we know that
The intersection point of both graphs is a common point for both functions, for which for the same input value, both functions will have the same output value.
so
the point of intersection is 
for an input value equal to 
the output value for both functions is 
therefore
<u>the answer is the option</u>
X= 4
<span>0.977
So we have a population with the mean being 0.8750 and the standard deviations being 0.0011. So let's see how many standard deviations we need to be off by to exceed the specifications.
Low end
(0.8725 - 0.8750)/ 0.0011 = -0.0025/0.0011 = -2.272727273
High end
(0.8775 - 0.8750)/ 0.0011 = 0.0025/0.0011 = 2.272727273
So we need to be within 2.272727273 deviations of the mean. Let's use a standard normal table to look up that value, which is 0.48848, which is half the percentage. So 0.48848 * 2 = 0.97696, rounding to 3 digits gives 0.977</span>
Answer:
A. 4.26 in^2
Step-by-step explanation:
Step 1: Find the area of the sector DBC. Here we have to use the formula.
Area of a sector = Central Angle/360 *π
The area of the sector DBC = (54/360)*3.14*
= 30.16
Step 2: Area of segment CFD = Area of the sector DBC - Area of the ΔCBD
= 30.17 - 25.9
Area of segment CFD = 4.27in^2
<span>The lowest common factor of 7 and 18 is 126 (this is a number that both 7 and 18 multiply into, in fact by each other). This means the 126th person will be the first to get both coupons.</span>
Answer with explanation:
Given : Sample mean =
Standard deviation : 
Sample size : n = 36
Claim : 
∴ 

Since the alternative hypothesis is two tail , then the test is two tail test.
By using a z statistic and a 0.05 level of significance. Reject
if z < -1.960 or is z> 1.960.
Then , the test static for population mean is given by :-


We reject
since
. We have statistically significant evidence at
to show that the mean amount of garbage per bin is not different from 50.