we have that the equation for his wife's savings is:

B = 65.95*N.

So, to find the total amount saved using both plans combined (T), we have to sum A and B:

T = A + B = 75 + 50.85*N + 65.95*N = 75 + 116.9*N

So the equation that relates T to N is:

T = 75 + 116.9*N

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A sample of blood pressure measurements is taken for a group of​ adults, and those values​ (mm Hg) are listed below. The values
Slav-nsk [51]

Answer:

Systolic on right

\hat{CV} =\frac{18.68}{127.5}=0.147

Systolic on left

\hat{CV} =\frac{12.65}{74.2}=0.170

So for this case we have more variation for the data of systolic on left compared to the data systolic on right but the difference is not big since 0.170-0.147 = 0.023.

Step-by-step explanation:

Assuming the following data:

Systolic (#'s on right) Diastolic (#'s on left)

117; 80

126; 77

158; 76

96; 51

157; 90

122; 89

116; 60

134; 64

127; 72

122; 83

The coefficient of variation is defined as " a statistical measure of the dispersion of data points in a data series around the mean" and is defined as:

CV= \frac{\sigma}{\mu}

And the best estimator is \hat {CV} =\frac{s}{\bar x}

Systolic on right

We can calculate the mean and deviation with the following formulas:

[te]\bar x = \frac{\sum_{i=1}^n X_i}{n}[/tex]

s= \frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}

For this case we have the following values:

\bar x = 127.5, s= 18.68

So then the coeffcient of variation is given by:

\hat{CV} =\frac{18.68}{127.5}=0.147

Systolic on left

For this case we have the following values:

\bar x = 74.2 s= 12.65

So then the coeffcient of variation is given by:

\hat{CV} =\frac{12.65}{74.2}=0.170

So for this case we have more variation for the data of systolic on left compared to the data systolic on right but the difference is not big since 0.170-0.147 = 0.023.

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Answer: the lines on the outside mean absolute value and the absolute value of -3 is 3

Step-by-step explanation:

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Suppose you choose a team of two people from a group of n &gt; 1 people, and your opponent does the same (your choices are allow
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Answer:

The number of possible choices of my team and the opponents team is

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Step-by-step explanation:

selecting the first team from n people we have \left(\begin{array}{ccc}n\\1\\\end{array}\right)  = n possibility and choosing second team from the rest of n-1 people we have \left(\begin{array}{ccc}n-1\\1\\\end{array}\right) = n-1

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Therefore, the total possibility is \frac{n(n-1)}{2}

Since our choices are allowed to overlap, the second team is \frac{n(n-1)}{2}

Possibility of choosing both teams will be

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1³ + 2³ + ........... + n³ =[\frac{n(n+1)}{2}] ^{2}

1³ + 2³ + ............ + (n-1)³ = [x^{2} \frac{n(n-1)}{2}] ^{2}

=\left[\begin{array}{ccc}n-1\\E\\i=1\end{array}\right] =   [\frac{n(n-1)}{2}]^{3}

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