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alukav5142 [94]
3 years ago
13

Why can you use cross products to solve the proportion 18/5 = x/100 for x ?

Mathematics
1 answer:
Tcecarenko [31]3 years ago
4 0

Answer:

Yes, you can use cross products to solve for x in this proportion. Also, x is 360.

Step-by-step explanation:

Yes, you can use cross products to solve for x in that proportion.

Here's how you do it :

\frac{18}{5} = \frac{x}{100}

We cross product, getting :

18 * 100 = 5 * x

1800 = 5x

Divide by 5 on both sides to get x by itself.

360 = x

Yes, you can use cross products to solve for x in this proportion. Also, x is 360.

Hope this helps, please mark brainliest if possible. Have a great day.

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The height of 44 students (in centimeter) are presented in the frequency distribution below. Complete the table, and then find t
JulijaS [17]

Given:

The height of 44 students (in centimeter) are presented in the given frequency distribution.

To find:

The median of the given data.

Solution:

The given intervals are inclusive first make the intervals exclusive by adding 0.5 in upper limit and by subtracting 0.5 from the lower limit.

The complete frequency distribution table is shown below:

Height              Frequency           L                            <cf

182 - 188                  1                  181.5 - 188.5            44

175 - 181                   3                 174.5 - 181.5             43

168 - 174                  6                 167.5 - 174.5            40

161 - 167                 15               160.5 - 167.5            34   Median class

154 - 160                 12                153.5 - 160.5           19

147 - 153                   4                146.5 - 153.5             7

140 - 146                   3                139.5 - 146.5             3

We have, n=44.

\dfrac{n}{2}=\dfrac{44}{2}

\dfrac{n}{2}=22

The 22th term is include in the class 160.5 - 167.5 because is cf is greater than 22 but the cf of o preceding class is less than 22. So, it is the median class is 160.5 - 167.5.

Formula for median:

Median=l+\dfrac{\dfrac{n}{2}-cf}{f}\times h

Where, l is the lower limit of median class, cf is the cumulative frequency of preceding class, f is the frequency of median class and h is the size of the class.

Now, l=160.5,f=15,cf=19,h=7.

Median=160.5+\dfrac{22-19}{15}\times 7

Median=160.5+\dfrac{3}{15}\times 7

Median=160.5+1.4

Median=161.9

Therefore, the median of the given data table is 161.9.

3 0
3 years ago
Parallel to y = 3x – 1 and passes through the point ( – 4, – 11)
s344n2d4d5 [400]
If the line is parallel, you know it’s going to have the same slope, which in this case is 3

Using the format y= mx + b, plug in your x and y values to find b (m is the slope)
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b = 1

You now have all the needed values to write the equation in slope-intercept form:

y = 3x + 1



7 0
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