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3 years ago

What is the least common multiple (LCM) of 3 and 4?

1 answer:

6 0

Answer:Least common multiple (LCM) of 3 and 4 is 12. Least common multiple or lowest common denominator (lcd) can be calculated in two way; with the LCM formula calculation of greatest common factor (GCF), or multiplying the prime factors with the highest exponent factor. Least Common Multiple of 3 and 4 with GCF Formula.

Step-by-step explanation:

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What is |32|? <br><br> PLEASE HELP!!!

fredd [130]

Answer:

the absolute value of 32 ( |32| ) is 32

Step-by-step explanation:

absolute value is asking how far away a number is from 0. so, 32 is 32 numbers away from 0.

hope this helps!

3 0

3 years ago

Read 2 more answers

Miguel measured a house and its lot and made a scale drawing. The backyard, which is 40 feet wide in real life, is 20 inches wid

viva [34]

Answer:

Can you please post a picture?!

Step-by-step explanation:

4 0

3 years ago

If the diameter of circle O is 20 and CE = 4, then determine the length of AB. Show how you arrived at your answer.

Aneli [31]

Answer: Length of AB is 16 cm

Step-by-step explanation:

Given: Diameter of a circle is 20 cm and CE = 4 cm

Now as shown in figure AO, BO, CO are radii

$\text { Radius } =\dfrac{\text {Diameter}}{2} = \dfrac{20}{2} = 10 cm$

Therefore

AO=BO=CO= 10 cm

Now

$OE = CO- CE \Rightarrow OE= 10 -4 = 6 cm$

Now in Δ OEB

∠OEB =90°

Therefore by Pythagoras theorem : In a right angle triangle the square of hypotenuse is equal to the sum of square of other two sides.

So we have

$BO^2=BE^2+EO^2\\\\\Rightarrow 10^2= BE^2 +6^2\\\\\Rightarrow BE^2= 100-36\\\\\Rightarrow BE^2 = 64 \\\\\Rightarrow BE= 8 cm$

Now as we know perpendicular to the chord from the center of a circle bisect the chord.

Therefore

$AB= 2\times BE = 2\times 8=16 cm$

Hence , the length of AB is 16 cm .

8 0

3 years ago

Suppose that you had the following data set. 500 200 250 275 300 Suppose that the value 500 was a typo, and it was suppose to be

hodyreva [135]

Answer:

$\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305$

$s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108$

$\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105$

$s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221$

The absolute difference is:

$Abs = |340.221-115.108|= 225.113$

If we find the % of change respect the before case we have this:

$\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%$

So then is a big change.

Step-by-step explanation:

The subindex B is for the before case and the subindex A is for the after case

Before case (with 500)

For this case we have the following dataset:

500 200 250 275 300

We can calculate the mean with the following formula:

$\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305$

And the sample deviation with the following formula:

$s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108$

After case (With -500 instead of 500)

For this case we have the following dataset:

-500 200 250 275 300

We can calculate the mean with the following formula:

$\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105$

And the sample deviation with the following formula:

$s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221$

And as we can see we have a significant change between the two values for the two cases.

The absolute difference is:

$Abs = |340.221-115.108|= 225.113$

If we find the % of change respect the before case we have this:

$\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%$

So then is a big change.

8 0

3 years ago

What is the first step in solving the inequality

polet [3.4K]

Answer:

nsjjdjdjdjdnnd

Step-by-step explanation:

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sjdjdjd

8 0

3 years ago