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jasenka [17]
2 years ago
13

Last year, Gilbert contributed $75 toward his 401(k) account each month. If his employer matched 16% of his contributions, what

was the total amount contributed to the 401(k) at the end of the year?
Mathematics
2 answers:
Levart [38]2 years ago
7 0

Answer:

$1044 was contributed.

Step-by-step explanation:

Last year, Gilbert contributed toward his 401(k) account each month = $75.00

Contribution in a year = $75.00 × 12 = $900.00

His employer matched 16% of his contribution

Employer's contribution = 16% of $900.00

\frac{16}{100} × 900

= 0.16 × 900 = $144.00

Total contribution = $900 + $144 = $1044

The total amount $1044 was contributed to the 401(k) at the end of the year.

Darya [45]2 years ago
5 0
$75 to his 401k each month......75(12) = 900 in a yr.
his employee matched 16% of his contributions..
16% of 900 =
0.16(900) = 144

so the total amount contributed was : 900 + 144 = $ 1044
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Answer:

Step-by-step explanation:

Givens

Circle:      3 grams       Number:  1  right hand side Number LHS: 2

Square:   2 grams        Number:  2 Number LHS: 5

Triangle: x                    Number:  5  Number LHS: 2

Formula

Mass RHS   = Mass LHS

Solution

3*1 + 2*2 + 5x = 3*2 + 5*2 + 2*x        Combine

3 + 2 + 5x = 6 + 10 + 2x

5 + 5x = 16 + 2x                                 Subtract 5 from both sides

5-5+5x = 16-5 + 2x                            Combine

5x = 11 + 2x                                       Subtract 2x from both sides

3x = 11                                               Divide by 3

x = 11/3

As you can see, I'm not getting any of the answers that you have provided. Ask your teacher how it was done. (The method used is correct).

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Travka [436]

Answer:

a) Point estimate of the population mean = 4.883

b) B.There is 95​% confidence that the population mean pH of rain water is between 4.646 and 5.120.

c) C.There is 99​% confidence that the population mean pH of rain water is between 4.549 and 5.217.

d) As the level of confidence​ increases, the width of the interval increases.

This makes sense since the margin of error increases as well.

Step-by-step explanation:

We have a sample for the pH of rain.

The mean of the sample is:

M=\dfrac{1}{12}\sum_{i=1}^{12}(5.3+5.72+4.38+4.8+5.02+...+4.56+4.68)\\\\\\ M=\dfrac{58.59}{12}=4.883

The sample standard deviation is:

s=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^{12}(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{11}\cdot [(5.3-4.883)^2+(5.72-4.883)^2+...+(4.68-4.883)^2]}\\\\\\

s=\sqrt{\dfrac{1}{11}\cdot [(0.17)+...+(0.1)+(0.04)]}\\\\\\s=\sqrt{\dfrac{1.526625}{11}}=\sqrt{0.1387841}\\\\\\s=0.373

a) The point estimation for the population mean is the sample mean and has a value of 4.883.

b) We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=4.883.

The sample size is N=12.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

s_M=\dfrac{s}{\sqrt{N}}=\dfrac{0.373}{\sqrt{12}}=\dfrac{0.373}{3.464}=0.108

The t-value for a 95% confidence interval is t=2.201.

The margin of error (MOE) can be calculated as:

MOE=t\cdot s_M=2.201 \cdot 0.108=0.237

Then, the lower and upper bounds of the confidence interval are:

LL=M-t \cdot s_M = 4.883-0.237=4.646\\\\UL=M+t \cdot s_M = 4.883+0.237=5.12

The 95% confidence interval for the mean is (4.646, 5.120).

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c) We have to calculate a 99% confidence interval for the mean.

The t-value for a 99% confidence interval is t=3.106.

The margin of error (MOE) can be calculated as:

MOE=t\cdot s_M=3.106 \cdot 0.108=0.334

Then, the lower and upper bounds of the confidence interval are:

LL=M-t \cdot s_M = 4.883-0.334=4.549\\\\UL=M+t \cdot s_M = 4.883+0.334=5.217

The 99% confidence interval for the mean is (4.549, 5.217).

C.There is 99​% confidence that the population mean pH of rain water is between 4.549 and 5.217.

d) When the confidence level is increased, the width also increases as it has to include more possible values for the true mean of the population.

As the level of confidence​ increases, the width of the interval increases.

This makes sense since the margin of error increases as well.

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