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3 years ago

I need help on question 18. i would appreciate it; please and thank you :)

Mathematics

1 answer:

MaRussiya [10]3 years ago

8 0

Hello!

First, let's take a look at the two angles together. We can see that the two angles, when the measures are combined, add up to form a line, DEF. We can therefore see that the two angles DEG and GEF must add up to 180 degrees, as the measure of a line is 180 degrees.

Therefore, we can create our equation.

(7x - 3) + (12x - 7) = 180

Now, solve for x.

7x - 3 + 12x - 7 = 180

19x = 190

x = 10

Finally, substitute x into the angle measures to find the integer measurements.

Angle DEG: 7x - 3

7 * 10 - 3

70 - 3

67 degrees

Angle GEF: 12x - 7

12 * 10 - 7

120 - 7

113 degrees

Just to be sure, add up the angle measurements to make sure they equal 180. 67 + 113 = 180, which means we did our math correctly.

Hope this helps!

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The answer you are looking for is no solution.

To find this, you first distribute the -5 into the parenthesis (-4r - 6). -5 x -4r = 20r and -5 x -6 = 30.Next, distribute the 5 into the parenthesis (6 - r). 5 x 6 = 30 and 5 x r = 5r.

The equation says 20r + 30 = 5r + 30. Now you solve for r.Subtract 5r from both sides. This leaves the equation to say 15r + 30 = 30. Then subtract 30 from both sides, to get 15r = 0.

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Step-by-step explanation:

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4 years ago

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So the answer for this case would be n=1337 rounded up to the nearest integer

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample

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$\sigma=1.1$ represent the population standard deviation

n represent the sample size

Solution to the problem

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

And on this case we have that ME =0.07 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

The critical value for 98% of confidence interval now can be founded using the normal distribution. And in excel we can use this formula to find it:"=-NORM.INV(0.01;0;1)", and we got $z_{\alpha/2}=2.326$ , replacing into formula (b) we got:

$n=(\frac{2.326(1.1)}{0.07})^2 =1336.006 \approx 1337$

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