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3 years ago

Of the following , which is closet to 4% of 73

Mathematics

1 answer:

Komok [63]3 years ago

7 0

You don't have any choices listed.
Anyway 4 % of 73 = 2.92

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When renting a limo for prom, the number of people varies inversely

lana [24]

Answer:

sorry but thx

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Which of these is NOT an integer

Tema [17]

Hi there!

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I believe your answer is:

Option A

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Here’s why:

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$\boxed{\text{Option A:}}\\\\3^{-5}\\\\\rightarrow \frac{1}{3^5}\\\\\rightarrow \frac{1}{243}\\\\\ \text{This is \textbf{NOT} an integer.}}$

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$\boxed{\text{Option B:}}\\\\-3^5\\\\\rightarrow -3 * -3 * -3 * -3 * -3 \\\\\rightarrow \boxed{-243}\\\\\\\text{This \textbf{IS} an integer.}$

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$\boxed{\text{Option C:}}\\\\\frac{1}{3}^{-5} \\\\\rightarrow \frac{1}{(\frac{1}{3})^5}\\\\\rightarrow \frac{1}{\frac{1}{243} }\\\\\rightarrow \boxed{243}\\\\\\\text{This \textbf{IS} an integer.}$

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$\text{Option \textbf{A} does not simplify into an integer.}$

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Hope this helps you. I apologize if it’s incorrect.

7 0

2 years ago

Peppy Pets charges a flat fee of $15 plus $3 per hour to keep a dog during the day. Happy Hounds charges flat fee of $21 plus $1

Ostrovityanka [42]

Answer:3 hours total

Step-by-step explanation:

15+(3*3)=21+(1*3) which equals out to 24$

6 0

3 years ago

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If a passanger is sellected at random, what is the probability that the passenger was female and survived

Jet001 [13]

??? Depends on how many people are there

3 0

3 years ago

What is the completely factored form of f(x)=x^3-7x^2+2x+4

xxMikexx [17]

$Solution, \mathrm{Factor}\:x^3-7x^2+2x+4:\quad \left(x-1\right)\left(x^2-6x-4\right)$

$Steps:$

$x^3-7x^2+2x+4$

$Use\:the\:rational\:root\:theorem,\\a_0=4,\:\quad a_n=1,\\\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:2,\:4,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1,\\\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:2,\:4}{1},\\\frac{1}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x-1,\\\left(x-1\right)\frac{x^3-7x^2+2x+4}{x-1}$

$\frac{x^3-7x^2+2x+4}{x-1}$

$\mathrm{Divide}\:\frac{x^3-7x^2+2x+4}{x-1},\\\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}x^3-7x^2+2x+4\mathrm{\:and\:the\:divisor\:}x-1\mathrm{\::\:}\frac{x^3}{x},\\\mathrm{Quotient}=x^2,\\\mathrm{Multiply\:}x-1\mathrm{\:by\:}x^2:\:x^3-x^2,\\\mathrm{Subtract\:}x^3-x^2\mathrm{\:from\:}x^3-7x^2+2x+4\mathrm{\:to\:get\:new\:remainder},\\\mathrm{Remainder}=-6x^2+2x+4,\\Therefore,\\\frac{x^3-7x^2+2x+4}{x-1}=x^2+\frac{-6x^2+2x+4}{x-1}$

$\mathrm{Divide}\:\frac{-6x^2+2x+4}{x-1},\\\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}-6x^2+2x+4\mathrm{\:and\:the\:divisor\:}x-1\mathrm{\::\:}\frac{-6x^2}{x}=-6x,\\\mathrm{Quotient}=-6x,\\\mathrm{Multiply\:}x-1\mathrm{\:by\:}-6x:\:-6x^2+6x,\\\mathrm{Subtract\:}-6x^2+6x\mathrm{\:from\:}-6x^2+2x+4\mathrm{\:to\:get\:new\:remainder},\\\mathrm{Remainder}=-4x+4,\\\mathrm{Therefore},\\\frac{-6x^2+2x+4}{x-1}=-6x+\frac{-4x+4}{x-1}$

$\mathrm{Divide}\:\frac{-4x+4}{x-1},\\\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}-4x+4\mathrm{\:and\:the\:divisor\:}x-1\mathrm{\::\:}\frac{-4x}{x}=-4,\\\mathrm{Quotient}=-4,\\\mathrm{Multiply\:}x-1\mathrm{\:by\:}-4:\:-4x+4,\\\mathrm{Subtract\:}-4x+4\mathrm{\:from\:}-4x+4\mathrm{\:to\:get\:new\:remainder},\\\mathrm{Remainder}=0,\\\mathrm{Therefore},\\\frac{-4x+4}{x-1}=-4$

$\mathrm{The\:Correct\:Answer\:is\:\left(x-1\right)\left(x^2-6x-4\right)}$

$\mathrm{Hope\:This\:Helps!!!}$

$\mathrm{-Austint1414}$

8 0

3 years ago