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Damm [24]
3 years ago
8

Pls help Lcm of 4 and 15? least common multiple

Mathematics
2 answers:
9966 [12]3 years ago
4 0

Answer:

60

Step-by-step explanation:

kakasveta [241]3 years ago
3 0
<h3>Answer:  60</h3>

=================================================

Explanation:

Multiply the two values to get 4*15 = 60

Then divide by the GCF 1 to get 60/1 = 60. The GCF being 1 means the result hasn't changed.

------

Another example would be: "Find the LCM of 6 and 8". We would first do 6*8 = 48, then divide by the GCF 2 to get 48/2 = 24. The LCM of 6 and 8 is 24.

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Which graph represents the solution to the given system?<br> Y=-x+5<br> y= 1/4 x+10
jeka57 [31]

Answer:

See attachment

Step-by-step explanation:

The given system is :

y =  - x + 5

y =  \frac{1}{4}x + 10

Let us equate both equations and solve for x.

- x + 5 =  \frac{1}{4}x + 10

- 4x + 20 = x + 40

20 - 40 =x + 4x

- 20 = 5x

x =  - 4

y =  -  - 4 + 5 = 9

The the two lines of this system will meet at (-4,9)

The graph is shown in the attachment.

7 0
3 years ago
An honest die is rolled. If the roll comes out even (2, 4, or 6), you will win $1; if the roll comes out odd (1,3, or 5), you wi
jenyasd209 [6]

Answer:

(a) 50%

(b) 47.5%

(c) 2.5%

Step-by-step explanation:

According to the honest coin principle, if the random variable <em>X</em> denotes the number of heads in <em>n</em> tosses of an honest coin (<em>n</em> ≥ 30), then <em>X</em> has an approximately normal distribution with mean, \mu=\frac{n}{2} and standard deviation, \sigma=\frac{\sqrt{n}}{2}.

Here the number of tosses is, <em>n</em> = 2500.

Since <em>n</em> is too large, i.e. <em>n</em> = 2500 > 30, the random variable <em>X</em> follows a normal distribution.

The mean and standard deviation are:

\mu=\frac{n}{2}=\frac{2500}{2}=1250\\\\\sigma=\frac{\sqrt{n}}{2}=\frac{\sqrt{2500}}{2}=25

(a)

To not lose any money the even rolls has to be 1250 or more.

Since, <em>μ</em> = 1250 it implies that the 50th percentile is also 1250.

Thus, the probability that by the end of the evening you will not have lost any money is 50%.

(b)

If the number of "even rolls" is 1250, it implies that the percentile of 1250 is 50th.

Then for number of "even rolls" as 1300,

1300 = 1250 + 2 × 25

        = μ + 2σ

Then P (μ + 2σ) for a normally distributed data is 0.975.

⇒ 1300 is at the 97.5th percentile.

Then the area between 1250 and 1300 is:

Area = 97.5% - 50%

        = 47.5%

Thus, the probability that the number of "even rolls" will fall between 1250 and 1300 is 47.5%.

(c)

To win $100 or more the number of even rolls has to at least 1300.

From part (b) we now 1300 is the 97.5th percentile.

Then the probability that you will win $100 or more is:

P (Win $100 or more) = 100% - 97.5%

                                   = 2.5%.

Thus, the probability that you will win $100 or more is 2.5%.

7 0
3 years ago
Select all figures for which there exists a direction such that all cross sections taken at that direction are congruent.
dimulka [17.4K]

Answer:

  • Rectangular Prism
  • Cube
  • Cylinder

Step-by-step explanation:

I know because I just did it on an assignement and it said these were the right answers.

4 0
3 years ago
In the triangle below, the value of a is 12.6.<br><br> True<br><br> False
aev [14]

Answer:

True

Step-by-step explanation:

Formula

Tan(x) = opposite / adjacent

Givens

x = 42

adjacent = 14

opposite = ?

Solution

Tan(42) = opp/14                     Find tan(42)

Tan(42) = 0.900404

0.900404 = opp/ 14                Multiply both sides by 14

14 * 0.900404 = opp/14 * 14   Simplify the left

12.6 rounded  = opp

a = 12.6


7 0
3 years ago
PLEASE HELP ME THIS.. I REALLY NEED<br> HELP
Colt1911 [192]

9514 1404 393

Answer:

  5 hours

Step-by-step explanation:

A quick way to look at this is to compare the difference in hourly charge to the difference in 0-hour charge.

The first day, the charge is $3 more than $12 per hour.

The second day, the charge is $12 less than $15 per hour.

The difference in 0-hour charges is 3 -(-12) = 15. The difference in per-hour charges is 15 -12 = 3. The ratio of these is ...

  $15/($3/h) = 5 h

The charges are the same after 5 hours.

__

If you write equations for the charges, they will look like ...

  y1 = 15 + 12(x -1)

  y2 = 3 + 15(x -1)

Equating these charges, we have ...

  15 +12(x -1) = 3 + 15(x -1)

  12x +3 = 15x -12 . . . . . . . . eliminate parentheses

  15 = 3x . . . . . . . . . . add 12-12x

  x = 15/3 = 5 . . . . . . divide by 3

You might notice that the math here is very similar to that described in words, above.

The charges are the same after 5 hours.

6 0
3 years ago
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