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3 years ago

What are two reasons a common measuring system is important?

1 answer:

4 0

<span>Measuring systems use units for comparative data, to measure things such as: volume, quantity, distance, scale, height, time etc. Across continents many different types of measuring units are used, for example car distance (Miles and Kilometers); common measuring systems ar important for us to compare data, create standardization for products and services, to be able to replicate measurements for experiments or producing medicines, food recipes and product making.</span>

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how do you use absolute value to find the distance between two points that have the same X-coordinates but different Y-coordinat

sdas [7]

So since x coordinates are same, we just have a horizontal distance to find (no fancy slanted lines)

so what you do is
|y1-y2|=distance

6 0

3 years ago

Find the % of change. Original price: $185 New price: $160

mojhsa [17]

Answer:

-13.51

Step-by-step explanation:

work out the difference (increase) between the two numbers you are comparing. Then: divide the increase by the original number and multiply the answer by 100. % increase = Increase ÷ Original Number × 100.,

so, -13.51

8 0

2 years ago

Any help would be greatly appreciated!

Aleks [24]

Answer:

the answer for that question is sqrt(49)

7 0

3 years ago

Read 2 more answers

Solve for x<br><br> 4x + 7 + 3x = 19 + x

IceJOKER [234]

7x-x=19-7
6x=12
x=2
Ask me if you have a question. Glad to help!! :))

6 0

3 years ago

The mean SAT score in mathematics, M, is 600. The standard deviation of these scores is 48. A special preparation course claims

kupik [55]

Answer:

Step-by-step explanation:

The mean SAT score is $\mu=600$ , we are going to call it \mu since it's the "true" mean

The standard deviation (we are going to call it $\sigma$ ) is

$\sigma=48$

Next they draw a random sample of n=70 students, and they got a mean score (denoted by $\bar x$ ) of $\bar x=613$

The test then boils down to the question if the score of 613 obtained by the students in the sample is statistically bigger that the "true" mean of 600.

- So the Null Hypothesis $H_0:\bar x \geq \mu$

- The alternative would be then the opposite $H_0:\bar x < \mu$

The test statistic for this type of test takes the form

$t=\frac{| \mu -\bar x |} {\sigma/\sqrt{n}}$

and this test statistic follows a normal distribution. This last part is quite important because it will tell us where to look for the critical value. The problem ask for a 0.05 significance level. Looking at the normal distribution table, the critical value that leaves .05% in the upper tail is 1.645.

With this we can then replace the values in the test statistic and compare it to the critical value of 1.645.

$t=\frac{| \mu -\bar x |} {\sigma/\sqrt{n}}\\\\= \frac{| 600-613 |}{48/\sqrt(70}}\\\\= \frac{| 13 |}{48/8.367}\\\\= \frac{| 13 |}{5.737}\\\\=2.266\\$

<h3>since 2.266>1.645 we can reject the null hypothesis.</h3>

6 0

4 years ago

Read 2 more answers