One angle of an isosceles triangle measures 42°what other angles could be in that triangle 42°

1 answer:

3 0

In isosceles triangles, all angles = 180 degrees and two angles are the same.

Therefore, the two remaining angles equal 69 degrees

180 - 42 = 138
138/2 = 69

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The graph of a linear system is shown.

Cloud [144]

Answer:

(- 3, 6 )

Step-by-step explanation:

The solution is at the point of intersection of the 2 lines

The lines intersect at (- 3, 6 ) ← solution

8 0

3 years ago

Assume that the one-way commute time of an UoU student from his house to school is a normally distributed random variable which

Marizza181 [45]

Answer:

$n=(\frac{1.960(10)}{5})^2 =15.36 \approx 16$

So the answer for this case would be n=16 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

$\mu$ population mean (variable of interest)

$\sigma=10$ represent the sample standard deviation

n represent the sample size

ME=5 represent the margin of error

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The margin of error is given by this formula:

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

And on this case we have that ME =5 and we are interested in order to find the value of n, if we solve n from equation (2) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (3)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got $z_{\alpha/2}=1.96$ , replacing into formula (3) we got: