if the ∡E = ∡F = 90°, that means for that quadrilateral of BEGF, the other angles are ∡B = ∡G = 90°, which means is a square, meaning all its sides are equal, therefore

$\bf y + 5 = 3y\implies 5=2y\implies \cfrac{5}{2}=y\implies 2\frac{1}{2}=y$

4 0

3 years ago

0.72 in scientific notation

boyakko [2]

Remember that scientific notation always contains a number between 0 and 10 that is multiplied by ten to some power. In this case, to get a number between 0 and 10 (7.2) we would move the decimal point once to the right. Since it is moving to ththe right, the exponent over the 10 will be negative, and it will be -1 because we moved the point once. So the answer is 7.2*10^-1

Hope this helps

4 0

3 years ago

Complete the statement by filling in the blanks: When constructing a confidence interval, if the level of confidence increases,

Yuki888 [10]

Answer:

$\hat p \pm z_{\alpha/2} SE_{\hat p}$

And for this case the margin of error would be:

$ME= z_{\alpha/2} SE_{\hat p}$

If the level of confidence increase we can conclude that the value of $z_{\alpha/2}$ would increase and the the confidence interval would be wider, since the margin of error increase.

c. Increase; wider

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".

Solution to the problem

Let's assume that we have a parameter of interest $p$ and we want to estimate the value of p with $\hat p$ and in general the confidence interval if the distribution of p is normal is given by:

$\hat p \pm z_{\alpha/2} SE_{\hat p}$

And for this case the margin of error would be:

$ME= z_{\alpha/2} SE_{\hat p}$

If the level of confidence increase we can conclude that the value of $z_{\alpha/2}$ would increase and the the confidence interval would be wider, since the margin of error increase.

c. Increase; wider

8 0

3 years ago