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xxMikexx [17]
3 years ago
15

I hate math can someone please help me

Mathematics
2 answers:
balu736 [363]3 years ago
5 0

To change the negative to a positive, just put the same number as into a fraction under 1 but without the negative sign:


11^-4 would become 1 / 11^4

lina2011 [118]3 years ago
3 0

Don't hate math! It can be scary, but if you crush big problems in smaller ones, everything gets simple :)

Take exponents, for example: I suppose you know how to deal with positive exponents: you multiply the base times itself as many times as the exponent commands: for example,

4^3 = 4 \times 4 \times 4 = 64

Because the base is 4 and the exponent is 3, so you multiply 4 by itself 3 times.

Now, negative exponents work exactly the same, except you have to perform all of these operations at the denominator. Here's the same example as before, but with negative exponent:

4^{-3} = \dfrac{1}{4^3} = \dfrac{1}{4\times 4 \times 4} = \dfrac{1}{64}

See? Exactly the same as before, except it happened at the denominator.

So, in your example, the negative exponent simply means that you have to perform the exponentiation at the denominator, so

11^{-4} = \dfrac{1}{11^4}

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8j = 4/5j

3/5i + 4/5j

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2 years ago
What is <br> 5(2x+8)-5x=5(5x+6?
yanalaym [24]

5(2x+8)-5x=5(5x+6)

Distribute

10x+40-5x=25x+30

combine like terms

5x+40=25x+30

-5x       -5x

subtract 5x from both sides

40=20x+30

-30      -30

subtract 30 from both sides

10=20x

÷20   ÷20

divide both sides by 20

1/2=x is your final answer


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3 years ago
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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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