Answer:
<h2>In this particular case,the target population of interest to the university administration constitutes the university students.</h2>
Step-by-step explanation:
- The university administration is interested to conduct a statistical study to identify the average or mean time taken by the students to find a vacant parking spot.
- Therefore,the research topic here is the average time taken by the university students to find parking spot. The administrator collects an inconspicuous sample of 240 samples from the target population of the study,which is the overall student population of the university.
- The sample collected by the university administration is used to observe the average or mean parking time by the university students.
Answer:
25 wooden boards
Step-by-step explanation:
Given that:
Width of wooden board = 6 inches
Number of boards required to build a fence of 150 inches long if there are no gaps :
The lenght of fence / width of wooden board
= 150 inches / 6 inches
= 25 wooden boards
The answer is: " 100 cultures " .
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Take the number of the day we are at (from the beginning; in this case, "9" (for "Day 9").
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Then add "1" to that number (in this case, " 9 + 1 = 10" .
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Then, take that value (in this case, "10"); and "square" that value ; (in this case; " 10² = 10 * 10 = 100 . This is the answer.
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The answer is: " 100 cultures " .
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<u>Note</u><u>:</u> We get this steps by noticing the trend from the information provided.
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Answer:
a)
b) 
c) 
Step-by-step explanation:
Part a
The significance level given is
and the degrees of freedom are given by:

Since we are conducting a right tailed test we need to find a critical value on the t distirbution who accumulates 0.1 of the area in the right and we got:

Part b
The significance level given is
and the degrees of freedom are given by:

Since we are conducting a left tailed test we need to find a critical value on the t distirbution who accumulates 0.01 of the area in the left and we got:

Part c
The significance level given is
and
and the degrees of freedom are given by:

Since we are conducting a two tailed test we need to find a critical value on the t distirbution who accumulates 0.025 of the area on each tail and we got:

Answer:

Explanation:
05/10
Simplifying
1/2
or 2^-1
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