HEEELLLLPPP!!!! PLZZZZ!!!! THANK U!!!<br> (5 screenshots included)

Order the following values from least (top) to greatest (bottom).

Blizzard [7]

Answer:

Step-by-step explanation:

0.002

0.02

2/10

22/100

200%

2.2

220/10

4 0

2 years ago

The formula for the area A of a triangle is A=(1/2)bh, where b is the length of the base and h is the height. Rearrange the form

zalisa [80]

Answer:

$b=\dfrac{2A}{h}$

Step-by-step explanation:

The formula for the area of a triangle is given by :

$A=\dfrac{1}{2}bh$ ...(1)

Where b is the length of the base and h is the height.

We need to find the value of b from the above formula.

Cross multiplying equation (1) we get :

$2A=bh$

Now dividing both sides by h. So,

$\dfrac{2A}{h}=\dfrac{bh}{h}\\\\b=\dfrac{2A}{h}$

So, the correct option is (D) i.e. $\dfrac{2A}{h}$ .

8 0

2 years ago

Aaron wants to go car shopping. He wants to buy different models all at the same price. He buys 3 Lincolns, 5 Fords, and 10 BMWs

ExtremeBDS [4]

The answer is $6.654, that’s the price for each car. You just divide 119.772 / the number of cars. Hope it helped.

8 0

2 years ago

Ryan flew from Wiley Post to Ponca City and back. Ryan maintained an average rate of 450 mph going to Ponca City and an average

Reil [10]

The answer is B, and here's why. Set up a table for "there" and "back" and use the distance = rate * time formula, like this:
               d             r            t
there       d           450         t
back        d           400        1-t

Let me explain this table to you. The distance is d, we don't know what it is, that's what we are actually looking for. We only know that if we go somewhere from point A to point B, then back again to point A, the distance there is the same as the distance back. Hence, the d in both spaces. There he flew 450 mph, back he flew 400 mph. If the total distance was 1 hour, he flew an unknown time there and one hour minus that unknown time back. For example, if he flew for 20 minutes there, one hour minus 20 minutes means that he flew 60 minutes - 20 minutes = 40 minutes back. See? Now, because the distance there = the distance back, we can set the rt in both equal to each other. If d = rt there and d = rt back and the d's are the same, then we can set the rt's equal to each other. 450t = 400(1-t) and
450t = 400 - 400t and 850t = 400. Solve for t to get t = .47058. Now, t is time, not the distance and we are looking for distance. So multiply that t value by the rate (cuz d = r*t) to get that the distance one way is
d = 450(.470580 and d = 211. 76 or, rounded like you need, 212.

4 0

2 years ago

In a random sample of 80 teenagers, the average number of texts handled in a day is 50. The 96% confidence interval for the mean

Nastasia [14]

Answer:

a) $\bar X =\frac{46+54}{2}=50$

And the margin of error is given by:

$ME= \frac{54-46}{2}= 4$

The confidence level is 0.96 and the significance level is $\alpha=1-0.96=0.04$ and the value of $\alpha/2 =0.02$ and the margin of error is given by:

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

We can calculate the critical value and we got:

$z_{\alpha/2} = 2.05$

And if we solve for the deviation like this:

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

And replacing we got:

$\sigma =4 *\frac{\sqrt{80}}{2.05} =17.45$

b) $ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=2.05 *\frac{17.45}{\sqrt{160}}=2.828$

And as we can see that the margin of error would be lower than the original value of 4, the margin of error would be reduced by a factor $\sqrt{2}$

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

$\mu$ population mean (variable of interest)

$\sigma$ represent the population standard deviation

n=80 represent the sample size

Solution to the problem

Part a

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

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

For this case we can calculate the mean like this:

$\bar X =\frac{46+54}{2}=50$

And the margin of error is given by:

$ME= \frac{54-46}{2}= 4$

The confidence level is 0.96 and the significance level is $\alpha=1-0.96=0.04$ and the value of $\alpha/2 =0.02$ and the margin of error is given by:

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

We can calculate the critical value and we got:

$z_{\alpha/2} = 2.05$

And if we solve for the deviation like this:

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

And replacing we got:

$\sigma =4 *\frac{\sqrt{80}}{2.05} =17.45$

Part b

For this case is the sample size is doubled the margin of error would be:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=2.05 *\frac{17.45}{\sqrt{160}}=2.828$

And as we can see that the margin of error would be lower than the original value of 4, the margin of error would be reduced by a factor $\sqrt{2}$

5 0

2 years ago

HEEELLLLPPP!!!! PLZZZZ!!!! THANK U!!! (5 screenshots included)