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3 years ago

5) A submarine is cruising at -40 meters (40 meters below the surface). It

Mathematics

1 answer:

Verdich [7]3 years ago

5 0

Answer:

-25

Step-by-step explanation:

You begin with -40, then decrease by 20 so your new value is -60, then increase by 35 so your final value is -25m... hope this helps (:

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Find the 10 th term of the following geometric sequence. 1,5, 25, 125, ...

IceJOKER [234]

Answer:

1953125

Step-by-step explanation:

5: 625

6: 3125

7: 15625

8: 78125

9: 390625

10: 1953125

Sequence

×5

6 0

3 years ago

Suppose that the sample standard deviation was s = 5.1. Compute a 98% confidence interval for μ, the mean time spent volunteerin

NISA [10]

Answer:

The 95% confidence interval would be given by (5.139;5.861)

Step-by-step explanation:

1) 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.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

2) Confidence interval

Assuming that $\bar X =5.5$ and the ranfom sample n=1086.

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:

$df=n-1=1086-1=1085$

Since the Confidence is 0.98 or 98%, the value of $\alpha=0.02$ and $\alpha/2 =0.01$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.01,1085)".And we see that $t_{\alpha/2}=2.33$

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

$5.5-2.33\frac{5.1}{\sqrt{1086}}=5.139$

$5.5+2.333\frac{5.1}{\sqrt{1086}}=5.861$

So on this case the 95% confidence interval would be given by (5.139;5.861) We are 98% confident that the mean time spent volunteering for the population of parents of school-aged children is between these two values.

5 0

3 years ago

Define table represents a group frequency discretion of the number of hours spent on the computer per week for 49 student. What

Scilla [17]

Observe the given data distribution table carefully.

The 5th class interval is given as,