Need help with question 15! please!!! This is due 02/2/22 which is tmr please!
1 answer:
Answer: x = 6
Step-by-step explanation:
Because we are given that the two figures are already similar, we can set up a proportional relationship to determine the value of x. Since the larger rectangle has a width of 21 and the smaller rectangle has a width of 7, those values can be used as the numerators in the proportion. Since the larger rectangle's length is x plus the additional 3 units on the smaller rectangle, x+3 and 3 can be used as the denominators in the proportion.
Next, we set up the proportion as follows:
Then, you cross multiply to get the equation:
After solving down algebraically, you get the value of x to be 6.
I hope this helps!
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Answer:
The 95% confidence interval for the population's mean length of vacation, in days, is (4.24, 4.76).
The 92% confidence interval for the population's mean length of vacation, in days, is (4.27, 4.73).
Step-by-step explanation:
We have the standard deviations for the sample, which means that the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 85 - 1 = 84
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 84 degrees of freedom(y-axis) and a confidence level of . So we have T = 1.989.
The margin of error is:
In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 4.5 - 0.26 = 4.24 days
The upper end of the interval is the sample mean added to M. So it is 4.5 + 0.26 = 4.76 days
The 95% confidence interval for the population's mean length of vacation, in days, is (4.24, 4.76).
92% confidence interval:
Following the sample logic, the critical value is 1.772. So
The lower end of the interval is the sample mean subtracted by M. So it is 4.5 - 0.23 = 4.27 days
The upper end of the interval is the sample mean added to M. So it is 4.5 + 0.23 = 4.73 days
The 92% confidence interval for the population's mean length of vacation, in days, is (4.27, 4.73).