Answer:
see explaination
Step-by-step explanation:
Here the null hypothesis is that the PCB survives against the alternate that the PCB 'does not survive'. The test says that the PCB will survice if it is classified as 'good'; or, it will not survive if it is classifies as 'bad'.
a. The Type II error is the error committed when a PCB which cannot actually survive is classified as 'good'.
b. Therefore P(Type II error) = P(The PCB is classified as 'good' | PCB does not survives) = 0.03.
Use equation

Question 1: Need to find A:

Question 2: Need to find t, use LOGARITHM:

A=35000
P=49339


So 2010+32 =
2042.
She walks 22/7 km in 5/4 hrs, so she walks (22/7)/(5/4) km, or 88/35 (2_18/35) km, in 1 hr.
Answer:
Here we have given two catogaries as degree holder and non degree holder.
So here we have to test the hypothesis that,
H0 : p1 = p2 Vs H1 : p1 not= p2
where p1 is population proportion of degree holder.
p2 is population proportion of non degree holder.
Assume alpha = level of significance = 5% = 0.05
The test is two tailed.
Here test statistic follows standard normal distribution.
The test statistic is,
Z = (p1^ - p2^) / SE
where SE = sqrt[(p^*q^)/n1 + (p^*q^)/n2]
p1^ = x1/n1
p2^ = x2/n2
p^ = (x1+x2) / (n1+n2)
This we can done in TI_83 calculator.
steps :
STAT --> TESTS --> 6:2-PropZTest --> ENTER --> Input all the values --> select alternative "not= P2" --> ENTER --> Calculate --> ENTER
Test statistic Z = 1.60
P-value = 0.1090
P-value > alpha
Fail to reject H0 or accept H0 at 5% level of significance.
Conclusion : There is not sufficient evidence to say that the percent of correct answers is significantly different between degree holders and non-degree holders.
<span>Total mass contains 320g copper. How many parts of 16g copper are there of the total mass of 320g copper? 320/16 = 20. There are 20 portions of 16g copper.
(320/16) * 3 = 60g
There are 60g of zinc in the compound.</span>