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kotegsom [21]
3 years ago
10

Kenny selected 3/10 of number of seashells in a box. Chloe selected 54/100 of the original number of seashells. What is the tota

l fraction of the original number of seashell that both students selected?
Mathematics
2 answers:
beks73 [17]3 years ago
7 0

Answer:

21/25

Step-by-step explanation:

disa [49]3 years ago
5 0

Answer:

84/100 Or Simplified 21/25

Step-by-step explanation:

3/10 x 10/10= 30/100

30/100+ 54/100= 84/100

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If Upper X overbar equals 62​, Upper S equals 8​, and n equals 36​, and assuming that the population is normally​ distributed, c
marishachu [46]

Answer:

The 99% confidence interval would be given by (58.373;65.627)    

We are 99% confident that the true mean for the variable of interest is between 58.373 and 65.627.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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\bar X=62 represent the sample mean

\mu population mean (variable of interest)

s=8 represent the sample standard deviation

n=36 represent the sample size  

Part a: Confidence interval

The confidence interval for the mean is given by the following formula:

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

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=36-1=35

Since the Confidence is 0.99 or 99%, the value of \alpha=0.01 and \alpha/2 =0.005, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,35)".And we see that t_{\alpha/2}=2.72

Now we have everything in order to replace into formula (1):

62-2.72\frac{8}{\sqrt{36}}=58.373    

62+2.72\frac{8}{\sqrt{36}}=65.627    

So on this case the 99% confidence interval would be given by (58.373;65.627)    

We are 99% confident that the true mean for the variable of interest is between 58.373 and 65.627.

3 0
3 years ago
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