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3 years ago

What is the interquartile range of the data set? 23, 35, 55, 61, 64, 67, 68, 71, 75, 94, 99

Mathematics

2 answers:

Marina86 [1]3 years ago

7 0

99 94 75 71 68 67 64 61 55 35 23

MaRussiya [10]3 years ago

4 0

The IQR=20 because the 3rd Q(75) minus the 1st Q(55) equals 20

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Use the Remainder Theorem to determine which of the roots are roots of F(x). Show your work.

Brums [2.3K]

Answer: x1=1 x2=-2 and x3=2

Step-by-step explanation:

1st x1=1 is 1 of the roots , so

F(1)=1-1-4+4=0 - true

So lets divide x^3-x^2-4x+4 by (x-x1), i.e (x^3-x^2-4x+4) /(x-1)=(x^2-4)

x^2-4 can be factorized as (x-2)*(x+2)

So x^3-x^2-4x+4=(x-1)*(x^2-4)=(x-1)(x-2)*(x+2)

So there are 3 dofferent roots:

x1=1 x2=-2 and x3=2

3 0

3 years ago

What set of numbers is arranged from largest to smallest? (Mind the blue on one of the answers)

nikklg [1K]

Answer:

Step-by-step explanation:

it orders the negative things in the top right correctly because the bigger the negative number in the top right the lower the number is aka the greater of a negative number there is

8 0

3 years ago

Read 2 more answers

Answer ASAP please

Vikki [24]

Answer:

h = 16 || solution after squaring: $x = -0.683$ or $x = -7.32$

steps by fieranswererft:

Given: $x^2+8x+5=0$

Take half of the x term and square it

$[8*\frac{1}{2} ]^2=16$

$x^2+8x+16=- 5+16$

$(x+4)^2=-5+16$

$(x+4)^2=11$

$x+4 = \pm \sqrt{11}$

$x = \pm \sqrt{11}-4$

$x = -0.683$ or $x = -7.32$

7 0

3 years ago

James’ college charges tuition based only on the number of credits a student takes. James paid $1218 to take 12 credits in the f

Zielflug [23.3K]

Answer:

16 credits

Step-by-step explanation:

Divide $1218 by 12 to find out how much 1 credit costs. It's $101.50

Because he paid $1624 total for spring, you'll divide that by $101.50 because it's per credit.

1624/101.5 = 16 credits.

hope this is good enough

8 0

4 years ago

Read 2 more answers

David wants to measure the width of a road

alexira [117]

Answer:

The width of the road = 25.359 m

Step-by-step explanation:

See the attached figure which represents the problem.

The length of AB = 40 m

let the width of the road w.

Construct PC perpendicular to AB, So, the measure of angle C = 90°

∠B = 60° and ∠A = 45°

Let the length of AC = x, so, the length of BC = 40-x

At ΔBCP which is a right triangle at C

tan B = opposite/adjacent = w/(40-x)

w = (40-x) * tan B ⇒(1)

At ΔACP which is a right triangle at C

tan A = opposite/adjacent = w/x

w = x * tan A ⇒(2)

from (1) and (2)

(40-x) * tan B = x * tan A

40 tan B - x *tan B = x tan A

40 tan B = x tan A + x *tan B

40 tan B = x (tan A + tan B)

x = (40 tan B)/(tan A + tan B) = (40 tan 60)/(tan 60 + tan 45) = 25.359 m

substitute at (2) with x

w = x tan A = 25.359 tan 45 = 25.359 m

So, The width of the road = 25.359 m

8 0

4 years ago