Complete Question
Polymer composite materials have gained popularity because they have high strength to weight ratios and are relatively easy and inexpensive to manufacture. However, their nondegradable nature has prompted development of environmentally friendly composites using natural materials. An article reported that for a sample of 10 specimens with 2% fiber content, the sample mean tensile strength (MPa) was 51.1 and the sample standard deviation was 1.2. Suppose the true average strength for 0% fibers (pure cellulose) is known to be 48 MPa. Does the data provide compelling evidence for concluding that true average strength for the WSF/cellulose composite exceeds this value? (Use α = 0.05.)
t=8.169
P-value= ?
Answer:
a) 
b) Hence,We FAil to reject the alternative hypothesis and accept that the true average strength for the WSF/ cellulose composite exceeds 48 MPa.
Step-by-step explanation:
From the question we are told that:
Sample size 
Mean 
Standard deviation 
Significance level is taken as 
t test statistics

Therefore

Critical point



Therefore
P-value from T distribution table

Conclusion

We Reject the Null Hypothesis 
Hence,We FAil to reject the alternative hypothesis and accept that the true average strength for the WSF/ cellulose composite exceeds 48 MPa.
Answer:
The amount charged for a hamburger is $11
Step-by-step explanation:
The function relating the amount d charged with the number of hamburgers is;
d = 3h + 8
The amount charged for a hamburger can be calculated by substituting the value 1 for h
Thus, we have;
d = 3(1) + 8 = $11
The value of 7 in 870,541 is 10,000
Let n be the 40%, and 100-n be the 10%. Then:
.4n+.1(100-n)=.22(100)
.3n+10=22
.3n=12
n=40
The candy distributor needs 40kg of 40%, and 60kg of 10%
☺☺☺☺