What are the vertices of ΔA'B'C produced by T⟨−3, 6⟩ (ΔABC) = ΔA'B'C?
1 answer:
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Answer:
Option C - Do not reject the null hypothesis. There is not sufficient evidence that the true diameter differs from 0.5 in
Step-by-step explanation:
We are given;
n = 15
t-value = 1.66
Significance level;α = 0.05
So, DF = n - 1 = 15 - 1 = 14
From the one-sample t - table attached, we can see that the p - value of 0.06 at a t-value of 1.66 and a DF of 14
Now, since the P-value is 0.06,it is greater than the significance level of 0.05. Thus we do not reject the null hypothesis. We conclude that there is not sufficient evidence that the true diameter differs from 0.5 in.
Answer:
Step-by-step explanation:
To aid in understanding, draw three circles...
100 were registered for both Math and English.
50 were registered for both Sport and English but not Math.
70 were registered for Math, Sport, and English.
From here we will be able to calculate for those that signed up for only English class: Total number of students that signed up for english class
a. none signed up only for Maths (refer to attached document)
b. How many signed up for neither Math nor English meaning they signed up for only sports which is 70 from the venn diagram in the attachment.
c. How many signed up for an English class and at least one other class?
From the venn diagram:
we could have an English class and maths class - 100
we could also have an a English class and sports - 50
Thus total is 150.
d. What is the total fee for all the students enrolled in classes
From the diagram, 120 (english only and sport only) enrolled for only one class = 120 x $100 = $12,000
From the diagram, 180 (english and sport, maths and sport and english and maths) enrolled for two classes = 180 x $150 = $27,000
From the diagram, only 70 enrolled for the three classes = 70 x $200 = $14,000
Total fees = $12,000 + $27,000 + $14,000 = $53,000
Answer:
or ×
Step-by-step explanation:
In this type of context "of" means to multiply
Answer:
26.11% of the test scores during the past year exceeded 83.
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
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:
It is known that for all tests administered last year, the distribution of scores was approximately normal with mean 78 and standard deviation 7.8. This means that .
Approximately what percentatge of the test scores during the past year exceeded 83?
This is 1 subtracted by the pvalue of Z when . So:
has a pvalue of 0.7389.
This means that 1-0.7389 = 0.2611 = 26.11% of the test scores during the past year exceeded 83.