Question 16..........
2 answers:
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Answer:
11. D. May 20
12. C. Every 70 bags
Step-by-step explanation:
11.
Mark on the calendar what days she has swimming and what days she has diving. (See picture below) Red is Swimming and blue is diving. For swimming, count by 5s. For diving, count by 3s. Find the next square that has both red and blue on it.
Therefore, she has both swimming and diving lessons on May 20th.
12.
Use least common multiples.
Count by 7s and 10s until there is a number that's the same.
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22....... You do not need to count by 2s. This is because you need to have a number that can also be counted to by 10s (a number with 10 as its factor), and 2 reaches all the multiples of 10. It is like 10 already includes 2.
7, 14, 21, 28, 35, 42, 49, 56, 63, 70
10, 20, 30, 40, 50, 60, 70
Therefore, every 70 bags will have all three items.
Answer:
The p-value of the test is 0.0228, which is less than the standard significance level of 0.05, which means that there is evidence that the proportion of freshmen who graduated in the bottom third of their high school class in 2001 has been reduced.
Step-by-step explanation:
Before solving this question, we need to understand the central limit theorem and subtraction of normal variables.
Central Limit Theorem
The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean and standard deviation
Subtraction between normal variables:
When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.
1999:
20 out of 100 in the bottom third, so:
2001:
10 out of 100 in the bottom third, so:
Test if proportion of freshmen who graduated in the bottom third of their high school class in 2001 has been reduced.
At the null hypothesis, we test if the proportion is still the same, that is, the subtraction of the proportions in 1999 and 2001 is 0, so:
At the alternative hypothesis, we test if the proportion has been reduced, that is, the subtraction of the proportion in 1999 by the proportion in 2001 is positive. So:
The test statistic is:
In which X is the sample mean, is the value tested at the null hypothesis, and s is the standard error.
0 is tested at the null hypothesis:
This means that
From the two samples:
Value of the test statistic:
P-value of the test and decision:
The p-value of the test is the probability of finding a difference of at least 0.1, which is the p-value of z = 2.
Looking at the z-table, the p-value of z = 2 is 0.9772.
1 - 0.9772 = 0.0228.
The p-value of the test is 0.0228, which is less than the standard significance level of 0.05, which means that there is evidence that the proportion of freshmen who graduated in the bottom third of their high school class in 2001 has been reduced.