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Tamiku [17]
3 years ago
8

Does claire have a net gain or a net loss what amount

Mathematics
1 answer:
ololo11 [35]3 years ago
7 0
<span>(Net) Income - Expenses = Net Gain/Loss Net gain is: when the remainder of money calculated is a positive number. Net loss is: when the remainder of money calculated is a negative number. Expenses are: money that is spent on a variety of things.</span>
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Use the drawing tools to form the correct answer on the number line.
Ainat [17]

Answer:

In the picture

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
7 less than the product of a number and 8, increased by a number multiplied by 2
nataly862011 [7]

The expression for 7 less than the product of a number and 8, increased by a number multiplied by 2 is 8n-7 +2n

Given :

7 less than the product of a number and 8, increased by a number multiplied by 2

We need to write the given statement in an expression

Lets 'n' be the unknown number

the product of a number and 8

Product means multiplication

the product of a number and 8 mean n times 8 that is 8n

7 less than the product of a number and 8, subtract 7 from 8n

Expression becomes 8n-7

This is increased by a number multiplied by 2

number multiplied by 2 is 2n

So the final expression for the given statement is 8n-7 +2n

Learn more : brainly.ph/question/1023996

3 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
Which monomial function has a minimum value? y=5x^10 y=-4x^11 y=-x^12 y=6x^7
strojnjashka [21]
<span>y=-x^12  has the minimum value.</span>
8 0
3 years ago
Read 2 more answers
Last year, Keiko had $20,000 to invest. She invested some of it in an account that paid %7 simple interest per year, and she inv
erma4kov [3.2K]

Answer:

P_2 = \$6,000\\P_1 = \$14,000

Step-by-step explanation:

The formula of simple interest is:

I = P_0rt

Where I is the interest earned after t years

r is the interest rate

P_0 is the initial amount

We know that the investment was $20,000 in two accounts

_______________________________________________

<u><em>For the first account</em></u> r = 0.07 per year.

Then the formula is:

I_1 = P_1r_1t

Where

P_1 is the initial amount in account 1 at a rate r_1 during t = 1 year

I_1 = P_1(0.07)(1)\\\\I_1 = 0.07P_1

<u><em>For the second account </em></u>r = 0.05 per year.

Then the formula is:

I_2 = P_2r_2t

Where

P_2 is the initial amount in account 2 at a rate r_2 during t = 1 year

Then

I_2 = P_2(0.05)(1)\\\\I_2 = 0.05P_2

We know that the final profit was I $1,280.

So

I = I_1 + I_2=1,280

Substituting the values I_1, I_2 and I we have:

1,280 = 0.07P_1 + 0.05P_2

As the total amount that was invested was $20,000 then

P_0 = P_1 + P_2 = 20,000

Then we multiply the second equation by -0.07 and add it to the first equation:

0.07P_1 + 0.05P_2 = 1.280\\.\ \ \ \ \ \ \ \ +\\-0.07P_1 -0.07P_2 = -1400\\-------------

-0.02P_2 = -120\\\\P_2 = 6,000

Then P_1 = 14,000

4 0
3 years ago
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