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3 years ago

How do I calculate 5 percent of 60 using fractions

Mathematics

2 answers:

Crazy boy [7]3 years ago

6 0

5%=5/100 60=60/1 5/100 x 60/1 = 300/100 = 3/1 = 3

irinina [24]3 years ago

5 0

60*5%=60*5/100=300/100=3
So the answer is 3

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It would take 150 minutes to fill a swimming pool using the water from 5 taps.

mina [271]

Answer: points for points

Step-by-step explanation:

Hi t sure what to do?

5 0

3 years ago

HELP!!!!!!!!!!!

Svetradugi [14.3K]

Using the mean concept, it is found that:

Relative to Sabrina's goal, her average swim time over the last five weeks is 0.1 hours.

-----------------------

The mean of a data-set is given by the <u>sum of all observations divided by the number of observations</u>.

In this problem:

The data-set is her swim time relative to her goal, which is: {1.25, -1, 2.25, 0, -2.}

5 observations.

Thus, the mean is:

$M = \frac{1.25 - 1 + 2.25 + 0 - 2}{5} = 0.1$

Relative to Sabrina's goal, her average swim time over the last five weeks is 0.1 hours.

A similar problem is given at brainly.com/question/24787716

7 0

3 years ago

In a G.P the difference between the 1st and 5th term is 150, and the difference between the

liubo4ka [24]

Answer:

Either $\displaystyle \frac{-1522}{\sqrt{41}}$ (approximately $-238$ ) or $\displaystyle \frac{1522}{\sqrt{41}}$ (approximately $238$ .)

Step-by-step explanation:

Let $a$ denote the first term of this geometric series, and let $r$ denote the common ratio of this geometric series.

The first five terms of this series would be:

$a$ ,
$a\cdot r$ ,
$a \cdot r^2$ ,
$a \cdot r^3$ ,
$a \cdot r^4$ .

First equation:

$a\, r^4 - a = 150$ .

Second equation:

$a\, r^3 - a\, r = 48$ .

Rewrite and simplify the first equation.

$\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}$ .

Therefore, the first equation becomes:

$a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150$ ..

Similarly, rewrite and simplify the second equation:

$\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}$ .

Therefore, the second equation becomes:

$a\, r\, \left(r^2 - 1\right) = 48$ .

Take the quotient between these two equations:

$\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}$ .

Simplify and solve for $r$ :

$\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}$ .

$8\, r^2 - 25\, r + 8 = 0$ .

Either $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ or $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ .

Assume that $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}$ .

Similarly, assume that $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}$ .

4 0

3 years ago

If y = 1/2(d + 1), find y when d = 7

nata0808 [166]

Answer:

Step-by-step explanation:

Plug in 7 for d:

y= 1/2(7+1)

y=1/2(8) .... Adding what's in the parenthesis (PEMDAS)

y=4 ..... Multiply 1/2 by 8

Hope this helps:)

5 0

3 years ago

Read 2 more answers

solve systems of equation. equation number 1: 9 x squared + 4 y squared equals 144. equation 2: x squared plus y squared equals

grigory [225]

$9x^{2} + 4y^{2} =144$
$x^{2} + y^{2}=24$

$9x^{2} + 4y^{2} =144$
$4x^{2} + 4y^{2}=96$
by the difference
5 $x^{2}$ =48
so x= $\sqrt{\frac{48}{5}}$ or x=- $\sqrt{\frac{48}{5}}$
and $y^{2}$ =24- $\frac{48}{5}$
then y= $\sqrt{\frac{72}{5}}$ or - $\sqrt{\frac{48}{5}}$

5 0

3 years ago