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2 years ago

A map has a scale of 1 in : 3 km. If the actual distance is 15 km, what is the distance on the map?

Mathematics

1 answer:

Black_prince [1.1K]2 years ago

4 0

Answer:

5 in.

Step-by-step explanation:

The ratio of map distance to real distance is 1 in : 3 km meaning that 1 inch on the map describes 15 km in real life.

15 km is 5 times 3 km, so the map distance is 5 times 1 in., or 5 in.

1 in. x 5 : 3 km x 5

This is a "school math" problem where things come out nicely! Another way to look at this is with equivalent fractions.

$\frac{\text{1 in}}{\text{5 km}} = \frac{\text{m}}{\text{15 km}}$

"cross multiply"

$5m = (15)(1)\\5m=15\\m=3 \text{ in.}$

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Dual-energy X-ray absorptiometry (DXA) is a technique for measuring bone health. One of the most common measures is total body b

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Answer:

(a)

$\bar x = 0.0075$

$s = 0.0049$

(b)

B. The sample is too small to make judgments about skewness or symmetry.

Step-by-step explanation:

Given:

$n = 8$

$\begin{array}{ccccccccc}{} & {S} & {u} & {b} &{j} & {e} & {c} & {t} & {s} &{Operator} & {1} & {2} & {3} & {4} & {5} & {6} & {7} & {8} & {1} & 1.326 & 1.337 & 1.079 & 1.229 & 0.936 & 1.009 & 1.179 & 1.289 & 2 & 1.323 & 1.322 & 1.073 & 1.233 & 0.934 & 1.019 & 1.184 & 1.304 \ \end{array}$

Solving (a):

First, calculate the difference between the recorded TBBMC for both operators:

The last row which represents the difference between 1 and 2 is calculated using absolute values. So, no negative entry is recorded.

The mean is then calculated as:

$\bar x = \frac{\sum x}{n}$

$\bar x = \frac{0.003 + 0.015 + 0.006 + 0.004 + 0.002 + 0.010 + 0.005 + 0.015}{8}$

$\bar x = \frac{0.06}{8}$

$\bar x = 0.0075$

Next, calculate the standard deviation (s).

This is calculated using:

$s = \sqrt{\frac{\sum (x - \bar x)^2}{n}}$

So, we have

$s = \sqrt\frac{(0.003 - 0.0075)^2 + (0.015 - 0.0075)^2+ (0.006- 0.0075)^2+ ....... + (0.005- 0.0075)^2+ (0.015- 0.0075)^2 }{8}$ $s = \sqrt\frac{0.00019}{8}$