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3 years ago

Measuring Circles: Question 1

Mathematics

1 answer:

andreev551 [17]3 years ago

7 0

Answer:10.51inches

Step-by-step explanation:

circumference=66

Radius=circumference ➗ (2xπ)

Radius=66 ➗ (2x3.14)

Radius=66 ➗ 6.28

Radius=10.51inches

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A triangle has a length of 6cm,8cm,and 10cm what is the area of this triangle in cm cubed

notsponge [240]

Answer: 24 cm^2 (24 centimeters squared)

Step-by-step explanation:

6^2 + 8^2 = 36 + 64 = 100 = 10^2

(1/2)*6*8 = 24 cm^2.

8 0

3 years ago

Find the volume of a sphere with radius 15 <br> Using the formula in the picture

telo118 [61]

Answer:

in terms of pi: 4500π units^3

decimal: 14137.16694 units^3 (ROUND TO THE NEAREST DECIMAL POINT BASED OFF OF THE QUESTION PROMPT!)

Step-by-step explanation:

FOR “IN TERMS OF PI” =

V = (4/3) (π) (15^3)

V = (4/3) (π) (3375)

V=4500π

FOR ACTUAL DECIMAL=

V = (4/3) (π) (15^3)

V = (4/3) (π) (3375)

V=4500π

V= 14137.16694 units^3

hope this helps! :)

(please mark brainliest if you can! ty :D)

4 0

3 years ago

Kevin Horn is the national sales manager for National Textbooks Inc. He has a sales staff of 40 who visit college professors all

Oduvanchick [21]

Answer:

Following are the solution to the given points:

Step-by-step explanation:

Given values:

$38, 40, 41, 45, 48, 48, 50, 50, 51, 51, 52, 52, 53, \\\\$ $54, 55, 55, 55, 56, 56, 57, 59, 59, 59, 62, 62, 62, \\ 63, 64, 65,66, 66, 67, 67, 69, 69, 71, 77, 78, 79, 79$

Total staff=40

In point a:

To calculate the median number first we arrange the value into ascending order and then collect the even numbers of calls that were also made. Its average of the middle terms is thus the median.

The midterms =55 and 59 so,

Median = $\frac{55+59}{2} = \frac{114}{2}=57$

In point b:

First-quarter $Q_1 = \frac{1}{4} \text{number of 8th call}$

$=\frac{1}{4} \times 30 th \\\\= 7.5 th$

The first quarterlies are $7.5th \ \ that \ is = (7+0.5)th\ \ term$

therefore the multiply of 0.5 by calculating the difference of the 7th and 8th term are:

$=0.5 \times 0= 0 \\\\\to Q_1 = 50+0=50 \\\\\to Q_3=\frac{3}{4} \text{number of 8th call} \\\\=\frac{3}{4} \times 30 th \\\\=22.5 \ th = (22+0.5)th \ \ term$

therefore the it is multiply by the 0.5 for the difference of the 22nd and 23rd term:

$= 0.5 \times 1=0.5 \\\\\to Q_3= 63+0.5=63.5$

In point c:

First decile $D_1 = \frac{1}{18} \text{number of 8th call}$

$= \frac{1}{10} \times 30 th \\\\= 3rd \ \ term\\\\\to D_1= 41 \\\\\to D_9= \frac{9}{10} \times 8\ th\ call\\\\=\frac{9}{10} \times 30 th \\\\=27th \ \ term\\\\\to D9 = 67$