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.
Answer:
$108
Step-by-step explanation:
Plug in each variable into the simple interest formula and solve from there
I = PRT
P=$900
R=6% (0.06)
T=2 years
The drawing is missing, so i have attached it.
Answer:
24 yards
Step-by-step explanation:
From the image, we can see that AB and DE are both adjacent sides of their respective triangles.
While, CE and BC are both opposite sides of their respective triangles
Now, from your side, x is the distance to the island.
So, from ratio of similar triangles, we can find x.
Thus;
8/x = 3/9
Cross multiply to get;
8 × 9 = 3x
3x = 72
x = 72/3
x = 24
Thus, distance from your side to the island is 24 yards
Answer: multiply each point by 3. You would need to multiply both the x and y value in the point to find the new dilated point.
Step-by-step explanation:
Answer:
Step-by-step explanation:
The question is
<em>We let , so the equation becomes:</em>
Where
Putting it in the quadratic formula, we have:
<u>Quadratic formula:</u>
Substituting we have:
<em>We let , so x is:</em>
<em></em>
<em>and</em>
<em></em>
The solutions of the equation is (rounded to 2 decimal places), and (rounded to 2 decimal places)