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3 years ago

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What is greater: 1) 0.48 or 6/15th? 2) 0.75 or 7/8th? 3) 7.025 or 7 1/8ths?

1 answer:

fgiga [73]3 years ago

4 0

1) 0.48 3 2) 7/8 3) 7 1/8

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A student wants to find the answer to the question “What is the most popular restaurant in Bedford?” He asks 20 randomly selecte

mojhsa [17]

Answer:

The survey results will be inaccurate since the entire sample of the population was eating at the same restaurant.

Step-by-step explanation:

When doing a survey you must make sure it is unbiased by asking a variety of people for example if you are wondering what is the favorite class at school you have to ask a certain amount of people from each class because if you ask only people from P.E the answer more than likely will be P.E or if u ask 3 people from Art and 17 from P.E your answer will properly be P.E. Make sense?

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2 years ago

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What is the difference of the polynomials [m²n²-7]-(mn+4)

AleksAgata [21]

You can do addition and substraction operations with polynomials similarly as with simple real numbers.

First of all you have to open brackets and then you can add numbers (-7-4=-11). Then the result is following:
$(m^2n^2-7)-(mn+4) =m^2n^2-7-mn-4=m^2n^2-mn-11$

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3 years ago

Sun Country Bus Lines has 80,000 shares of stock outstanding. What would the dividend per share of stock be from a dividend decl

Evgen [1.6K]

The answer is $1.37
and if this is under it-

How many shares of stock does Julie Norris hold if her share of the dividend is $6.85?

The answer is 5

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3 years ago

PLEASE HELP ME WITH THIS QUESTION

valentina_108 [34]

Set 7x-4 and 31 equal to each other
7X-4=31
Add four to both sides
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X=3
The answer is D

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3 years ago

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Which of the following are solutions to the equation below?

sladkih [1.3K]

Answer:

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$x=3,\:x=-\frac{1}{2}$

Step-by-step explanation:

considering the equation

$2x^2\:-\:4x\:-\:3\:=\:x$

solving

$2x^2\:-\:4x\:-\:3\:=\:x$

$2x^2-4x-3-x=x-x$

$2x^2-5x-3=0$

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=2,\:b=-5,\:c=-3:\quad x_{1,\:2}=\frac{-\left(-5\right)\pm \sqrt{\left(-5\right)^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}$

solving

$x=\frac{-\left(-5\right)+\sqrt{\left(-5\right)^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}$

$x=\frac{5+\sqrt{\left(-5\right)^2+4\cdot \:2\cdot \:3}}{2\cdot \:2}$

$x=\frac{5+\sqrt{49}}{2\cdot \:2}$

$x=\frac{5+7}{4}$

$x=3$

also solving

$x=\frac{-\left(-5\right)-\sqrt{\left(-5\right)^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}$

$x=\frac{5-\sqrt{\left(-5\right)^2+4\cdot \:2\cdot \:3}}{2\cdot \:2}$

$x=\frac{5-\sqrt{49}}{4}$

$x=-\frac{2}{4}$

$x=-\frac{1}{2}$

Therefore,

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$x=3,\:x=-\frac{1}{2}$

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3 years ago