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3 years ago

What is the least common multiple of 11 and 19

Mathematics

2 answers:

Mariulka [41]3 years ago

8 0

The least common multiple would be 209

Sauron [17]3 years ago

6 0

The answer would be A) 209.

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A company had 41 employees in order 980 uniforms for them if they want to give each employee same number of uniforms how many mo

Tema [17]

Answer:

They would have to order 4 more uniforms in order to distribute an equal amount to each employee

Step-by-step explanation:

First we have to calculate the number of maximum uniforms that can be given to each employee equally

For this we simply divide the number of uniforms by the number of employees and look only at the whole number

980/41 = 23.92 = 23

we don't round we just take the decimals

now we multiply the number of maximum uniforms that we can give each one by the number of employees

23 * 41 = 943

to the 980 uniforms we subtract the 943

980 - 943 = 37

Calculate how much is left to 37 to reach 41

41 - 37 = 4

This means that they would have to order 4 more uniforms in order to distribute an equal amount to each employee

3 0

3 years ago

Unit 1: Scientific Notation Constructed Response Assignment

balu736 [363]

brainly.com/question/5027087

Let C, S and H be the diameters of cell, sand and hair

Diameter of a cell: $C=1 \cdot 10^{-5}m$

Diameter of a grain of sand: $S=2 \cdot10^{-4}m$

Part A:

$H=0.000025=2.5 \cdot 10^{-5}$ meters, because it takes 5 movements right of the decimal dot, until there is only one digit to its left.

Part B:

we are comparing $S=2 \cdot10^{-4} m$ to $C=1 \cdot10^{-5} m$ ,

$\frac{S}{C}= \frac{2 \cdot 10^{-4}}{1 \cdot 10^{-5}}=\frac{20 \cdot 10^{-5}}{1 \cdot 10^{-5}}=20$

thus S is 20 times larger than C

Part C

$1m=1 \cdot10^9nm$

thus,

$H=2.5 \cdot 10^{-5}m=2.5 \cdot 10^{-5} \cdot 1 \cdot 10^9nm=2.5 \cdot 10^{-5+9}=2.5\cdot10^4nm$

Part D.

$1m=1\cdot 10^9nm\\\\1nm=\displaystyle{ \frac{1m}{10^9}=1\cdot 10^{-9}m}$

thus 300 nm are written in meters as follows:

$300nm= 300\cdot 1\cdot 10^{-9}m=3\cdot 10^2 \cdot 10^{-9}m=3\cdot 10^{-7}m$

5 0

3 years ago

One week, seventy-two percent of the animals that cam to a veterinary hospital were dogs. If there were 230 animals total, how m

pentagon [3]

Answer:

165.6

Step-by-step explanation:

165.6 is 72% of 230

3 0

3 years ago

What's the perimeter of a rectangular hexagon with one side measures 3 inches

MakcuM [25]

Answer:

6 sides of the hexagon * one side length of 3 inches = 18

Step-by-step explanation:

3 0

2 years ago

Find the range, median, first and third quartiles, and interquartile range for each data set OF THE NUMBERS 52 72 96 21 58 40 75

Andrej [43]

Answer:

Range: 75 Median: 58 1Q: 40 3Q: 75 Interquartile: 35

4 0

3 years ago