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BARSIC [14]
3 years ago
11

How would I solve this problem? We have to classify the angle pair and find the value of x.

Mathematics
1 answer:
Irina18 [472]3 years ago
7 0
Well the 2 angles are supplementary.
You would set up the equation like this:
(9x-7)+(7x-5)= 180
16x-12=180
+12
16x= 192
Divide both sides by 16
X= 12
Hope this helps
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⦁ Show (2.3)(5.06) as a fraction multiplication problem and explain why the answer is in thousandths (three decimal places). 
astra-53 [7]

we have

(2.3)(5.06)

we know that

2.3=\frac{23}{10}

5.06=\frac{506}{100}

so

(2.3)(5.06)=(\frac{23}{10})(\frac{506}{100})

=\frac{11,638}{1,000}

=11.638

the answer is in thousandths, because the denominator of the multiplication of the two fractions is one thousand

8 0
3 years ago
Read 2 more answers
Help me with my homework idk how to do it
QveST [7]

Answer:

You have to make an equation

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If you need more help I can assist you

Step-by-step explanation:

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8 0
3 years ago
Add;200g+0.04kg=-----------dag
enyata [817]

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7 0
3 years ago
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The principal randomly selected six students to take an aptitude test. their scores were: 81.4 78.6 71.3 77.8 76.9 77.7 determin
Firdavs [7]
The confidence interval is 77.28\pm 2.23.

We first find the mean.  Add together all of the data points and divide by 6, the number of data points; the mean is 77.28.

Next we find the standard deviation.  Find the difference between each data point and the mean; square it; find the sum; divide by the number of data points; take the square root.  The standard deviation is 3.32.

To find the margin of error, we calculate the z-score associated with this level of confidence.  100-90 = 10% = 0.1; 0.1/2 = 0.05; 1-0.05 = 0.95.  Using a z-table (http://www.z-table.com) we see that this is between two scores, 1.64 and 1.65; we will use 1.645.

The margin of error is given by
z * (σ/√n) = 1.645*(3.32/√6) = 2.23.

Thus the confidence interval is 77.28 +/ 2.23.
3 0
3 years ago
Write an equation to represent each
Soloha48 [4]
N6+5=n-10
n6=n-15
n5=-15
n=-3
hope this helps:)
6 0
2 years ago
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