<span>Step 1: Find Q1.Q1 is represented by the left hand edge of the “box” (at the point where the whisker stops). In the above graph, Q1 is approximately at 2.6. ...Step 2: Find Q3. ...
<span>Step 3: Subtract the number you found in step 1 from the number you found in step 3.</span></span>
Answer:
(c) H0 should be rejected
Step-by-step explanation:
Null hypothesis (H0): population mean is equal to 5
Alternate hypothesis (Ha): population mean is greater than 5
Z = (sample mean - population mean) ÷ (sd/√n)
sample mean = 5.3, population mean = 5, sd = 1, n = 500
Z = (5.3 - 5) ÷ (1/√500) = 0.3 ÷ 0.045 = 6.67
Using the normal distribution table, for a one tailed test at 0.01 significance level, the critical value is 2.326
Conclusion:
Since 6.67 is greater than 2.326, reject the null hypothesis (H0)
I believe the answer would be 300. One eighth of the class would be 100 and 1/2 of the class would be 400. 800-(400+100)=300. Hope this helped!
5)Substituting any of the following equations; i chose bottom into top:
-2x+5=x+2
=>-3x=-3
=>x=1
Sub the x in any of the eqs to find y so
Y=1+2
=>Y=3
ANS: (1,3)
6) substituting the top equation into the bottom one:
3x+2x+4=9
=>5x=13
=>x=13/5 or 2.6
Sub the x in any of the eqs to find the y so
Y=5.2+4
=>y=9.2
ANS: (2.6, 9.2)
Corrected Question
Sue travels by bus or walks when she visits the shops. The probability that she catches the bus to the shops is 0.4. The probability she catches the bus from the shops is 0.7. Show the probability that Sue walks at one way is 0.72
Answer:
(Proved)
Step-by-step explanation:
Sue travels by bus or walks when she visits the shops.
Let the event that she catches the bus to the shop=A
Let the event that she catches the bus from the shop=B
P(A)=0.4
P(B)=0.7
Both A and B are independent events.
Therefore,Probability that she catches the bus to and from the shop:
P(A∩B) = 0.4 X 0.7= 0.28
Probability Sue walks at least one way 
Hence, the probability that Sue walks at least one way is 0.72.
