Answer:
1800 miles
Step-by-step explanation:
No. of miles driven by Mr. Thomas in May = 75
It is given that miles driven in July is 6 times of miles driven by Mr. Thomas in May(75 miles).
Thus
No. of miles driven by Mr. Thomas in July = 6 * No. of miles driven by Mr. Thomas in May = 6*75 = 450 miles.
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Another condition given that miles driven in June is 4 times of miles driven by Mr. Thomas in July(450miles as calculated above).
Thus
No. of miles driven by Mr. Thomas in June = 4 * No. of miles driven by Mr. Thomas in July = 4* 450 miles = 1800 miles.
No. of miles driven by Mr. Thomas in June is 1800 miles.
Answer: 15.6 m
Step-by-step explanation:
Two legs of right angle in right triangle given are 12m and 10m.
For the third side use Pythagoras theorem:
C^2= a^2+b^2
C=√[(12^2)+(10^2)]
C=√[144+100]
C=√[244]= 15.6 m
Answer:
a) 0.4121
b) $588
Step-by-step explanation:
Mean μ = $633
Standard deviation σ = $45.
Required:
a. If $646 is budgeted for next week, what is the probability that the actual costs will exceed the budgeted amount?
We solve using z score formula
= z = (x-μ)/σ, where
x is the raw score
μ is the population mean
σ is the population standard deviation.
For x = $646
z = 646 - 633/45
z = 0.22222
Probability value from Z-Table:
P(x<646) = 0.58793
P(x>646) = 1 - P(x<646) = 0.41207
≈ 0.4121
b. How much should be budgeted for weekly repairs, cleaning, and maintenance so that the probability that the budgeted amount will be exceeded in a given week is only 0.16? (Round your answer to the nearest dollar.)
Converting 0.16 to percentage = 0.16 × 100% = 16%
The z score of 16%
= -0.994
We are to find x
Using z score formula
z = (x-μ)/σ
-0.994 = x - 633/45
Cross Multiply
-0.994 × 45 = x - 633
-44.73 = x - 633
x = -44.73 + 633
x = $588.27
Approximately to the nearest dollar, the amount should be budgeted for weekly repairs, cleaning, and maintenance so that the probability that the budgeted amount will be exceeded in a given week is only 0.16
is $588
X= 10
I= 1
10+10+1+1+1= 23
I hope this helps!
