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1 year ago

The interquartile range is the length that contains the middle ___ (what?) of the values in a data set

Mathematics

1 answer:

mina [271]1 year ago

7 0

Answer:

Step-by-step explanation:

The interquartile range is the difference between Q3 and Q1.

Since Q3 = 75% mark and Q1 = 25% mark, this range is representing the data between those 2 marks:

IQR = 75% - 25% = 50%

The blank is 50%

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HELP MATH ASAP! MARKING BRANLIST

qwelly [4]

Answer:

-4/3

Step-by-step explanation:

Use the slope formula:

$\frac{y2-y1}{x2-x2}$

-4 - 12/ 13 - 1

-16/12

Simplifies to -4/3

8 0

2 years ago

Daniel has a 3/4 pound bag of dog

Leya [2.2K]

Answer: 6 days

Step-by-step explanation:

if you make them into common denominators, it would be 6/8, so it would last 6 days

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2 years ago

Read 2 more answers

Y=-2x+5<br> X | Y<br> ____<br> 5<br> -1<br> 2<br> 0

Y_Kistochka [10]

Answers:

Step-by-step explanation:

Plug in the x into the equation and solve.

<h3>Value 1: X = 5</h3>

Y = -2(5) + 5

Y = -10 + 5

Y = -5

<h3>Value 2: X = -1</h3>

Y = -2(-1) + 5

Y = 2 + 5

Y = 7

<h3>Value 3: X = 2</h3>

Y = -2(2) + 5

Y = -4 + 5

Y = 1

<h3>Value 4: X = 0</h3>

Y = -2(0) + 5

Y = 0 + 5

Y = 5

I'm always happy to help :)

8 0

2 years ago

Allometric relations often can be modeled by f(x)= x^b, where a and b are constants. One study showed that for a male fiddler

dolphi86 [110]

Answer:

a) $f(2)= 0.448 (2)^{1.21}= 1.036 grams$

b) $0.5 = 0.448 x^{1.21}$

Dividing both sides by 0.448 we got:

$\frac{0.5}{0.448} = x^{1.21}$

We can appy the exponent $\frac{1}{1.21}$ in both sides of the equation and we got:

$(\frac{0.5}{0.448})^{\frac{1}{1.21}} = x= 1.095grams$

Step-by-step explanation:

For this case we know the following function:

$f(x) = 0.448 x^{1.21}$

The notation is: x is the weight of the crab in grams, and the output f(x) is the weight of the claws in grams.

Part a

For this case we just need to replace x = 2 gram in the function and we got:

$f(2)= 0.448 (2)^{1.21}= 1.036 grams$

Part b

For this case we know tha value for $f(x) =0.5$ and we want to find the value of x who satisfy this condition:

$0.5 = 0.448 x^{1.21}$

Dividing both sides by 0.448 we got:

$\frac{0.5}{0.448} = x^{1.21}$

We can appy the exponent $\frac{1}{1.21}$ in both sides of the equation and we got:

$(\frac{0.5}{0.448})^{\frac{1}{1.21}} = x= 1.095grams$

8 0

3 years ago

What is the value of n in 9x27+2x31-28=n

rosijanka [135]

243 + 62-28 = n
243 + 34 = n
277 = n
n= 277

6 0

3 years ago

Read 2 more answers