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3 years ago

Please help! i literally suck at geometry

Mathematics

1 answer:

sdas [7]3 years ago

8 0

Don’t say like that!Think you can and you will achieve! :) :)

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The grades received on a test by twenty students were 100, 55, 75, 80, 65, 65, 95, 90, 80, 45, 40, 50, 85, 85, 85, 80, 80, 70, 6

Delvig [45]

To find the average of a set of numbers you add them all together. Then divide by the amount of numbers

8 0

4 years ago

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There are 10 cranberry and 15 strawberry juices. You randomly pick two juices. What's the probability you'll get two strawberry

lorasvet [3.4K]

Answer:

7/20

Explanation:

The theoretical probability, P (A), of an event, A, is:

P(A) = number of outcomes of event A / total number of possible outcomes

For 10 cranberry and 15 strawberry juices, the number of outcomes for two strawberry juices are:

Combination of 15 strawberry juices chosen in two:

$C^{15}_{2}=\frac{15!}{(15-2)!(2!)}=\frac{15!}{13!2!}=\frac{15\times 14}{2}=105$

The total number of possible outcomes is:

Combination of 25 juices chosen in two:

$C^{25}_{2}=\frac{25!}{(25-2)!(2!)}=\frac{25!}{23!2!}=\frac{25\times 24}{2}=300$

Thus, the probability of randomly picking two strawberry juices is:

P (two strawberry juices) = 105/300 = 7/20

Note that you can obtain it as the product of the probabilities that the first juice is a strawberry juice and the second juice is also strawberry juice:

$15/25\times 14/24=(15\times 25)/(14\times 24)=210/600=7/20$

5 0

3 years ago

Plot the points A(9, 11) and B(–3, –5). Find midpoint M of AB. Then show that AM = MB and AM + MB =AB

Alex_Xolod [135]

Answer:

The midpoint is (3, 3).

Step-by-step explanation:

We are given the two points A(9, 11) and B(-3, -5).

The midpoint is given by:

$\displaystyle M=\Big(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\Big)$

So:

$\displaystyle M = \Big( \frac{9+(-3) }{2}, \frac{ 11+(-5) }{2} \Big) = (3,3)$

The midpoint is (3, 3).

We want to show that AM = MB.

We can use the distance formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2$

The distance between A(9, 11) and M(3, 3) will then be:

$AM=\sqrt{(9-3)^2+(11-3)^2}=\sqrt{6^2+8^2}=\sqrt{100}=10$

And the distance between B(-3, -5) and M(3, 3) will be:

$MB = \sqrt{ (3-(-3))^2 + (3-(-5))^2 } = \sqrt{(6)^2+(8)^2} = \sqrt{ 100 } = 10$

So, AM = MB = 10.

Since AM = MB = 10, AM + MB = 10 + 10 = 20.

So, we want to prove that AB = 20.

By the distance formula:

$AB=\sqrt{(9-(-3))^2+(11-(-5))^2}=\sqrt{12^2+16^2}}=\sqrt{400}=20\stackrel{\checkmark}{=}20$

4 0

3 years ago

What is 87.6 rounded to the nearest one?

IgorC [24]

88 would be the answer

7 0

4 years ago

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Please give me the answer to 8p+5=77?

denis23 [38]

The answer is P=9
because to isolate the variable subtract 5 from both sides then divide 72 by 8 to get 9

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3 years ago

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