Solve k−7≤0. Graph the solution
1 answer:
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Using the z-table, the probability that a student taking this test will finish in 100 minutes or less is 0.0824 or 8.24%.
For a normally distributed set of data, given the mean and standard deviation, the probability can be determined by solving the z-score and using the z-table.
First, solve for the z-score using the formula below.
z-score = (x – μ) / σ
where x = individual data value = 100
μ = mean = 125
σ = standard deviation = 18
z-score = (100 – 125) / 18
z-score = (-25) / 18
z-score = -1.39
Find the probability that corresponds to the z-score in the z-table. (see attached images)
-1.39 - (-1.3) : -1.4 - (-1.3) = x - 0.0968 : 0.0808 - 0.0968
-0.09 : -0.1 = x - 0.0968 : -0.016
x - 0.0968 = -0.09(-0.016)/-0.1
x = -0.0144 + 0.0968
x = 0.0824
x = 0.0824
Hence, the probability that a student taking this test will finish in 100 minutes or less is 0.0824 or 8.24%.
Learn more about probability here: brainly.com/question/26822684
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Image showing the arena ticket prices is missing, so i have attached it.
Answer:
$10000 more money was spent when the tickets first went on sale than after the first 2 weeks
Step-by-step explanation:
We are told that when the full season tickets first went on sale, that 2000 full season tickets were sold for section N.
Now, from the area ticket prices table attached, we can see that full season tickets for Section N costs $20
Thus, amount of money spent when the tickets first went on sale = 2000 × 20 = $40000
We are told that 2 weeks after the tickets first went on sale, they sold 1500 tickets. Thus, amount spent after 2 weeks release = 1500 × 20 = $30000
Difference in amount spent at the beginning and after 2 weeks = $40000 - $30000 = $10000
Thus, $10000 more money was spent when the tickets first went on sale than after the first 2 weeks
