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Korolek [52]
3 years ago
11

which pair of ratios forms a proportion? hint: check for cross products. a. 6/9=9/12 b. 8/16=10/21 c. 6/15=10/25 d. 4/14=6/27

Mathematics
1 answer:
Anna35 [415]3 years ago
6 0

Answer:

  c. 6/15 = 10/25

Step-by-step explanation:

Using the hint, we can form the cross products of the offered choices:

a.  6(12) vs 9(9) ⇒ 72 vs 81 . . . not a proportion

b.  8(21) vs 16(10) ⇒ 168 vs 160 . . . not a proportion

c.  6(25) vs 15(10) ⇒ 150 vs 150 . . . IS A PROPORTION

d.  4(27) vs 14(6) ⇒ 108 vs 84 . . . not a proportion

__

The cross products are the same for choice C, so that is a pair of ratios that form a proportion:

  6/15 = 10/25

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The answer 323.4 is pound.

Step-by-step explanation:

First 25% and 8% together which is 33%.Then find out 33% of 980.33/100×980=323.4 pound.

So he saved 323.4 pound.

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4 years ago
A set of data with a mean of 53, and a standard deviation of 6.2 is normally distributed. Find the value of a data point that is
Evgesh-ka [11]

The value of a data point that is -2 standard deviations from the mean

so, 40.6 to 65.4.

<h3>What is the empirical rule?</h3>

According to the empirical rule, also known as the 68-95-99.7 rule, the percentage of values that lie within an interval with 68%, 95%, and 99.7% of the values lies within one, two, or three standard deviations of the mean of the distribution.

P(\mu - \sigma < X < \mu + \sigma)  \approx 68\%\\P(\mu - 2\sigma < X < \mu + 2\sigma)  \approx 95\%\\P(\mu - 3\sigma < X < \mu + 3\sigma)  \approx 99.7\%

From the empirical rule, we know that for 95% we are in 2 standard deviations of the mean. so:

= 53- 6.2- 6.2

= 40.6

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53 + 6.2 + 6.2

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Learn more about the empirical rule here:

brainly.com/question/13676793

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How do you get an equation to solve for all 3 sides of a triangle?
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In your solving toolbox (along with your pen, paper and calculator) you have these 3 equations:

1. The angles always add to 180°:

A + B + C = 180°

When you know two angles you can find the third.

 

2. Law of Sines (the Sine Rule):

Law of Sines

When there is an angle opposite a side, this equation comes to the rescue.

Note: angle A is opposite side a, B is opposite b, and C is opposite c.

 

3. Law of Cosines (the Cosine Rule):

Law of Cosines

This is the hardest to use (and remember) but it is sometimes needed  

to get you out of difficult situations.

It is an enhanced version of the Pythagoras Theorem that works  

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With those three equations you can solve any triangle (if it can be solved at all).

Six Different Types (More Detail)

There are SIX different types of puzzles you may need to solve. Get familiar with them:

1. AAA:

This means we are given all three angles of a triangle, but no sides.

AAA Triangle

AAA triangles are impossible to solve further since there are is nothing to show us size ... we know the shape but not how big it is.

We need to know at least one side to go further. See Solving "AAA" Triangles .

 

2. AAS

This mean we are given two angles of a triangle and one side, which is not the side adjacent to the two given angles.

AAS Triangle

Such a triangle can be solved by using Angles of a Triangle to find the other angle, and The Law of Sines to find each of the other two sides. See Solving "AAS" Triangles.

 

3. ASA

This means we are given two angles of a triangle and one side, which is the side adjacent to the two given angles.

ASA Triangle

In this case we find the third angle by using Angles of a Triangle, then use The Law of Sines to find each of the other two sides. See Solving "ASA" Triangles .

 

4. SAS

This means we are given two sides and the included angle.

SAS Triangle

For this type of triangle, we must use The Law of Cosines first to calculate the third side of the triangle; then we can use The Law of Sines to find one of the other two angles, and finally use Angles of a Triangle to find the last angle. See Solving "SAS" Triangles .

 

5. SSA

This means we are given two sides and one angle that is not the included angle.

SSA Triangle

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6. SSS

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