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RSB [31]
4 years ago
9

object weighs 8,000 grams, how many kilograms does it weigh? A) 8 kilograms B) 80 kilograms C) 800 kilograms D) 80,000 kilograms

Mathematics
1 answer:
givi [52]4 years ago
8 0

Answer:

A) 8 kilograms

Step-by-step explanation:

"kilo-" is a prefix meaning 1000. So, 8 kilo-grams = 8 thousand grams = 8,000 grams.

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The graph of a linear function passes through the point (1, 1) and has a slope of -3. What is the equation of the function in sl
Zigmanuir [339]
Use the point-slope formula
y-y1=m(x-x1)
y-1=-3(x-1)
y-1=-3x+3
y=-3x+4

5 0
3 years ago
How to solve 42j+18-19j=-28
Kryger [21]

Answer:

j=-2

Step-by-step explanation:

42j+18-19j=-28

Combine like terms

23j +18=-28

Subtract 18 from each side

23j +18-18 = -28 -18

23j =-46

Divide each side by 23

23j/23 = -46/23

j = -2

5 0
3 years ago
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An incomplete distribution is given below:Variable You are given that the median value is 70 and the total number of items is 20
expeople1 [14]

Answer:

The missing frequencies are x = 8 and y = 43.

Step-by-step explanation:

Median Value =70

Then the median Class =60-80

Let the missing frequencies be x and y.

Given: Total Frequncy = 200 , Median = 46

\left|\begin{array}{c|ccccccc}Value&0-20&20-40&40-60&60-80&80-100&100-120&120-140\\Frequency&12&30&x&66&y&27&14\\$Cumu.Freq&12&42&42+x&108+x&108+x+y&135+x+y&149+x+y\end{array}\right|

From the table

\sum f_i =149+x+y  

Here, n = 200

n/2 = 100

Lower Class Boundary of the median class, l=60

Frequency of the median class(f) =66

Cumulative Frequency before the median class, f=42+x

Class Width, h=10

Median = l + \dfrac{\dfrac{n}{2} - c.f }{f} \times h

70 = 60+ \dfrac{100- 42+x }{66}\times 10\\70 = 60+ \dfrac{58+x }{66}\times 10\\70-60=\dfrac{58+x }{66}\times 10\\10*66=10(58+x)\\58+x=66\\x=66-58\\x=8

200=149+x+y

200=149+8+y

y=200-(149+8)

y=43

Hence, the missing frequencies are x = 8 and y = 43.

8 0
3 years ago
Solve B=1/4m(s+z) for s
ElenaW [278]

Answer:

s = \frac{-mz+4B}{m}

Step-by-step explanation:

Step 1: Distribute.

  • B = \frac{1}{4}m * s + \frac{1}{4}m * z
  • B = \frac{1}{4}ms + \frac{1}{4}mz
  • \frac{1}{4}ms + \frac{1}{4}mz = B

Step 2: Subtract 1/4(mz) from both sides.

  • \frac{1}{4}ms + \frac{1}{4}mz - \frac{1}{4}mz = B - \frac{1}{4}mz
  • \frac{1}{4}ms = -\frac{1}{4}mz + B

Step 3: Divide both sides by m/4.

  • \frac{1}{4}ms/\frac{m}{4}=(\frac{-1}{4}mz + B)/\frac{m}{4}
  • s = \frac{-mz+4B}{m}
6 0
3 years ago
If a train is moving 72 km/h on a road of 360 km, how much time will it take to arrive?
zloy xaker [14]
It should 5 hours. You do 360 divided by 72 and get 5 
6 0
3 years ago
Read 2 more answers
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