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victus00 [196]
2 years ago
11

Sum of 3y and 5 is 47

Mathematics
2 answers:
Dennis_Churaev [7]2 years ago
8 0
3y + 5 = 47
- 5 -5
——————
3y = 42
———— 42/3 = 14
3y

y = 14


CHECK YOUR WORK:

y = 14 , so 3(14) + 5 should equal 47!
Viefleur [7K]2 years ago
6 0

subtract 5 from both sides to get 3y=42

divide both sides by 3 to get y = 14

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The scale factor of figure JKLMN to figure PQRST is 3:2. If KL = 9 cm and MN = 15 cm, what is the length of side QR?
Alecsey [184]
The length of side QR is 6 cm
5 0
3 years ago
A girl has 3 different swimsuits, 4 different shorts and 3 different pairs of flip-flops. In how many different ways can this gi
In-s [12.5K]

Answer:

Option D is the correct answer.

Explanation:

 A girl has 3 different swimsuits, 4 different shorts and 3 different pairs of flip-flops.

 Different ways can this girl wear a swimsuit, a pair of short and a pair of flip-flops = 3P₁ x 4P₁ x 3P₁

               = 3 x 4 x 3

               = 36

  Different ways can this girl wear a swimsuit, a pair of short and a pair of flip-flops = 36

Option D is the correct answer.

4 0
2 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
4(x+3)=3(3x-1)
zhuklara [117]
24=26 is the answer... want mhe 2 show you?
5 0
3 years ago
P(-2) = 1P(-1) =3 P(1) = 0 P(2) = 4
ratelena [41]

p(-1) p(-2) p(1) p(6)= 4

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2 years ago
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