Daniel is planning a 6-night road trip. It will cost $310 for transportation, $275 for food, and $70 per night for a hotel room.
2 answers:
Answer:
13 hours
Step-by-step explanation:
Number of days Daniel is planning for 6-night road trip
Cost of transportation =$ 310
Cost for food=$ 275
Cost per night for hotel= $70
Total number of nights= 6
Total cost for Hotel= 6×$70= $420
Total cost for a trip= $ 310+ $ 275+420
Total cost for a trip= $1005
Savings of Daniel= $ 745
Total money to be needed= $1005-$745
Total money to be needed= $260
Cost of tutoring= $20
Number of hours he needs to do tutoring=
Number of hours he needs to do tutoring=13 hours
Hence, the correct answer is 13 hours
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Answer:
The interval that describes how long it takes for passengers to board the middle 95% of the time is between 40.16 minutes and 55.84 minutes.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
.
Which interval describes how long it takes for passengers to board the middle 95% of the time?
This is between the 2.5th percentile and the 97.5th percentile.
So this interval is the value of X when Z has a a pvalue of 0.025 and the value of X when Z has a pvalue of 0.975
Lower Limit
Z has a pvalue of 0.025 when . So
Upper Limit
Z has a pvalue of 0.975 when . So
The interval that describes how long it takes for passengers to board the middle 95% of the time is between 40.16 minutes and 55.84 minutes.
Answer:
a) 32
b)
Step-by-step Explanation:
Initial pattern has 7 sticks.
Second one has 7+5 sticks.
Third has 7+5+5 sticks.
.
.
.
Sixth has 7+5+5+5+5+5=32 sticks.
$n^{th}$ has $7+ 5(n-1)$ sticks.