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4 years ago

Which events are dependent?

Mathematics

2 answers:

RSB [31]4 years ago

8 0

It is the 2nd answer because both are dependent on what you do.

Zarrin [17]4 years ago

5 0

B, because those are a completely random circumstance that DEPENDS on different times that something is removed/spun/rolled

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What is the value of N in this proportion 16/40= 8/n

Eduardwww [97]

Answer:

n = 20

Step-by-step explanation:

$\frac{16}{40} = \frac{8}{n}$

Cross multiplication would result in:

16 * n = 8 * 40

16n = 320

n = 320 / 16

n = 20

Therefore <u>the value of "n" in this proportion is 20.</u>

Hope this helps!

5 0

3 years ago

Please help out, using your knowledge of equilateral and isosceles triangles, find the value of x

EleoNora [17]

Answer:

can not see the picture

Step-by-step explanation:

7 0

3 years ago

What is the equation of the line that passes through the point (-7, 2) and has a<br> slope of -1?

lorasvet [3.4K]

Answer: y= -1x - 5

Step-by-step explanation:

The equation has to be in slope-intercept form which uses the formula

y=mx+b where m is the slope and b is the y-intercept.

So we given the slope and an x and y coordinate so we will plot those numbers into the formula and solve for the y intercept to write the equation.

the x coordinate is -7 and the y coordinate is 2

2 = -1(-7) + b

2 = 7 + b

-7 -7

b = -5 the y intercept is -5

Now the equation will be y= -1x - 5

3 0

3 years ago

4. The product of 32 and n is equal to 59.13.

Ivanshal [37]

Answer:

pretty sure it is 1.8478125

Step-by-step explanation:

5 0

3 years ago

Oney four tenths of the print area of the newspaper carried new. The rest was filled with advertisements. What percentage of the

Nikitich [7]

Answer: The required percentage is 75% of print area filled with advertisement.

Step-by-step explanation:

Since we have given that

Part of the print area of the newspaper carried new = $\dfrac{1}{4}$

Rest was filled with advertisements = $1-\dfrac{1}{4}=\dfrac{4-1}{4}=\dfrac{3}{4}$

So, Percentage of the print area was filled with advertisement is given by

$\dfrac{3}{4}\times 100\\\\=75\%$

Hence, the required percentage is 75% of print area filled with advertisement.

3 0

3 years ago