Need to use a the table given to find the rate of change in fruit flies per day over the given intervals. Not sure how to use th
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First let's find the number of the elements in the sample space, that is the total number of codes that can be produced.
The first digit is any of {0, 1, 2...,9}, that is 10 possibilities
the second digit is any of the remaining 9, after having picked one.
and so on...
so in total there are 10*9*8*7 = 5040 codes.
a. What is the probability that the lock code will begin with 5?
Lets fix the first number as 5. Then there are 9 possibilities for the second digit, 8 for the third on and 7 for the last digit.
Thus, there are 1*9*8*7=504 codes which start with 5.
so
P(first digit is five)=
b. <span>What is the probability that the lock code will not contain the number 0?
from the set {0, 1, 2...., } we exclude 0, and we are left with {1, 2, ...9}
from which we can form in total 9*8*7*6 codes which do not contain 0.
P(codes without 0)=n(codes without 0)/n(all codes)=(9*8*7*6)/(10*9*8*7)=6/10=0.6
Answer:
0.1 ; 0.6</span>
The first digit is any of {0, 1, 2...,9}, that is 10 possibilities
the second digit is any of the remaining 9, after having picked one.
and so on...
so in total there are 10*9*8*7 = 5040 codes.
a. What is the probability that the lock code will begin with 5?
Lets fix the first number as 5. Then there are 9 possibilities for the second digit, 8 for the third on and 7 for the last digit.
Thus, there are 1*9*8*7=504 codes which start with 5.
so
P(first digit is five)=
b. <span>What is the probability that the lock code will not contain the number 0?
from the set {0, 1, 2...., } we exclude 0, and we are left with {1, 2, ...9}
from which we can form in total 9*8*7*6 codes which do not contain 0.
P(codes without 0)=n(codes without 0)/n(all codes)=(9*8*7*6)/(10*9*8*7)=6/10=0.6
Answer:
0.1 ; 0.6</span>
Answer:
a) H0 : u = 28.5%
H1 : u < 28.5%
b) critical value = - 1.645
c) test statistic Z= - 1.41
d) Fail to reject H0
e) There is not enough evidence to support the professor's claim.
Step-by-step explanation:
Given:
P = 28.5% ≈ 0.285
X = 210
n = 800
Level of significance = 0.05
a) The null and alternative hypotheses are:
H0 : u = 28.5%
H1 : u < 28.5%
b) Given a 0.05 significance level.
This is a left tailed test.
The critical value =
The critical value = -1.645
c) Calculating the test statistic, we have:
Z = -1.41
d) Decision:
We fail to reject null hypothesis H0, since Z = -1.41 is not in the rejection region, <1.645
e) There is not enough evidence to support the professor's claim that the proportion of obese male teenagers decreased.