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Stephen’s lunch bill is currently at $8.33. Stephen orders a fruit salad for take-out, and wants to leave $2.25 as a tip for his

KatRina [158]

Answer:

$4.42

Step-by-step explanation:

by combining $8.33 and $2.25 you get a combined total of $10.58 and if you subtract that from $15 you get $4.42 as the amount of money remaining.

7 0

3 years ago

PLs Help! Pls help! Pls Help!

LiRa [457]

70% of 60,000,000 is 42,000,000

5 0

3 years ago

Between 8 a.m. and 3 p.m. on a summer day, the outdoor temperature increased by 23 degrees. Between 3 p.m. and 10 p.m., the temp

AleksAgata [21]

Answer:

46

Step-by-step explanation:

46

5 0

3 years ago

PLEASE ANSWER ASAP FOR BRAINLEST!!!!!!!!!!!!!!!!!!!!!!!

Doss [256]

Answer:

12 x 9 (12 on the top and bottom and 9 on the sides)

Step-by-step explanation:

100 km = 2 units, or 2 boxes

450 x 2/100 = 900/100 = 9

(not sure if this would help or not)

4 0

3 years ago

A cellular phone company monitors monthly phone usage. The following data represent the monthly phone use in minutes of one part

Sergeu [11.5K]

Answer:

The standard deviation increased but there was no change in the interquantile range

Step-by-step explanation:

We are given the following data in the question:

320, 411, 348, 537, 420, 449, 462, 403, 454, 517, 515, 358, 438, 541, 387, 368, 502, 437, 431, 428.

n = 20

a) Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{8726}{20} = 436.3$

Sum of squares of differences =

13525.69 + 640.09 + 7796.89 + 10140.49 + 265.69 + 161.29 + 660.49 + 1108.89 + 313.29 + 6512.49 + 6193.69 + 6130.89 + 2.89 + 10962.09 + 2430.49 + 4664.89 + 4316.49 + 0.49 + 28.09 + 68.89 = 75924.2

$S.D = \sqrt{\frac{75924.2}{19}} = 63.21$

Sorted Data = 320, 348, 358, 368, 387, 403, 411, 420, 428, 431, 437, 438, 449, 454, 462, 502, 515, 517, 537, 541

$IQR = Q_3 - Q_1\\Q_3 = \text{upper median},\\Q_1 = \text{ lower median}$

$Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}$

$Median = \frac{431 + 437}{2} = 434$

$Q_1 = \frac{387 + 403}{2} = 395\\\\Q_3 = \frac{462 + 502}{2} = 482$

IQR = $482 - 395 = 87$

b) After changing the observation

0, 411, 348, 537, 420, 449, 462, 403, 454, 517, 515, 358, 438, 541, 387, 368, 502, 437, 431, 428

$Mean =\displaystyle\frac{8406}{20} = 420.3$

Sum of squares of differences =

176652.09 + 86.49 + 5227.29 + 13618.89 + 0.09 + 823.69 + 1738.89 + 299.29 + 1135.69 + 9350.89 + 8968.09 + 3881.29 + 313.29 + 14568.49 + 1108.89 + 2735.29 + 6674.89 + 278.89 + 114.49 + 59.29 = 247636.2

$S.D = \sqrt{\frac{247636.2}{19}} = 114.16$

Sorted Data = 0, 348, 358, 368, 387, 403, 411, 420, 428, 431, 437, 438, 449, 454, 462, 502, 515, 517, 537, 541

$Median = \frac{431 + 437}{2} = 434$

$Q_1 = \frac{387 + 403}{2} = 395\\\\Q_3 = \frac{462 + 502}{2} = 482$

IQR = $482 - 395 = 87$

Thus. the standard deviation increased but there was no change in the interquantile range.

5 0

3 years ago