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3 years ago

When might the inter-quartile range be better for describing a data set than the range?

2 answers:

7 0

I think the answer is C. if the data has outliers

if the data has outliers it means that it has extreme values

, for example 1, 4000, 4500, 5000, 5500, 6000, 10000000000

1 and 10000000000 are considered as the outlier, which is not fair to be used as a range of measure

Hope this helps

Olenka [21]3 years ago

4 0

When might the inter-quartile range be better for describing a data set than the range?

Answer: First we have to understand that a interquartile is the distance between the first and third quartiles of a data set. It is the upper quartile minus lower quartile. Out of all the options shown above the one that represents when it might be better to use for describing a data set than the range is answer choice C) if the data has outliers.

I hope it helps, Regards.

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Prakash bought a new car at the dealership for $27,000. it is estimated that the value of the car will decrease 7% each year. wh

Dennis_Churaev [7]

The exponential function models the value v of the car after t years is V = 27000 * (0.93)^t

<h3>How to determine the exponential model?</h3>

The given parameters are:

Initial value, a = $27,000

Depreciation rate, r = 7%

The value of the car is then calculated as:

V = a * (1 -r)^t

Substitute known values

V = 27000 * (1 - 7%)^t

Evaluate the difference

V = 27000 * (0.93)^t

Hence, the exponential function models the value v of the car after t years is V = 27000 * (0.93)^t

Read more about exponential function at:

brainly.com/question/11464095

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1 year ago

I need help with 16 and 17

vredina [299]

16 is 40 because you just need to add 16 to 14 and to 10 which is equal 40

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2 years ago

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Please help with 7 and 8

Serhud [2]

For #7 - part a: greatest common factor (GCF) is 5.

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3 years ago

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Lester’s car can go 15.4 miles on 1 gallon of gas. How far can he go on 0.7 gallon?

MAVERICK [17]

Answer:

He can go 10.78 miles

Step-by-step explanation:

This problem can be solve by creating a proportion, knowing that 15.4 miles is to one gallon the same as "x" miles (our unknown) is to 0.7 gallons. This can be set in math terms as the equation that relates these two ratios of miles per gallon:

$\frac{15.4\,miles}{1\,gallon} =\frac{x }{0.7\,gallon}$

and to solve for the unknown number of mile, we multiply both sides of the equation by "0.7 gallon" in order to isolate our unknown:

$\frac{15.4\,miles}{1\,gallon} =\frac{x}{0.7\,gallon}\\\frac{15.4\,miles\,*\,0.7\,gallon}{1\,gallon} =x \\x=10.78\,miles$

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3 years ago

A jeweler wants to make 14 grams of an alloy that is precisely 75% gold.. The jeweler has alloys that are 25% gold, 50% gold, &a

Goryan [66]

Given that the jeweler has alloys that are 25% gold, 50% gold, and 82% gold.

As he wants to make 14 grams of an alloy by adding two different alloys that is precisely 75% gold, so one alloy must have a percentage of gold more than 75%.

One alloy is 82% gold and, the second can be chosen between 25% gold, 50% gold, so there are two cases.

Case 1: 82% gold + 50% gold

Let x grams of 82% gold and y grams of 50% gold added to make x+y=14 grams of 75% gold, so

75% of 14 = 82% of x + 50% of y

$\Rightarrow 75/100 \times 14 = 82/100 \times x + 50/100 \times y \\\\$

$\Rightarrow 75/100 \times 14 = 82/100 \times x + 50/100 \times (14-x)$ [as x+y=14]

$\Rightarrow 75 \times 14 = 82 \times x + 50 \times (14-x) \\\\\Rightarrow 75 \times 14 = 82 \times x + 50 \times14-50\times x \\\\\Rightarrow 75 \times 14 = 32 \times x + 50 \times14 \\\\\Rightarrow 32 \times x =75 \times 14 - 50 \times14 \\\\$

$\Rightarrow x =(25 \times 14)/32=10.9375$ grams

and $y = 14-x= 14-10.9375=3.0625$ grams.

Hence, 10.9375 grams of 82% gold and 3.0625 grams of 50% gold added to make 14 grams of 75% gold.

Case 2: 82% gold + 25% gold

Let x grams of 82% gold and y grams of 25% gold added to make x+y=14 grams of 75% gold, so

75% of 14 = 82% of x + 25% of y

$\Rightarrow 75/100 \times 14 = 82/100 \times x + 25/100 \times y \\\\\Rightarrow 75/100 \times 14 = 82/100 \times x + 25/100 \times (14-x) \\\\ \Rightarrow 75 \times 14 = 82 \times x + 25 \times (14-x) \\\\\Rightarrow 75 \times 14 = 82 \times x + 25 \times14-25\times x \\\\\Rightarrow 75 \times 14 = 57 \times x + 25 \times14 \\\\\Rightarrow 57 \times x =75 \times 14 - 25 \times14 \\\\$

$\Rightarrow x =(50 \times 14)/57=12.28$ grams

and $y = 14-x= 14-12.28=1.72$ grams.

Hence, 12.28 grams of 82% gold and 1.72 grams of 50% gold added to make 14 grams of 75% gold.

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2 years ago