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2 years ago

the lee family eats at a restaurant for a total bill of 45$ they decided to give a tip of 18% how much tip will they give

Mathematics

1 answer:

Virty [35]2 years ago

4 0

They will tip $8.10.

How to Do it/ Work:

18 * 0.45 = 8.1

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in a random sample of 28 people, the mean commute time to work was 31.2 minutes and the standard deviation was 7.3 minutes. assu

antoniya [11.8K]

Answer:

The margin of error of u is of 3.8.

The 99% confidence interval for the population mean u is between 27.4 minutes and 35 minutes.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 28 - 1 = 27

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 27 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.99}{2} = 0.995$ . So we have T = 2.7707

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.7707\frac{7.3}{\sqrt{28}} = 3.8$

In which s is the standard deviation of the sample and n is the size of the sample.

The margin of error of u is of 3.8.

The lower end of the interval is the sample mean subtracted by M. So it is 31.2 - 3.8 = 27.4 minutes

The upper end of the interval is the sample mean added to M. So it is 31.2 + 3.8 = 35 minutes

The 99% confidence interval for the population mean u is between 27.4 minutes and 35 minutes.

3 0

2 years ago

Help me please math

Dominik [7]

13+9=22

5+18=23

22x23=506

½x5x13=32.5

9x5=45

½x9x18=81

506-32.5-45-81=347.5

5 0

2 years ago

Put the following 3x + y < 8 into linear equation, Slope Intercept formmat<br>

Verdich [7]

We have to get the y alone. So we'll subtract 3x on both sides.

Our new expression is y<-3x+8

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sorry if this doesn't help

5 0

2 years ago

NEED ASAP

Gala2k [10]

Answer:

(- 2, 4 )

Step-by-step explanation:

Given endpoints (x₁, y₁ ) and (x₂, y₂ ) , then the midpoint is

( $\frac{x_{1}+x_{2} }{2}$ , $\frac{y_{1}+y_{2} }{2}$ )

Here (x₁, y₁ ) = A (4, 6 ) and (x₂, y₂ ) = B (- 8, 2 )

midpoint = ( $\frac{4-8}{2}$ , $\frac{6+2}{2}$ ) = ( $\frac{-4}{2}$ , $\frac{8}{2}$ ) = (- 2, 4 )

6 0

2 years ago

Read 2 more answers

Find the area of the region that lies inside the first curve and outside the second curve.

marishachu [46]

Answer:

Step-by-step explanation:

From the given information:

r = 10 cos( θ)

r = 5

We are to find the the area of the region that lies inside the first curve and outside the second curve.

The first thing we need to do is to determine the intersection of the points in these two curves.

To do that :

let equate the two parameters together

So;

10 cos( θ) = 5

cos( θ) = $\dfrac{1}{2}$

$\theta = -\dfrac{\pi}{3}, \ \ \dfrac{\pi}{3}$

Now, the area of the region that lies inside the first curve and outside the second curve can be determined by finding the integral . i.e

$A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} (10 \ cos \ \theta)^2 d \theta - \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ 5^2 d \theta$

$A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} 100 \ cos^2 \ \theta d \theta - \dfrac{25}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ d \theta$

$A = 50 \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix} \dfrac{cos \ 2 \theta +1}{2} \end {pmatrix} \ \ d \theta - \dfrac{25}{2} \begin {bmatrix} \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}}$

$A =\dfrac{ 50}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix} {cos \ 2 \theta +1} \end {pmatrix} \ \ d \theta - \dfrac{25}{2} \begin {bmatrix} \dfrac{\pi}{3} - (- \dfrac{\pi}{3} )\end {bmatrix}$

$A =25 \begin {bmatrix} \dfrac{sin2 \theta }{2} + \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{\dfrac{\pi}{3}} \ \ - \dfrac{25}{2} \begin {bmatrix} \dfrac{2 \pi}{3} \end {bmatrix}$

$A =25 \begin {bmatrix} \dfrac{sin (\dfrac{2 \pi}{3} )}{2}+\dfrac{\pi}{3} - \dfrac{ sin (\dfrac{-2\pi}{3}) }{2}-(-\dfrac{\pi}{3}) \end {bmatrix} - \dfrac{25 \pi}{3}$

$A = 25 \begin{bmatrix} \dfrac{\dfrac{\sqrt{3}}{2} }{2} +\dfrac{\pi}{3} + \dfrac{\dfrac{\sqrt{3}}{2} }{2} + \dfrac{\pi}{3} \end {bmatrix}- \dfrac{ 25 \pi}{3}$

$A = 25 \begin{bmatrix} \dfrac{\sqrt{3}}{2 } +\dfrac{2 \pi}{3} \end {bmatrix}- \dfrac{ 25 \pi}{3}$

$A = \dfrac{25 \sqrt{3}}{2 } +\dfrac{25 \pi}{3}$

The diagrammatic expression showing the area of the region that lies inside the first curve and outside the second curve can be seen in the attached file below.

Download docx

7 0

2 years ago