Which expression is equivalent to 3x+7+2x

Answers

Answer 1

Answer:

5x + 7

Step-by-step explanation:

Add 3x and 2 x

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Evaluate: 8x6 + (12-4) divided by 2

Answers

It’s is 28 I will explain how to do it first times 8x6 then subtract 12 and 4 and the two numbers u get add them and then divide it by 2

(10 points) Starting salaries of 64 college graduates who have taken a statistics course have a mean of $42,500 with a standard deviation of $6,800. Find an 90% confidence interval for ????. (NOTE: Do not use commas or dollar signs in your answers. Round each bound to three decimal places.) Lower-bound: 41101.87 Upper-bound: 43898.13

Answers

Answer:

41101.750 to 43898.250

Step-by-step explanation:

Using this formula X ± Z (s/√n)

Where

X = 42500 --------------------------Mean

S = 6800----------------------------- Standard Deviation

n = 64 ----------------------------------Number of observation

Z = 1.645 ------------------------------The chosen Z-value from the confidence table below

Confidence Interval Z

80%. 1.282

85% 1.440

90%. 1.645

95%. 1.960

99%. 2.576

99.5%. 2.807

99.9%. 3.291

Substituting these values in the formula

Confidence Interval (CI) = 42500 ± 1.645(6800/√64)

CI = 42500 ± 1.645(6800/8)

CI = 42500 ± 1.645(850)

CI = 42500 ± 1398.25

CI = 42500+1398.25 ~. 42500-1398.25

CI = 43898.25 ~ 41101.75

In other words the confidence interval is from 41101.750 to 43898.250

Final answer:

To find a 90% confidence interval for the mean starting salary, use the formula CI = sample mean ± (Z * sample standard deviation / √n).

Explanation:

To find a 90% confidence interval for the mean starting salary, we will use the formula:

CI = sample mean ± (Z * sample standard deviation / √n)

Given that the sample mean is $42,500, the sample standard deviation is $6,800, and the number of college graduates is 64, we can substitute these values into the formula to calculate the confidence interval. The lower-bound is $41,101.87 and the upper-bound is $43,898.13.

Learn more about Confidence Interval here:

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I would like to create a rectangular vegetable patch. The fencing for the east and west sides costs $4 per foot, and the fencing for the north and south sides costs only $2 per foot. I have a budget of $176 for the project. What are the dimensions of the vegetable patch with the largest area I can enclose?

Answers

x = E/W dimension
y = N/S dimension

4x + 4x + 2y + 2y = 64
8x + 4y = 64
4y = 64 - 8x
y = 16 - 2x

Area = xy = x(16 - 2x) = 16x - 2x^2

Maximum of y = ax^2 + bx + c is when x = -b / 2a

so x = -16 / -4 = 4

Calculate two iterations of Newton's Method to approximate a zero of the function using the given initial guess. (Round your answers to three decimal places.) f(x) = x9 − 9, x1 = 1.6

Answers

Answer:

Iteration 1: $x_(2)=1.446$

Iteration 2: $x_(3)=1.337$

Step-by-step explanation:

Formula for Newton's method is,

$x_(n+1)=x_n-(f\left(x_n\right))/(f'\left(x_n\right))$

Given the initial guess as $x_(1)=1.6$ , therefore value of n = 1.

Also, $f\left(x\right)=x^(9)-9$ .

Differentiating with respect to x,

$(d)/(dx)\left(f\left(x\right)\right)=(d)/(dx)\left(x^9-9\right)$

Applying difference rule of derivative,

$(d)/(dx)\left(f\left(x\right)\right)=(d)/(dx)\left(x^9\right)-(d)/(dx)\left(9\right)$

Applying power rule and constant rule of derivative,

$(d)/(dx)\left(f\left(x\right)\right)=\left(9x^(9-1)\right)-0$

$(d)/(dx)\left(f\left(x\right)\right)=9x^(8)$

Substituting the value,

$x_(1+1)=x_1-(f\left(x_1\right))/(f'\left(x_1\right))$

$x_(2)=1.6-(f\left(1.6\right))/(f'\left(1.6\right))$

Calculating the value of $f\left(1.6\right)$ and $f'\left(1.6\right)$

Calculating $f\left(1.6\right)$

$f\left(1.6\right)=\left(1.6\right)^(9)-9$

$f\left(1.6\right)=59.71947674$

Calculating $f'\left(1.6\right)$ ,

$f'\left(1.6\right)=9\left(1.6\right)^(8)$

$f'\left(1.6\right)=386.5470566$

Substituting the value,

$x_(2)=1.6-(59.71947674)/(386.5470566)$

$x_(2)=1.446$

Therefore value after second iteration is $x_(2)=1.446$

Now use $x_(2)=1.446$ as the next value to calculate second iteration. Here n = 2

Therefore,

$x_(2+1)=x_2-(f\left(x_2\right))/(f'\left(x_2\right))$

$x_(3)=1.446-(f\left(1.446\right))/(f'\left(1.446\right))$

Calculating the value of $f\left(1.446\right)$ and $f'\left(1.446\right)$

Calculating $f\left(1.446\right)$

$f\left(1.446\right)=\left(1.446\right)^(9)-9$

$f\left(1.446\right)=18.63851065$

Calculating $f'\left(1.446\right)$ ,

$f\left(1.446\right)=9\left(1.446\right)^(8)$

$f\left(1.446\right)=172.0239252$

Substituting the value,

$x_(3)=1.446-(18.63851065)/(172.0239252)$

$x_(3)=1.337$

Therefore value after second iteration is $x_(3)=1.337$

Final answer:

To calculate two iterations of Newton's Method, use the formula xn+1 = xn - f(xn)/f'(xn). Given an initial guess of x1 = 1.6 and the function f(x) = x9 - 9, calculate f(xn) and f'(xn) at x1 and then use the formula to find x2 and x3.