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3 years ago

Which of these values are farthest from zero 0.75 2/3 and 0.25

Mathematics

1 answer:

leva [86]3 years ago

8 0

.75 is farthest because 2/3 is .66 (infinitely) and .25 is smaller than both

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Q7.

VLD [36.1K]

Answer:

Step-by-step explanation:

Let c = sum of ages of all children.

mean age of all children = c/15 = 7

c = 15 * 7 = 105

The sum of the ages of all children is 105.

Let b = sum of ages of the 9 boys.

mean age of boys = b/9 = 5

b = 9 * 5 = 45

The sum of the ages of the boys is 45.

The sum of the ages of the girls is c - b.

c - b = 105 - 45 = 60

The number of girls is 15 - 9 = 6

mean age of girls = (sum of ages of the girls)/(number of girls) =

= 60/6 = 10

3 0

3 years ago

On the playground the teacher observed bicycles and tricycles. she counted 72 wheels and 28 total cycles. How many of them were

Andreyy89

Answer:

16 tricycles

12 bicycles

Step-by-step explanation:

If all were bicycles, there would be 28·2 = 56 wheels. Each additional wheel indicates the presence of a tricycle instead of a bicycle.

There are 72 -56 = 16 tricycles. The remaining 28-16 = 12 are bicycles.

3 0

3 years ago

Find the product <br>(3a+3b)(3a-2b)

vitfil [10]

Answer:

9a^2 + 3ab + 6b^2

6 0

3 years ago

Suppose that you have $6000 to invest. Which investment yields the greater return over four years: 8.25% compounded quarterly or

WARRIOR [948]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$6000\\
r=rate\to 8.25\%\to \frac{8.25}{100}\to &0.0825\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{quarterly, thus four}
\end{array}\to &4\\
t=years\to &4
\end{cases}
\\\\\\
A=6000\left(1+\frac{0.0825}{4}\right)^{4\cdot 4}\implies A=6000(1.020625)^{16}\\\\
-------------------------------\\\\$

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$6000\\
r=rate\to 8.3\%\to \frac{8.3}{100}\to &0.083\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{semiannually, thus two}
\end{array}\to &2\\
t=years\to &4
\end{cases}
\\\\\\
A=6000\left(1+\frac{0.083}{2}\right)^{2\cdot 4}\implies A=6000(1.0415)^8$

compare them away.

4 0

3 years ago

Jayla recently accepted a new job earning $15.50 per hour. Last month, Jayla earned $2294 after four weeks of work. How many hou

V125BC [204]

Assuming the pay last month was from Jayla's new job, it represents pay for weekly hours of ...

$\dfrac{\$2294}{4\,week}\times\dfrac{1\,hours}{\$15.50}=37\,\dfrac{hours}{week}$

Jayla worked an average of 37 hours each week.

4 0

3 years ago