Hailey simplified the expression (12xy74x4y3)2 using the following steps:

Answer: (97.98, 112.020)

Step-by-step explanation: We will create a 95% confidence interval for the average weight of melons.

Given the information, we determine that the critical value for the interval needs to be retrieved from a t distribution table due to the sample size being below 30 (specifically, 20), and we are provided with the sample standard deviation (s = 15 lb).

The parameters provided are:

Sample mean = x = 105 lb

Sample standard deviation = s = 15 lb

Sample size = n = 20

To establish the 95% confidence interval, we indicate that the level of significance is 5%.

The formula for the confidence interval is:

u = x + tα/2 × s/√n... for the upper limit

u = x - tα/2 × s/√n... for the lower limit.

tα/2 represents the critical value for the test (which will be determined using the t distribution table).

To derive tα/2, we look for the value based on the degrees of freedom (sample size - 1) against the significance level for a two-tailed test (α/2 = 0.025%) in a t distribution table.

For the upper limit, we calculate:

u = 105 + 2.093×15/√20

u = 105 + 2.093× (3.3541)

u = 105 + 7.020

u = 112.020.

<pfor the="" lower="" limit="" we="" find:="">

u = 105 - 2.093×15/√20

u = 105 - 2.093× (3.3541)

u = 105 - 7.020

u = 97.98

Confidence interval (97.98, 112.020)

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To calculate this, a specific formula will be necessary. Years = log (total/principal) / [n * log (1 + rate / n)]. Part A) For Calvin: $400 at 5% monthly results in $658.80; Time =? Monthly compounding, n = 12. Thus, Years = log(658.80/400) / [12 * log(1+ (.05/12))]. Subsequently, Years = log( 1.647) / (12 * log(1.0041666667)). Then, Years = 0.21669359917 / 12 * 0.0018058008777. Thus resulting in Years = 0.21669359917 / 0.0216696105. Ultimately, Years ≈ 9.999884362. Part B) For Makayla: $300 at 6% quarterly yields $613.04; Time=? Quarterly compounding, n = 4. Therefore, Years = log(613.04/300) / [4 * log (1 +.06/4)]. This results in Years = log(2.0434666667) / (4 * log(1.015)). Years thus equals 0.31036755784 / (4 * 0.0064660422492), resulting in Years ≈ 11.9999044949. The approximate difference is about 3 years.

Sample Answer: To determine the difference between 10 and 13, you apply the concept of additive inverses. Juan is lacking $3 to buy his supplies.