First, find the 40% discount.
40% of 195 is 78
195 x 0.40 = 78
Subtract that from 195
195 - 78 = 117
Now, find 66 2/3% of 117, since it is an additional reduction after the original discount.
2/3 = 0.667, so 66 2/3% is 66.667%
117 x 0.66667 = 78.00039
And subtract that from 117
117 - 78.00039 = 38.99961, which in dollars would be $38.99. (hundredths, etc. of a cent are usually just dropped; no rounding involved.)
The discount price would then be $38.99
Answer:
Step-by-step explanation:
Since the results for the standardized test are normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = test reults
µ = mean score
σ = standard deviation
From the information given,
µ = 1700 points
σ = 75 points
We want to the probability that a student will score more than 1700 points. This is expressed as
P(x > 1700) = 1 - P(x ≤ 1700)
For x = 1700,
z = (1700 - 1700)/75 = 0/75 = 0
Looking at the normal distribution table, the probability corresponding to the z score is 0.5
P(x > 1700) = 1 - 0.5 = 0.5
Given that,
Sample size= 83
Mean number= 39.04
Standard deviation= 11.51
We know the critical t-value for 95% confidence interval which is equal to 1.989.
We also know the formula for confidence interval,
CI=( mean number - critical t-value*standard deviation/(sample size)^(1/2), mean number + critical t-value*standard deviation/(sample size)^(1/2))
So, we have
CI= (39.04 - 1.989*11.51/83^(1/2), 39.04 + 1.989*11.51/83^(1/2)
CI= (39.04 - 2.513,39.04 + 2.513)
CI= (36.527,41.553)
Therefore, 95% confidence interval for these data is (36.527,41.553), and this result interpret that the true value for this survey sample lie in the interval (36.527,41.553).
Step-by-step explanation:
- 6x-2=8x+11
- 6x-8x=11+2
- -2x=13
- x=13/-2
- x=-6.5
