Answer:
where is the shape of your question
Honestly idk this question I just need to do this to ask questions :(
Answer:
Step-by-step explanation:
10%=325
1%=32.5
325-32.5=292.5
3250+3542.5=3542.50£
So 9% increase.
A sextant<span> is a </span>doubly reflecting navigation instrument<span> that measures the </span>angular distance<span> between two visible objects. The primary use of a sextant is to measure the angle between an </span>astronomical object<span> and the </span>horizon<span> for the purposes of </span>celestial navigation<span>.
The estimation of this angle, the altitude, is known as </span>sighting<span> or </span>shooting<span> the object, or </span>taking a sight<span>. The angle and the time when it was measured, can be used to calculate a </span>position line<span> on a nautical or aeronautical </span>chart. F<span>or example; sighting the </span>Sun<span> at </span>noon <span>or </span>Polaris<span> at night (in the Northern Hemisphere) to estimate </span>latitude<span>. Sighting the height of a landmark can give a measure of </span>distance off<span> and held horizontally.
A sextant can measure angles between objects for a </span>position on a chart.<span> A sextant can also be used to measure the </span>lunar distance<span> between the moon and another celestial object (such as a star or planet) in order to determine </span>Greenwich Mean Time<span> and hence </span>longitude<span>.
Hope this helps!
<em>~ ShadowXReaper069</em></span>
a. Find the probability that an individual distance is
greater than 214.30 cm
We find for the value of z score using the formula:
z = (x – u) / s
z = (214.30 – 205) / 8.3
z = 1.12
Since we are looking for x > 214.30 cm, we use the
right tailed test to find for P at z = 1.12 from the tables:
P = 0.1314
b. Find the probability that the mean for 20 randomly
selected distances is greater than 202.80 cm
We find for the value of z score using the formula:
z = (x – u) / s
z = (202.80 – 205) / 8.3
z = -0.265
Since we are looking for x > 202.80 cm, we use the
right tailed test to find for P at z = -0.265 from the tables:
P = 0.6045
c. Why can the normal distribution be used in part (b),
even though the sample size does not exceed 30?
I believe this is because we are given the population
standard deviation sigma rather than the sample standard deviation. So we can
use the z test.
