Start off with initially calculating those numbers to find out how far above or below the airplane is from the altitude of 3000 feet.

To clear things up, when an aircraft dives, the altitude of how high it is traveling decreases. When an aircraft climbs, the altitude of the aircraft increases. When it mentions that the aircraft starts to dive, you need to subtract. When it climbs, you need to add.

As the pilot initially dives the plane 988 feet and rapidly climbs 748 feet, you'd need to form an equation which contributes those numbers. It can be illustrated as
3000 - 988 + 748, which is equal to 2760.
The pilot then dives again 885 feet before making another climb of 1099 feet. Adding on to the previous answer we received, this can be represented as 2760 - 885 + 1099, which equals to 2974.

You can conclude that the airplane is now travelling at an altitude of 2974 feet. To find out how far above or below the airplane is from an altitude of 3000 feet, simply subtract those numbers.

3000 - 2974 = 26

The aircraft is travelling <u>26 feet</u> below an altitude of 3000 feet.

6 0

2 years ago

If Mary deposits $275 in principal at an interest rate of 3.2 percent,how much interest will she earn in one year

Free_Kalibri [48]

0.032*$275 = $8.80 . . . . interest on Mary's account for the year.

3 0

2 years ago

You randomly select 16 restaurants and measure the temperature of the coffee sold at each. The sample mean temperature is 162 de

stepan [7]

Answer:

b) (156.7, 167.3).

Step-by-step explanation:

Sample size (n) = 16 restaurants

Mean coffee temperature (X) = 162 degrees

Standard deviation (s) = 10 degrees

degrees of freedom (n-1) = 15

t-score for a 95% confidence interval (t) = 2.131415

Assuming a normal distribution, the confidence interval is given by:

$X \pm t*\frac{s}{\sqrt{n} }\\162 \pm 1.960*\frac{10}{\sqrt{16} }$

The lower (L) and upper (U) bounds of the confidence interval are:

$L=162 - 5.3\\L=156.7\\U=162 + 5.3\\U=167.3$

The confidence interval is b) (156.7, 167.3).

6 0

2 years ago