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3 years ago

Two examples that have a gcf of 8

Mathematics

2 answers:

timama [110]3 years ago

8 0

12 & 24
16 & 24
4 & 16
16 & 32

hope this helps.

STALIN [3.7K]3 years ago

8 0

An example would be 24 and 16. Their GCF would be 8.

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balandron [24]

Answer:

The first step is to determine the average

$x= \frac{0,6+0,7+ 0,7+0,3+0,4+ 0,5+ 0,4 +0,2}{8}\\\ x= 0,475$

The exercise says it’s a normal distribution: (n=8)

$s^{2}= \frac{1}{n-1} ((0,6.x)^2+2(0,7-x)^2+(0.3-x)^2+2(0.4-x)^2+(0.5-x)^2+(0.2-x)^2)\\\ \frac{1}{7}*0,235 = 0,0336 \\\\ s= 0183$

According to the exercise, the mean is equal to 0,5 then the value of t of the distribution can be obtained

$t= \frac{x-u}{\frac{s}{\sqrt{n} }}\\\ t= \frac{0,475 - 0,5}{0,183}\\\ t= -0,3860$

The variable t has 7 grade to liberty, we calculate the p-value as:

$2* P(t < - 0,3860)= \\\ 2* 0,3554 = 0,7108$

This value is very high, therefore the hypothesis is not rejected

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3 years ago

On the coordinate grid, the graph of is shown. It is a translation of y= ∛x-1+3 is shown. It is a translation of y= ∛x

lisov135 [29]

If you want to know the domain of a function through its graph, I simply have to "scan" the graph of the function from left to right along the X axis and see the ranges of values of the x for which it exists a value on the Y axis. In this graph it is observed that for any value of X there will be a value of Y, so that the domain of the graph function is all the real values of X

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kotegsom [21]

In the given question, it is required to solve an inequality. The equation that needs to be solved is:
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3 years ago

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A cannonball is shot into the air. Its height can be described by the equation H = -5t² + 40t, where “h” is the height in feet a

DedPeter [7]

Answer:

t = 8 seconds

Step-by-step explanation:

Since this is a quadratic equation, there will be two values for which H = 0, (the ball is on the ground), so we'll just solve the equation and the first value will be the moment when the cannonball is launched, and the other will be when the cannonball falls to the ground.

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t(40 - 5t) = 0

the equation is true whenever t = 0 or t = 8, so we know that at the zeroth second, the ball was being launched, and at the 8th second, the ball was on the ground.

Here is how the graph looks like:

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Svetllana [295]

False. Obtuse is always greater than 90°.

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