Answer:
To solve (4/9)/x = 12, multiply both sides by x to get (4/9) = 12x. Then, divide both sides by 12. The answer is 27.
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Answer:
Given that the frog jumps every 10 seconds
(using digits from a random number table)
- It requires 7 jumps with 2 in the reverse direction (either left or right) for the frog to get off the board in 60 seconds.
- Alternatively, 3 jumps in the same direction will also lead to the frog being off the board.
- Furthermore, it would take 5 jumps with one in the opposite direction within the time limit of 60 seconds to leave the board.
Step-by-step explanation:
A frog positioned right at the center of a 5ft long board is 2.5 ft away from either edge.
Every 10 seconds, the frog jumps left or right.
If the frog's jumps are LLRLRL, it will remain on the board at the leftmost square.
If it jumps as LLRLL, it will jump off the board after fifty seconds.
Given that the frog jumps every 10 seconds
(using digits from a random number table)
- It requires 7 jumps with 2 in reverse direction (either left or right) for the frog to get off the board in 60 seconds.
- Alternatively, 3 jumps in the same direction will also lead to the frog being off the board.
- Furthermore, it would take 5 jumps with one in the opposite direction within the time limit of 60 seconds to leave the board.
Mel descends 50-35=15 feet in 5-2=3 seconds. Thus, her descent rate is 15/3=5 ft/sec.
Victor descends 60-50=10 feet in 4-1=3 seconds. Hence, his descent rate is 10/3=3⅓ ft/sec. Mel is faster, traveling at 5 ft/sec.
An acute angle measures under 90°. An angle bisector is a ray that divides an angle into two equal neighboring angles. For example, if you have an angle of 270°, which exceeds a semicircle, it divides into two angles of 135° each. In this instance, the resulting angles are not acute; rather, they are obtuse.
Answer:
At the α = 0.10 level, there is no substantial evidence indicating that the average vertical jump for students at this school differs from 15 inches.
Step-by-step explanation:
A hypothesis test is necessary to verify the assertion that the average vertical jump of students diverges from 15 inches.
The null and alternative hypotheses are:

The significance level is set at 0.10.
The sample mean recorded is 17, and the sample standard deviation is 5.37.
The degrees of freedom are calculated as df=(20-1)=19.
The t-statistic is:

The two-tailed P-value corresponding to t=1.67 is P=0.11132.
<pSince this P-value exceeds the significance level, the result is not significant. Therefore, the null hypothesis remains unchallenged.
At the α = 0.10 level, there is no compelling evidence that the average vertical jump of students at this school deviates from 15 inches.