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3 years ago

11

a cleaning compony charges 125$ To visit a home. and 40$ to clean each room if the total bill was $365 how many rooms were clean

ed

1 answer:

FinnZ [79.3K]3 years ago

7 0

Answer:

6 rooms were cleaned

Step-by-step explanation:

365-125=240

240/40=6 rooms

PLEASE MARK BRAINLIEST

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The estimated daily living costs for an executive traveling to various major cities follow. The estimates include a single room

Alexandra [31]

Answer:

$\bar x = 260.1615$

$\sigma = 70.69$

The confidence interval of standard deviation is: $53.76$ to $103.25$

Step-by-step explanation:

Given

$n =20$

See attachment for the formatted data

Solving (a): The mean

This is calculated as:

$\bar x = \frac{\sum x}{n}$

So, we have:

$\bar x = \frac{242.87 +212.00 +260.93 +284.08 +194.19 +139.16 +260.76 +436.72 +355.36 +.....+250.61}{20}$

$\bar x = \frac{5203.23}{20}$

$\bar x = 260.1615$

$\bar x = 260.16$

Solving (b): The standard deviation

This is calculated as:

$\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}$

$\sigma = \sqrt{\frac{(242.87 - 260.1615)^2 +(212.00- 260.1615)^2+(260.93- 260.1615)^2+(284.08- 260.1615)^2+.....+(250.61- 260.1615)^2}{20 - 1}}$ $\sigma = \sqrt{\frac{94938.80}{19}}$

$\sigma = \sqrt{4996.78}$

$\sigma = 70.69$ --- approximated

Solving (c): 95% confidence interval of standard deviation

We have:

$c =0.95$

So:

$\alpha = 1 -c$

$\alpha = 1 -0.95$

$\alpha = 0.05$

Calculate the degree of freedom (df)

$df = n -1$

$df = 20 -1$

$df = 19$

Determine the critical value at row $df = 19$ and columns $\frac{\alpha}{2}$ and $1 -\frac{\alpha}{2}$

So, we have:

$X^2_{0.025} = 32.852$ ---- at $\frac{\alpha}{2}$

$X^2_{0.975} = 8.907$ --- at $1 -\frac{\alpha}{2}$

So, the confidence interval of the standard deviation is:

$\sigma * \sqrt{\frac{n - 1}{X^2_{\alpha/2} }$ to $\sigma * \sqrt{\frac{n - 1}{X^2_{1 -\alpha/2} }$

$70.69 * \sqrt{\frac{20 - 1}{32.852}$ to $70.69 * \sqrt{\frac{20 - 1}{8.907}$

$70.69 * \sqrt{\frac{19}{32.852}$ to $70.69 * \sqrt{\frac{19}{8.907}$

$53.76$ to $103.25$

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AD= 36 cm. Points C, B∈AD, such that AB:BC:CD=2:3:4. Find the distance of midpoints of the segments AB and CD.

brilliants [131]

Answer:

The distance of the midpoints of AB and CD is 24 cm

Step-by-step explanation:

The given information are;

The length of AD = 36 cm.

C and B are points on AD

The ratio of AB:BC:CD 2:3:4

The distance of the midpoints of segments AB and CD required

Therefore, we have the following proportion of the total length, 36

Segment AB = 2/(2 + 3 + 4) = 2/9×36 = 8 cm

Segment BC = 3/(2 + 3 + 4) = 3/9×36 = 12 cm

Segment CD = 4/(2 + 3 + 4) = 4/9×36 = 16 cm

The x-coordinate of the midpoint of segment AB = 4 cm

The x-coordinate of the midpoint of segment CD = 8 + 12 + 16/2 = 28 cm

Which gives;

The distance of midpoint of segment AB from A = 4 cm

The distance of midpoint of segment CD from A = 28 cm

And the distance of the midpoints of AB and CD = 24 - 4 = 24 cm.

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