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Pachacha [2.7K]
3 years ago
13

How many pounds of each type of fruit did she buy? she bought

Mathematics
1 answer:
il63 [147K]3 years ago
8 0

Answer:

1 pound

Step-by-step explanation:

becuase ; the question ask about each type of fruit

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Javier is comparing two checking accounts. One has a monthly fee of $10 and a per-check fee of $0.10, and the other has a monthl
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When Javier writes 40 on both accounts they cost him the same amount of money so it should be 41.

Javier needs to write 41 checks for the second bank to beat better

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Please explain your answer as well, THANKS!!!
dusya [7]

Answer:

11,232,000

Step-by-step explanation:

The key to this problem is knowing "repetition is NOT ALLOWED".

<u>For the first 3 letters slot:</u>

For the 1st slot, we can pick any of the 26 alphabets

For the 2nd slot, we can pick any of the 25 remaining alphabets (since 1 is already taken)

For the 3rd slot, we can pick any of the 24 remaining letters (since 2 is already taken)

<u>For the three digits slot:</u>

For the 1st slot, we can pick any one of the 10 digits

For the 2nd slot, we can pick any one of the 9 digits left, since 1 is taken already

For the 3rd slot, we can pick any one of the 8 digits left, since 2 is already taken

The number of license plates, thus, would be all of these multiplied together:

26 * 25 * 24 * 10 * 9 * 8 = 11,232,000

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3 years ago
Solve 10/6 + 3/6- 5/6 mentally​
Alinara [238K]

Answer:

4/3

Step-by-step explanation:

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Select the correct symbol
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Answer:

C

Step-by-step explanation:

\sqrt{33} =5.74...\\\frac{16}{3} =5.33...

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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
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