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2 years ago

Add x^3 - 4x^2 + 1 to 3x^2 + x Show Your work!

Mathematics

1 answer:

Sladkaya [172]2 years ago

6 0

$\qquad \qquad\huge \underline{\boxed{\sf Answer}}$

Let's solve ~

$\qquad \sf \dashrightarrow \: (x {}^{3} - 4x {}^{2} + 1) + (3 {x}^{2} + x)$

$\qquad \sf \dashrightarrow \: {x}^{3} - 4 {x}^{2} + 1 + 3 {x}^{2} + x$

$\qquad \sf \dashrightarrow \: {x}^{3} - 4 {x}^{2} + 3x {}^{2} + x + 1$

$\qquad \sf \dashrightarrow \: {x}^{3} - {x}^{2} + x + 1$

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. (0.5 point) We simulate the operations of a call center that opens from 8am to 6pm for 20 days. The daily average call waiting

SashulF [63]

Answer:

The 95% t-confidence interval for the difference in mean is approximately (-2.61, 1.16), therefore, there is not enough statistical evidence to show that there is a change in waiting time, therefore;

The change in the call waiting time is not statistically significant

Step-by-step explanation:

The given call waiting times are;

24.16, 20.17, 14.60, 19.79, 20.02, 14.60, 21.84, 21.45, 16.23, 19.60, 17.64, 16.53, 17.93, 22.81, 18.05, 16.36, 15.16, 19.24, 18.84, 20.77

19.81, 18.39, 24.34, 22.63, 20.20, 23.35, 16.21, 21.73, 17.18, 18.98, 19.35, 18.41, 20.57, 13.00, 17.25, 21.32, 23.29, 22.09, 12.88, 19.27

From the data we have;

The mean waiting time before the downsize, $\overline x_1$ = 18.7895

The mean waiting time before the downsize, s₁ = 2.705152

The sample size for the before the downsize, n₁ = 20

The mean waiting time after the downsize, $\overline x_2$ = 19.5125

The mean waiting time after the downsize, s₂ = 3.155945

The sample size for the after the downsize, n₂ = 20

The degrees of freedom, df = n₁ + n₂ - 2 = 20 + 20 - 2 = 38

df = 38

At 95% significance level, using a graphing calculator, we have; $t_{\alpha /2}$ = ±2.026192

The t-confidence interval is given as follows;

$\left (\bar{x}_{1}- \bar{x}_{2} \right )\pm t_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}$

Therefore;

$\left (18.7895- 19.5152 \right )\pm 2.026192 \times \sqrt{\dfrac{2.705152^{2}}{20}+\dfrac{3.155945^2}{20}}$

(18.7895 - 19.5125) - 2.026192*(2.705152²/20 + 3.155945²/20)^(0.5)

The 95% CI = -2.6063 < μ₂ - μ₁ < 1.16025996668

By approximation, we have;

The 95% CI = -2.61 < μ₂ - μ₁ < 1.16

Given that the 95% confidence interval ranges from a positive to a negative value, we are 95% sure that the confidence interval includes '0', therefore, there is sufficient evidence that there is no difference between the two means, and the change in call waiting time is not statistically significant.

6 0

3 years ago

Use the arc length formula to find the length of the curve y = 4x − 5, −1 ≤ x ≤ 2. check your answer by noting that the curve is

professor190 [17]

Ok, here we go. Pay attention. The formula for the arc length is $AL= \int\limits^b_a { \sqrt{1+( \frac{dy}{dx})^2 } } \, dx$ . That means that to use that formula we have to find the derivative of the function and square it. Our function is y = 4x-5, so y'=4. Our formula now, filled in accordingly, is $AL= \int\limits^2_1 { \sqrt{1+4^2} } \, dx$ (that 1 is supposed to be negative; not sure if it is til I post the final answer). After the simplification we have the integral from -1 to 2 of $\sqrt{17}$ . Integrating that we have $AL= \sqrt{17}x$ from -1 to 2. $2 \sqrt{17}-(-1 \sqrt{17} )$ gives us $3 \sqrt{17}$ . Now we need to do the distance formula with this. But we need 2 coordinates for that. Our bounds are x=-1 and x=2. We will fill those x values in to the function and solve for y. When x = -1, y=4(-1)-5 and y = -9. So the point is (-1, -9). Doing the same with x = 2, y=4(2)-5 and y = 3. So the point is (2, 3). Use those in the distance formula accordingly: $d= \sqrt{(2-(-1))^2+(3-(-9))^2}$ which simplifies to $d= \sqrt{9+144}= \sqrt{153}$ . The square root of 153 can be simplified into the square root of 9*17. Pulling out the perfect square of 9 as a 3 leaves us with $3 \sqrt{17}$ . And there you go!

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3 years ago

4×7×25=tell what multiplication strategy to use

Snezhnost [94]

I think u use multiplication

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3 years ago

Determine the interest rate if $2300 invested at simple interest for 2 years earns $402.50 interest

Degger [83]

We would need to use the simple interest formula given by:

I = Prn

In this case, I = $402.50
P = $2300
N = 2 years

Sub these values into the equation,

402.50 = 2300 x r x 2
402.50 = 4600r

Divide both sides by 4600 to get r,

r = 0.0875

Multiply 0.0875 by 100 to get the percentage rate,

0.0875 x 100 = 8.75%

Hope this helped!

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3 years ago

The sum of first three terms of a finite geometric series is - and their product is -. [Hint: Use , a, and ar to represent the f

Veseljchak [2.6K]

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3 years ago

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