You made two deposits to your bank account this month. One deposit was $17.92, and the second deposit was $15.33. Your balance a
2 answers:
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Answer:
The answer is below
Step-by-step explanation:
a) Cost of hiring the bicycle for 14 days is:
Cost = 65 * 14 days + 44 = 954 riyals
b) The number of days the bicycle is hired is to be the same as the number of days the bicycle is to be hired. Let us assume that the bicycle and helmet is to be hired for x days, hence:
Cost of bicycle = 65 * x + 14 = 65x + 44
Cost of helmet = 12.5 * x = 12.5x
Total cost = 65x + 44 + 12.5x = 77.5x + 44
Since 750 riyals is available, hence:
750 = 77.5x + 44
77.5x = 706
x = 9.11
Therefore the maximum number of days is 11. The total cost for 11 days is:
Total cost = 77.5(9) + 44 = 741.5 riyals
Money not spent = 750 riyals - 741.5 riyals = 8.5 riyals
Answer:
statistic value t= 5.43
p-value: < 0.00001.
Step-by-step explanation:
Hello!
To obtain information over the corrosion-resistance properties of a certain type of steel conduit a random sample of 45 specimens was taken and buried for two years.
The study variable is:
X: Max. penetration of a steel conduit.
The data of the sample
n= 45
sample mean X[bar]= 53.4
sample standard deviation S= 4.2
The conduits are manufactured to have a true average penetration of at most 50 mills, symbolically: μ ≤ 50
The hypothesis is:
H₀: μ ≤ 50
H₁: μ > 50
α: 0.05
To choose the corresponding statistic to use to study the population mean, the variable must have a normal distribution. There is no available information to check this, so I'll just assume that the variable has a normal distribution and, since the population variance is unknown and the sample is small, the statistic to use is a Student t.
Under the null hypothesis, the critical region and the p-value are one-tailed.
Critical value:
Rejection rule:
Reject the null hypothesis when t ≥ 1.68
t=
t= 5.43
The calculated value is greater than the critical value, thedecision is to rject the null hypothesis.
p-value:
P(t ≥ 5.43) = 1 - P(t < 5.43) = < 0.00001.
The p-value is less than α so the decision is to reject the null hypothesis.
Since the null hypothesis was rejected, then the population average of the penetration of the conduits specimens is greater than 50 mils. It is not recommendable to use these conduits.
I hope it helps!