Answer:
The measure of angle JKL is 97 degrees.
Step-by-step explanation:
Given:
A triangle JKL .
Measure of exterior angle K = 83 degrees
We have to find the measure of
.
In
we see that the exterior angle is in linear pair with
.
Note:
Linear pair angles sum is 180 degree.
Let
be 'x' degrees .
So,
⇒ 
⇒ <em>Subtracting 83 from both sides.</em>
⇒ 
⇒ 
⇒
degrees
The measure of angle JKL in the triangle is of 97 degrees.
90The information that you'd need in determining sample size when the standard deviation is unknown is:
z-value for the given level of confidence ⇒ The level of confidence is 90% and the z-value by convention is 1.645
The margin error (E) = 4%
The proportion of the population that you'd expect to vote for Yi. If this information isn't provided, use 50%. We call this value, p, the probability of success.
The probability of not success is q = 1 - 0.5 = 0.5
STEP 1:
Half the margin error = 0.04 ÷ 2 = 0.02
STEP 2:
Multiply the probability of success by the probability of fail = 0.5 × 0.5 = 0.25
STEP 3:
Divide z-score by half of E = 1.645 ÷ 0.02 = 82.25
Then square the answer = 82.25² = 6765 rounded to the nearest integer)
STEP 4:
Multiply the answer from STEP 2 by the answer from STEP 4
0.25 × 6765 = 1691 (rounded to the nearest integer)
Answer: Minimum Sample Size is 1691
Price before tax = $18
Tax = 7% x $18 = 0.07 x 18 = $1.26
Price after tax = $18 + $1.26 = $19.26
Answer: $19.26
Answer:
The number of rainfalls is 
The answer to the second question is no it will not be valid this because from the question we are told that the experiment require one pH reading per rainfall so getting multiply specimens(used for the pH reading) from one rainfall will make the experiment invalid.
Step-by-step explanation:
from the question we are told that
The standard deviation is 
The margin of error is 
Given that the confidence level is 95% then we can evaluate the level of significance as



Next we will obtain the critical value of
from the normal distribution table , the value is 
Generally the sample size is mathematically represented as
![n = [\frac{Z_{\frac{\alpha }{2} * \sigma }}{ E} ]^2](https://tex.z-dn.net/?f=n%20%20%3D%20%20%5B%5Cfrac%7BZ_%7B%5Cfrac%7B%5Calpha%20%7D%7B2%7D%20%2A%20%20%5Csigma%20%7D%7D%7B%20E%7D%20%5D%5E2)
substituting values
![n = [\frac{1.96 * 0.5 }{ 0.1} ]^2](https://tex.z-dn.net/?f=n%20%20%3D%20%20%5B%5Cfrac%7B1.96%20%2A%200.5%20%7D%7B%200.1%7D%20%5D%5E2)

The answer to the second question is no the validity is null this because from the question we are told that the experiment require one pH reading per rainfall so getting multiply specimens(used for the pH reading) from one rainfall will make the experiment invalid