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miss Akunina [59]
3 years ago
8

The mean life expectancy of a certain type of light bulb is 895 hours with a standard deviation of 18 hours. What is the approxi

mate standard deviation of the sampling distribution of the mean for all samples with n = 40?
Mathematics
1 answer:
Aleks [24]3 years ago
6 0

Answer:

The approximate standard deviation of the sampling distribution of the mean for all samples with n = 40 is 2.8460.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

What is the approximate standard deviation of the sampling distribution of the mean for all samples with n = 40?

We have that \sigma = 18

So

s = \frac{18}{\sqrt{40}} = 2.8460

The approximate standard deviation of the sampling distribution of the mean for all samples with n = 40 is 2.8460.

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Find the values of a and b that complete the mapping diagram.
Rufina [12.5K]

Value of a and b that completes the mapping diagram will be a=0 and b=1

• Mapping diagram simply means a function that has a special type of relation where the elements of the domain are paired with that of the range.

• Mapping diagram consists of two parallel columns, the first column depicts the domain of the function whereas the second column depicts the range.

• Lines are drawn from first column to second column i.e. from domain to range, to represent the relation between 2 elements.

• Mapping is quite similar to making a flowchart, showing input and output values.

• Point to be noted that every mapping is a relation but every relation is not a mapping.

Value of a and b that completes the mapping diagram will be a=0 and b=1

Learn more about mapping

brainly.com/question/2328150

#SPJ9

6 0
1 year ago
An employee joined a company in 2009 with a starting salary of $50,000. Every year this employee receives a raise of $1000 plus
stepladder [879]

Answer:

(a) The required recurrence relation for  the salary of the employee of n years after 2009 is a_n=1.05a_{n-1}+1000.

(b)The salary of the employee will be $83421.88 in 2017.

(c) \therefore a_n=70,000 . \ 1.05^n-20,000

Step-by-step explanation:

Summation of a G.P series

\sum_{i=0}^n r^i= \frac{r^{n+1}-1}{r-1}

(a)

Every year the salary is increasing 5% of the salary of the previous year plus $1000.

Let a_n represents the salary of the employee of n years after 2009.

Then a_{n-1} represents the salary of the employee of (n-1) years after 2009.

Then a_n= a_{n-1}+5\%.a_{n-1}+1000

             =a_{n-1}+0.05a_{n-1}+1000

             =(1+0.05)a_{n-1}+1000

            =1.05a_{n-1}+1000

The required recurrence relation for  the salary of the employee of n years after 2009 is a_n=1.05a_{n-1}+1000.

(b)

Given, a_0=\$50,000

a_n=1.05a_{n-1}+1000

Since 2017 is 8 years after 2009.

So, n=8.

∴ a_8

=1.05 a_7+1000

=1.05(1.05a_6+1000)+1000

=1.05^2a_6+1.05\times 1000+1000

=1.05^2(1.05a_5+1000)+1.05\times 1000+1000

=1.05^3a_5+1.05^2\times 1000+1.05\times 1000+1000

=1.05^3(1.05a_4+1000)+1.05^2\times 1000+1.05\times 1000+1000

=1.05^4a_4+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000

=1.05^4(1.05a_3+1000)+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000

=1.05^5a_3+1.05^4\times1000+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000

=1.05^5(1.05a_2+1000)+1.05^4\times1000+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000

=1.05^6a_2+1.05^51000+1.05^4\times1000+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000

=1.05^6(1.05a_1+1000)+1.05^51000+1.05^4\times1000+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000

=1.05^7a_1+1.05^6\times1000+1.05^51000+1.05^4\times1000+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000

=1.05^7(1.05a_0+1000)+1.05^6\times1000+1.05^51000+1.05^4\times1000+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000

=1.05^8a_0+1.05^7\times1000+1.05^6\times1000+1.05^51000+1.05^4\times1000+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000

=1.05^8a_0+(1.05^7+1.05^6+1.05^5+1.05^4+1.05^3+1.05^2+1.05+1)1000

=1.05^8 \times 50,000+\frac{1.05^8-1}{1.05-1}\times 1000

=1.05^8\times 50,000+20,000(1.58^8-1)

=70,000\times 1.05^8-20,000

≈$83421.88

The salary of the employee will be $83421.88 in 2017.

(c)

Given, a_0=\$50,000

a_n=1.05a_{n-1}+1000

We successively apply the recurrence relation

a_n=1.05a_{n-1}+1000

    =1.05^1a_{n-1}+1.05^0.1000

   =1.05^1(1.05a_{n-2}+1000)+1.05^0.1000

   =1.05^2a_{n-2}+1.05^1.1000+1.05^0.1000

   =1.05^2(1.05a_{n-3}+1000)+(1.05^1.1000+1.05^0.1000)

   =1.05^3a_{n-3}+(1.05^2.1000+1.05^1.1000+1.05^0.1000)

                    ...............................

                   .................................

  =1.05^na_{n-n}+\sum_{i=0}^{n-1}1.05^i.1000

 =1.05^na_0+1000\sum_{i=0}^{n-1}1.05^i

 =1.05^n.50,000+1000.\frac{1.05^n-1}{1.05-1}

 =1.05^n.50,000+20,000.(1.05^n-1)

 =(50,000+20,000)1.05^n-20,000

 =70,000 . \ 1.05^n-20,000

\therefore a_n=70,000 . \ 1.05^n-20,000

6 0
3 years ago
What are the outliers for the data set?<br> {70, 67, 66, 61, 66, 58, 68, 71, 80, 78, 72}
Ket [755]

Answer:

65 and 80

Step-by-step explanation:

Because an outlier is greater than and lesser than everything in a set

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Write the function whose graph is the graph of y=√x​, but is shifted to the left 7 units .
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Step-by-step explanation:

2x + 7 = 15

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Solve the equation. Show work for partial credit.
Natali5045456 [20]

Answer:

x = \frac{8}{5}

Step-by-step explanation:

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\frac{8}{x} = 5 ( multiply both sides by x )

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\frac{8}{5} = x

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