Y=(kx)/z
15=(k10)/4
15=(5k/2)
15=k(5/2)
ties both sides by 2/5 to clear fraction
6=k
y=(6x)/z
x=20
z=6
y=(6*20)/6
y=20
Answer:
700
Step-by-step explanation:
Given:
Daily evening newspapers in 2015 = 389
Percentage decrease from 2000 to 2015 = 46.5%
To find:
Number of newspapers in 2000 = ?
Solution:
Let the number of newspapers in 2000 = 
When 46.5% of the total newspapers in 2000 (
) is subtracted from the total newspapers in 2000 (
), it will be equal to the number of newspapers in 2015.
Making the equation as per the given statement:

Therefore, the answer is:
There were <em>727 </em>newspapers in 2000.
Answer:
33.33%
Step-by-step explanation:
We need to calculate the <u>unit selling price and cost of each cosmetics.</u>
If a person bought some cosmetics from wholesale market at the rate of Rs 360 per dozen., then for 1 cosmetics, we will say;
x = 1 cosmetic
since 360 = 12 cosmetic
cross multiply
12x = 360
x = 360/12
x = 30
Hence the unit cost price of the cosmetics will be Rs. 30
Similarly, if he sells it at Rs 80 a pair, then he sold one cosmetic at 80/2 = Rs. 40 (a pair is 2 cosmetics)
Selling price per unit = Rs. 40
Cost price per unit = Rs. 30
percent gain = SP-CP/CP * 100%
percent gain = 40-30/30 * 100
percent gain = 10/30 * 100
percent gain = 100/3
percent gain = 33.33%
Hence the percentage gain is 33.33%
Answer:
The 99% confidence interval for the proportion of readers who would like more coverage of local news is (0.3685, 0.4315).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.

In which
z is the zscore that has a pvalue of
.
For this problem, we have that:

99% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
The lower limit of this interval is:

The upper limit of this interval is:

The 99% confidence interval for the proportion of readers who would like more coverage of local news is (0.3685, 0.4315).