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11 days ago

Time intervals measured with a stopwatch typically have an uncertainty of about 0.2 seconds, due to human reaction time at the s

tart and stop moments. What is the percent uncertainty of a hand-timed measurement of (a) 5.5 s, (b) 55 s, (c) 5.5 min?Time intervals measured with a stopwatch typically have an uncertainty of about 0.2 seconds, due to human reaction time at the start and stop moments. What is the percent uncertainty of a hand-timed measurement of (a) 5.5 s, (b) 55 s, (c) 5.5 min?

Mathematics

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Dev has 9 shells. Zoe has 55 shells, Zoe gives some shells to Dev. Now zoe has 3 times as many shells as dev. How many shells do

zzz [12365]

Zoe gives 7 shells to Dev, leaving her with 48. Dev then has 16 shells in total. Step-by-step explanation: Denote the initial shell count for Dev as A and Zoe’s as B. Based on the scenario where Zoe transfers x shells to Dev while then possessing three times the number of shells relative to him, the resulting references show that Zoe provided 7 shells, leaving her count at 48 while Dev's total became 16.

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1 month ago

An experiment was conducted to record the jumping distances of paper frogs made from construction paper. Based on the sample, th

tester [12383]

Answer:

$9.9676 - 2.326*0.5904 =8.594$

$9.9676 + 2.326*0.5904 =11.341$

Step-by-step explanation:

Notation

$\bar X$ is the sample mean

$\mu$ indicates the population mean (the variable of interest)

s signifies the sample standard deviation

n denotes the sample size

Solution to the problem

The mean's confidence interval is derived from the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In this instance, the 9% confidence interval corresponds to:

$8.8104 \leq \mu \leq 11.1248$

We can determine the mean using the following:

$\bar X = \frac{8.8104 +11.1248}{2}= 9.9676$

Additionally, the margin of error can be calculated as:

$ME= \frac{11.1248- 8.8104}{2}= 1.1572$

The margin of error for this situation is expressed as:

$ME = t_{\alpha/2}\frac{s}{\sqrt{n}} = t_{\alpha/2} SE$

Next, we find the standard error:

$SE = \frac{ME}{t_{\alpha/2}}$

The critical value for a 95% confidence interval using a normal standard distribution is roughly 1.96, and substituting gives us:

$SE = \frac{1.1572}{1.96}= 0.5904$

For the 98% confidence interval, the significance corresponds to $\alpha=1-0.98= 0.02$ and $\alpha/2 = 0.01$ with a critical value of 2.326, yielding a confidence interval of:

$9.9676 - 2.326*0.5904 =8.594$

$9.9676 + 2.326*0.5904 =11.341$

8 0

2 months ago

Maci made $170 grooming dogs one day with her mobile dog grooming business. She charges $60 per appointment and earned $50 in ti

tester [12383]

Answer:

I've got no idea; I just forgot the answer, but I did my best.

Step-by-step explanation:

6 0

3 months ago

Read 2 more answers