The Cute Kitten Shelter has $486$ cute kittens at the start of Monday. On each day, $\frac13$ of the kittens available at the st
2 answers:
Answer:
216
Step-by-step explanation:
We are told that 1/3 of the available kittens are adopted at the start of the day and no new kittens are added to the shelter. This means 2/3 of the available kittens will remain at the shelter at the end of every day.
Lat us find number of kittens left at the end of Monday.
Now we will find 2/3 of 324 to find kittens left at the shelter at the end of Tuesday.
Therefore, the number of kittens left at The Cute Kitten Shelter at the end of Tuesday will be 216.
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Answer: 225
<u>Step-by-step explanation:</u>
n₁ = 2(1) - 1
= 2 - 1
= 1
n₁₅ = 2(15) - 1
= 30 - 1
= 29
= 15 * 15
= 225
Answer:
From the given data, there is not enough evidence to prove that there is a statistically significant difference between the two population means
Step-by-step explanation:
The number of younger moms, in the study = 837
The average weight gain of younger moms, ₁ = 30.67 pounds
The standard deviation of the weight gain of younger moms, s₁ = 14.69 pounds
(30.67 - 28.52)/√((14.69^2)/837 + 13²/143))
The number of younger moms, in the study = 143
The average weight gain of mature moms, ₂ = 28.52 pounds
The standard deviation of the weight gain of mature moms, s₂ = 13 pounds
The test statistic for the difference in two populations is given as follows;
Therefore, we get;
The test statistic ≈ (1.7919)
Using a graphing calculator, we get;
The critical-t = ±1.971379, p = 0.07459697
Therefore, given that the test statistic, (1.7919), < critical-t (0.07459697), we fail to reject the null hypothesis, therefor, the given data does not provide convincing evidence that there is a significant difference between the two population means