17/2 = 8.5 hours per week
8.5 x 5 = 42.5 hours in 5 weeks
Complete Question
According to the Bureau of Labor Statistics, citizens remain unemployed for an average of 15.9 weeks before finding their next job (June, 2008). Suppose you want to show that Louisiana has been effective in getting their unemployed back to work sooner. You take a random sample of 50 citizens who were unemployed six months earlier and ask them to report the duration. You find that the average time spent unemployed was 13.4 weeks with a sample standard deviation of the time unemployed is 6.7 weeks.
1 Which of the following statements is the correct alternative hypothesis?
2 The test statistic for testing the hypothesis is
a. -2.64
b. -2.32
c. -2.11
d. -1.28
e. none of these are correct
Answer:
1
The alternative hypothesis 
2
The test statistics
Step-by-step explanation:
From the question we are told that
The population mean value for time citizens remain unemployed is 
The sample size is n = 50
The sample standard deviation is 6.7 weeks.
The sample mean value for time citizens remain unemployed is 
The null hypothesis is 
The alternative hypothesis 
Generally test statistics is mathematically represented as
=> 
=>
C, the median is 23 when you add 13 + 10. Then you divide 23 and 2.
Let:
v = Olga's speed
t = the time Anna spent going to school (3 km/hour)
t/2 = the time Anna spent running back home with speed of 6 km/hour
So, t + t/2 = 3t/2 is the total time
In the meantime distance = speed*time and the both made the same distance we can put it as:
3t + 6*(t/2) = (3t/2)*v
6t = (3/2)tv divide both sides by t
(3/2)v = 6
v = 2*6/3
v = 4 km/hour
Olga's speed was 4 km/hour.
This question is incomplete, the complete question is;
Let X denote the time in minutes (rounded to the nearest half minute) for a blood sample to be taken. The probability mass function for X is:
x 0 0.5 1 1.5 2 2.5
f(x) 0.1 0.2 0.3 0.2 0.1 0.1
determine;
a) P( X < 2.5 )
B) P( 0.75 < X ≤ 1.5 )
Answer:
a) P( X < 2.5 ) = 0.9
b) P( 0.75 < X ≤ 1.5 ) = 0.5
Step-by-step explanation:
Given the data in the question;
The probability mass function for X is:
x 0 0.5 1 1.5 2 2.5
f(x) 0.1 0.2 0.3 0.2 0.1 0.1
a) P( X < 2.5 )
P( X < 2.5 ) = p[ x = 0 ] + p[ x = 0.5 ] + p[ x = 1 ] + p[ x = 1.5 ] + p[ x = 2 ]
so
P( X < 2.5 ) = 0.1 + 0.2 + 0.3 + 0.2 + 0.1
P( X < 2.5 ) = 0.9
b) P( 0.75 < X ≤ 1.5 )
P( 0.75 < X ≤ 1.5 ) = p[ x = 1 ] + p[ x = 1.5 ]
so
P( 0.75 < X ≤ 1.5 ) = 0.3 + 0.2
P( 0.75 < X ≤ 1.5 ) = 0.5