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snow_tiger [21]
3 years ago
13

First to answer this correctly will get a brainlliest.

Mathematics
1 answer:
irina [24]3 years ago
7 0
The answer is y=mx+b y=(-2)-(1/2)
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How is the interquartile range calculated? HELP WITH APEX PLEASE!
AfilCa [17]

By definition the interquartile range or IQR is obtained when the 25th percentile or Q_{1} is subtracted from the 75th percentile or Q_{3}.

Thus, from the above definition, the formula for IQR is:

IQR=Q_{3}-Q_{1}

As we can see from the provided options that the definition of IQR matches the expression provided in Option A and thus, Option A is the correct answer.

8 0
3 years ago
Read 2 more answers
The perimeter of a rectangular table is 20 feet. The width of the table is 3 feet. What is the length of the table
lyudmila [28]

Answer:

7 ft

Step-by-step explanation:

Perimeter of a rectangle is denoted by the formula: P=2l+2w , where l is the length and w is the width.

We know that P = 20 and that w = 3. So, let's plug these values in:

20 = 2l+2*3=2l+6

Isolate the variable l by subtracting 6 from both sides and dividing by 2:

2l=20-6=14

l=7

Thus, the answer is 7 feet.

Hope this helps!

8 0
2 years ago
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Consider the sets below. U = {x | x is a real number} A = {x | x is an odd integer} R = {x | x = 3, 7, 11, 27} Is R ⊂ A? yes, be
mr_godi [17]

Answer:

yes, becoz all the elements of set R is in set A

8 0
3 years ago
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What two numbers equal 9 and multiply to 16
Artemon [7]
So 3x3 is equal to 9 but i do not understand what you mean and multiply to 16
4 0
3 years ago
Show that f(x) f(y) - f(x+y), where f(x)= [cos x - sin x o] [sin x cos x o] [o o 1]
Naddik [55]

Answer:

The answer is below

Step-by-step explanation:

Show that f(x) f(y) = f(x+y)

From trigonometric:

sin(x + y) = sinxcosy + cosxsiny

sin(x - y) = sinxcosy - cosxsiny

cos(x + y) = cosxcosy - sinxsiny

cos(x - y) = cosxcosy + sinxsiny

f(x)=\left[\begin{array}{ccc}cosx&-sinx&0\\sinx&cosx&0\\0&0&1\end{array}\right] ,f(y)=\left[\begin{array}{ccc}cosy&-siny&0\\siny&cosy&0\\0&0&1\end{array}\right] \\\\\\f(x)f(y)=\left[\begin{array}{ccc}cosxcosy-sinxsiny&-cosxsiny-sinxcosy&0\\sinxcosy+cosxsiny&-sinxsiny+cosxcosy&0\\0&0&1\end{array}\right] \\\\\\f(x)f(y)=\left[\begin{array}{ccc}cos(x+y)&-sin(x+y)&0\\sin(x+y)&cos(x+y)&0\\0&0&1\end{array}\right] \\\\\\

f(x+y)=\left[\begin{array}{ccc}cos(x+y)&-sin(x+y)&0\\sin(x+y)&cos(x+y)&0\\0&0&1\end{array}\right] \\\\\\Therefore\ f(x)f(y)=f(x+y)

4 0
3 years ago
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