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anyanavicka [17]
3 years ago
15

The measure of dispersion which is not measured in the same units as the original data is the

Mathematics
1 answer:
Ahat [919]3 years ago
7 0
D the answer is d
stop cheating
You might be interested in
. A clothing manufacturer needs 2.4 yards of fabric to make a coat and 1.6 yards of fabric to make a matching crop pants.
Sedaia [141]
Convert to an equation y in terms of x:
Y = (-2.4x + 48)/1.6
X Intercept is where y = 0
0 = -2.4x/1.6 + 48/1.6
X = 20

Y intercept is where x = 0
Y = (-2.4*0)/1.6 + 48/1.6
Y = 30

X and y are how coats and pants are made respectfully so the x intercept is where y = 0, meaning no pants are made, only jackets. The y intercept is the opposite, it’s how many pants can be made if no jackets are made.
4 0
2 years ago
18. The Tennis for Champs company is looking into new ways to package tennis balls. The packaging engineer at the company is exp
9966 [12]

Answer:

Amount of empty space in the cylinder = 395.64 cm³

Step-by-step explanation:

Since the cylinder contains 3 tennis balls each measuring 6 cm in diameter, we can say that the diameter of the cylinder is 6 cm because the balls are on top of each other. Since we have 18 cm is the height of the cylinder,then;

Formula for volume of a cylinder is;

V = πr²h

We are to use π = 3.14

Thus, V_cylinder = 3.14 × (6/2)² × 18

V_cylinder = 508.68 cm³

Volume of a tennis ball is;

V_tennis ball = (4/3)πr³

V_tennis ball = (4/3) × 3.14 × (6/2)³

V_tennis ball = 37.68 cm³

Thus, volume of 3 tennis balls = 3 × 37.68 = 113.04 cm³

Amount of empty space in the cylinder = V_cylinder - V_3 tennnis ball = 508.68 - 113.04 = 395.64 cm³

8 0
2 years ago
Using the slope formula, find the slope of the line through the given points <br> 8,5 and 6,1
zheka24 [161]

8,5 and 6,1

The slope is 2


8 0
3 years ago
PLEASE HELP, 46 PTS!!!! ANSWER QUICKLY AND WITH FULL REPONCE AND SHOW UR WORK!!
kramer
Please limit your posts to one or two problems each.  Your demands ("show work," "answer quickly," "give a full response") should apply to YOU, not to people whom you hope will do your work for you.

<span>2. A pool company is trying out several new drains. Drain A empties a pool at a rate of 2 gal/min. Drain B empties a pool at a rate of 5 gal/min. One pool has 108 gal of water in it. Write an equation for each drain that shows the amount of time it takes to empty the pool. Then solve each equation showing your work.</span>

Drain A:  rate is 2 gal/min.  We do not know which pool has 108 gallons of water in it.  If it's Pool A, then the time required to empty Pool A is

108 gallons  
------------------- = 54 minutes
2 gallons/min 

If it's Pool B, then we can represent the amount of water in Pool B by b.  Then the time required to empty Pool B is 

   b (gallons)
------------------ = (b/5) minutes
5 gallons/min
5 0
3 years ago
The estimated daily living costs for an executive traveling to various major cities follow. The estimates include a single room
Alexandra [31]

Answer:

\bar x = 260.1615

\sigma = 70.69

The confidence interval of standard deviation is: 53.76 to 103.25

Step-by-step explanation:

Given

n =20

See attachment for the formatted data

Solving (a): The mean

This is calculated as:

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

So, we have:

\bar x = \frac{242.87 +212.00 +260.93 +284.08 +194.19 +139.16 +260.76 +436.72 +355.36 +.....+250.61}{20}

\bar x = \frac{5203.23}{20}

\bar x = 260.1615

\bar x = 260.16

Solving (b): The standard deviation

This is calculated as:

\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}

\sigma = \sqrt{\frac{(242.87 - 260.1615)^2 +(212.00- 260.1615)^2+(260.93- 260.1615)^2+(284.08- 260.1615)^2+.....+(250.61- 260.1615)^2}{20 - 1}}\sigma = \sqrt{\frac{94938.80}{19}}

\sigma = \sqrt{4996.78}

\sigma = 70.69 --- approximated

Solving (c): 95% confidence interval of standard deviation

We have:

c =0.95

So:

\alpha = 1 -c

\alpha = 1 -0.95

\alpha = 0.05

Calculate the degree of freedom (df)

df = n -1

df = 20 -1

df = 19

Determine the critical value at row df = 19 and columns \frac{\alpha}{2} and 1 -\frac{\alpha}{2}

So, we have:

X^2_{0.025} = 32.852 ---- at \frac{\alpha}{2}

X^2_{0.975} = 8.907 --- at 1 -\frac{\alpha}{2}

So, the confidence interval of the standard deviation is:

\sigma * \sqrt{\frac{n - 1}{X^2_{\alpha/2} } to \sigma * \sqrt{\frac{n - 1}{X^2_{1 -\alpha/2} }

70.69 * \sqrt{\frac{20 - 1}{32.852} to 70.69 * \sqrt{\frac{20 - 1}{8.907}

70.69 * \sqrt{\frac{19}{32.852} to 70.69 * \sqrt{\frac{19}{8.907}

53.76 to 103.25

8 0
2 years ago
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