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vovikov84 [41]
3 years ago
13

Help I tested solving this problem multiple times and it is incorrect

Mathematics
1 answer:
Veseljchak [2.6K]3 years ago
5 0
Area=rsquared(pi)
24/2=12(12)=144(pi)
Around 452.16
I hope that is correct.
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Round of 389,422 in 1,000
Katen [24]
So,

To round the nearest thousand, look at the number in the hundredths place.
If n ≥ 5, round up
If n < 5, round down

4 < 5, therefore round down.

389,422 --> 389,000
4 0
3 years ago
Calcule a área da região sombreada da figura abaixo.<br> 120°<br> 4 cm<br> B
masya89 [10]

Answer:

Area of segment = 9.8 cm² (Approx.)

Step-by-step explanation:

Given:

Angle θ = 120°

Radius of circle = 4 cm

Find:

Area of segment

Computation:

Area of segment = [θ/360][π][r]² - [1/2][r²][sinθ]

Area of segment = [120/360][3.14][4]² - [1/2][4²][sin120]

Area of segment = [0.333][3.14][16] - [0.5][16][0.866]

Area of segment = 16.7299 - 6.28

Area of segment = 9.8019

Area of segment = 9.8 cm² (Approx.)

8 0
3 years ago
The number of defects in the first five cars to come through a new production line are 9, 7, 10, 4, and 6, respectively. If the
garri49 [273]

Answer:

D. I and III only

Step-by-step explanation:

Number of first fie defects are given as 9,7,10,4 and 6.

First we have to arrange the above data in ascending order which is 4,6,7,9,10

  • Now if we consider the defect in sixth car to be 3 then our data will look like:    3,4,6,7,9,10

So the mean of above data would be \frac{3+4+6+7+9+10}{6} = 6.5

and Median of the above data would be = \frac{3^{rd} obs + 4^{th}  obs}{2} = \frac{6+7}{2} = 6.5

Hence mean and median number of defects are same if the sixth car has 3 defects.

  • Now if we consider the defect in sixth car to be 12 then our data will look like:    4,6,7,9,10,12

So the mean of above data would be \frac{4+6+7+9+10+12}{6} = 8

and Median of the above data would be = \frac{3^{rd} obs + 4^{th}  obs}{2} = \frac{7+9}{2} = 8

Hence mean and median number of defects are same if the sixth car has 12 defects.

  • But Now if we consider the defect in sixth car to be 7 then our data will look like:    4,6,7,7,9,10        

So the mean of above data would be \frac{4+6+7+7+9+10}{6} = 7.167    

and Median of the above data would be =  \frac{3^{rd} obs + 4^{th}  obs}{2} = 7

   Hence mean and median number of defects are not same if the sixth car has 7 defects.      

Therefore option D is correct.                                                                                  

8 0
3 years ago
A hammer is 1 foot long a car is 15 feet long and shovel is 4 feet long which statement is correct? Choose all that apply’s
Arturiano [62]

Answer:

Tationia Rolon

Step-by-step explanation:

Here

3 0
3 years ago
1. b+4= 2.4 2. 3.6 = 1.2g 3. 3.75x = 3 4. w-9 = -4.1 5. -7.4 = 8.8 + p 6. 2.9 = y - 6.3 7. A+5.7 = 1.5 8. -3.5 = V/0.2 9. Z/9 =
Marrrta [24]
1. b=2.4-4
b= -1.6

2. 1.2g/1.2 = 3.6/1.2
g=3

3. 3.75x/3.75 = 3/3.75
x= 0.8

4. w = -4.1 + 9
w= 4.9

5. p = -7.4 - 8.8
p= -16.2

6. y = 2.9 + 6.9
y = 9.8

7. A= 1.5 - 5.7
A= -4.2

8. V/0.2 * 0.2 = -3.5 * 0.2
V= -0.7

9. Z/9 * 9 = 1.6 * 9
Z= 14.4

10. -9.9t/-9.9 = -9.9/-9.9
t= 1
7 0
3 years ago
Read 2 more answers
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