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scZoUnD [109]
3 years ago
12

If you had 26/12 dollars how much money would that be in dollars and cents

Mathematics
1 answer:
lbvjy [14]3 years ago
6 0

Answer:

$2.17

Step-by-step explanation:

12 goes into 26 2 times.

with 2 left over

2/12 = 1/6

1/6 of a dollar is

divide 100/6 = 16.67

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Carlo puts tins into small boxes and into large boxes. He puts 6 tins into each small box. He puts 20 tins into each large box.
meriva

Answer: Carlo is incorrect

Step-by-step explanation:

3000/(2+3) = 600 (each ratio)

2:3 = 2(600):3(600) = 1200:1800

1200/6 = 200 small boxes

1800/20 = 90 large boxes

290 boxes altogether (large and small)

(large boxes/boxes altogether)x100

(90/290)x100 = 31%

31%>30%

Therefore Carlo is incorrect

8 0
3 years ago
Please help! Consider the given circle with the shaded sector xy and central angle 225. The circumference is 30 units
OleMash [197]

\textit{circumference of a circle}\\\\ C=2\pi r~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ C=30\pi \end{cases}\implies 30\pi =2\pi r\implies \cfrac{30\pi }{2\pi }=r\implies \boxed{15=r} \\\\[-0.35em] ~\dotfill

\textit{arc's length}\\\\ s = \cfrac{\theta \pi r}{180}~~ \begin{cases} r=radius\\ \theta =\stackrel{degrees}{angle}\\[-0.5em] \hrulefill\\ r=15\\ \theta =225 \end{cases}\implies \widehat{XY}=\cfrac{(225)\pi (15)}{180}\implies \boxed{\widehat{XY}\approx 58.90} \\\\[-0.35em] ~\dotfill\\\\ \textit{area of a circle}\\\\ A=\pi r^2\qquad \qquad A=\pi (15)^2\implies A=225\pi \implies \boxed{A\approx 706.86}

6 0
3 years ago
PLZZZZZZZZZZ HELP
MrRissso [65]
<h2>Answer:</h2>

The correct answer is:   Option: A

A.   85, 78, 80, 108, 46, 66, 68, 82, 72, 68

<h2>Step-by-step explanation:</h2>

It is given that:

The whisker ranges from 66 to 85.

The box ranges from 68 to 82 with the vertical bar inside the box at 75.

  • Hence, the minimum value of the box plot is: 66
  • Maximum value of box plot is: 85

Also,

  • The first quartile or the lower quartile i.e. Q_1 is: 68
  • The middle quartile or median i.e. Q_2 is: 75
  • The upper quartile or third quartile i.e. Q_3 is: 82

Also, it is given that:

One dot mark above forty-six.

One dot mark above one hundred and eight.

A)

On arranging the data in increasing order after removing outliers 46 and 108 we have:

 66    68    68    72    78    80    82     85  

Hence, from this data we have:

Minimum value=66

maximum value=85

Q_1=68\\\\Q_2=75\\\\\\Q_3=81

As all the values matches the value defined.

Hence, this option is correct.

B)

On arranging the data in increasing order after removing 46 and 108 we have:

66    68    68    70   70   80    82     85

Hence, from this data we have:

Minimum value=66

maximum value=85

Q_1=68\\\\Q_2=70\\\\\\Q_3=81

As the value of Q_2=70\neq 75.

Hence, this option is incorrect.

C)

On arranging the data in increasing order after removing 46 and 108 we have:

66    68    72    78   80   80    82     85

Hence, from this data we have:

Minimum value=66

maximum value=85

Q_1=70\\\\Q_2=79\\\\\\Q_3=81

As the value of  Q_1,Q_2\ and Q_3 do not match.

Hence, this option is incorrect.

D)

On arranging the data in increasing order after removing 46 and 108 we have:

66    68    68    72   78   80    80     85

Hence, from this data we have:

Minimum value=66

maximum value=85

Q_1=68\\\\Q_2=79\\\\\\Q_3=80

As the value of  Q_3=80\neq 82 do not match.

Hence, this option is incorrect.

4 0
3 years ago
Read 2 more answers
If a = 2 cm and r = 3.4 cm, find the area of the regular pentagon.
cricket20 [7]

Answer:

Area = 3.4\ cm^{2}

Step-by-step explanation:

Assume a pentagon has equal length where a be the side of the pentagon and r be the apothem of the pentagon.

Given:

Sides of pentagon a=2\ cm

apothem of pentagon r=3.4\ cm

The area of the pentagon formula is given below.

A = \frac{1}{2}\times bh

Where b = length of the base or side

And h = height of apothem

Now, we substitute side and apothem length in above formula.

A = \frac{1}{2}\times 2\times 3.4

A = 3.4\ cm^{2}

Therefore, the area of the pentagon is 3.4\ cm^{2}

3 0
3 years ago
E xex2 + y2 + z2 dv, where e is the portion of the unit ball x2 + y2 + z2 ≤ 1 that lies in the first octant.
Jlenok [28]
\displaystyle\iiint_{\mathcal E}xe^{x^2+y^2+z^2}\,\mathrm dV
=\displaystyle\int_{\varphi=0}^{\varphi=\pi/2}\int_{\theta=0}^{\theta=\pi/2}\int_{\rho=0}^{\rho=1}\rho\cos\theta\sin\varphi e^{\rho^2}\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi
=\displaystyle\left(\int_{\rho=0}^{\rho=1}\rho^3e^{\rho^2}\,\mathrm d\rho\right)\left(\int_{\theta=0}^{\theta=\pi/2}\cos\theta\,\mathrm d\theta\right)\left(\int_{\varphi=0}^{\varphi=\pi/2}\sin^2\varphi\,\mathrm d\varphi\right)=\frac\pi8
7 0
3 years ago
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