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3 years ago

Please answer quick!!!

Mathematics

2 answers:

sergey [27]3 years ago

6 0

Answer:

$\huge\boxed{IQR = 30}$

Step-by-step explanation:

Q1 = 130 (Left hand edge of the box)

Q3 = 100 (Right Hand edge of the box)

Interquartile Range = Q3-Q1

IQR = 130-100

IQR = 30

Sidana [21]3 years ago

4 0

Answer:

A. 30

Step-by-step explanation:

The interquartile range for a box and whiskers plot, is the value from the right side of the box minus the value of the left side of the box.

In this case at the far right side of the box it is at 130, at the far left side of the box it is at 100.

130-100=30

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When Brad was born, his Grandma put, $1,500 in a new account for him. Since then, the balance of the account, has grown by 6.5%

marin [14]

Answer:

$969.04

Step-by-step explanation:

now at age of 18 Brad has 1500 * 1,065^18

at age of 21 he will have 1500 * 1,065^21

if he waits the difference will be 1500 * 1,065^21 - 1500 * 1,065^18 = $969.04

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3 years ago

Alison saves 29.26 each month how many months will it take her to save enough money to buy a stereo for 339.12

AleksandrR [38]

29.26x=339.12
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Therefore it wil take her 11.6 months

5 0

3 years ago

Read 2 more answers

Annette [7]

Answer:

x = 12√1.5

Step-by-step explanation:

Law of sines.

12/sin45 = x/sin60

x = 12(½√3) / (½√2)

x = 12√3/√2

x = 12√1.5

7 0

3 years ago

(25)2×100÷23+[24÷(13−5)]

hram777 [196]

Answer:

100*100/23+(24/8)

100*100/23+3

100*4.37+3

43.7+3

46.7

Step-by-step explanation:

7 0

3 years ago

What is the solution for the equation StartFraction 5 Over 3 b cubed minus 2 b squared minus 5 EndFraction = StartFraction 2 Ove

wolverine [178]

Answer:

The solutions are:

$b=0,\:b=4$

Step-by-step explanation:

Considering the expression

$\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}$

Solving the expression

$\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}$

$\mathrm{Apply\:fraction\:cross\:multiply:\:if\:}\frac{a}{b}=\frac{c}{d}\mathrm{\:then\:}a\cdot \:d=b\cdot \:c$

$5\left(b^3-2\right)=\left(3b^3-2b^2-5\right)\cdot \:2$

$5b^3-10=6b^3-4b^2-10$

$\mathrm{Switch\:sides}$

$6b^3-4b^2-10=5b^3-10$

$6b^3-4b^2-10+10=5b^3-10+10$

$6b^3-4b^2=5b^3$

$\mathrm{Subtract\:}5b^3\mathrm{\:from\:both\:sides}$

$6b^3-4b^2-5b^3=5b^3-5b^3$

$b^3-4b^2=0$

$Using\:the\:Zero\:Factor\:Principle:$ $if\:\mathrm ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

So,

$b=0,b-4=0$

$b=0,b=4$

Therefore, the solutions are:

$b=0,\:b=4$

4 0

3 years ago

Read 2 more answers