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Verdich [7]
3 years ago
9

If 50% of 120% of 30% of a number is 108,what is the number

Mathematics
2 answers:
Diano4ka-milaya [45]3 years ago
8 0

Answer:

600

Step-by-step explanation:

Of means multiply and is means equals

50% * 120% * 30% *N= 108

Change to decimal form

.50 * 1.20 * .30 * N = 108

Multiply

.18 N = 108

Divide each side by .18

.18N /.18 = 108/.18

N =600

miv72 [106K]3 years ago
5 0

Hello from MrBillDoesMath!

Answer:

The number is 600

Discussion:

Percent means "per 100" or divided by 100. The Question is equivalent to

50 % * 120% * 30% * x = 108

where x is the number to be found. This is equivalent to

(50/100)* (120/100) * (30/100) * x = 108   => simplify fractions

(1/2)       * (12/10)  *    ( 3/10) * x      = 108   =>  as 3 * 12 = 36

36/ (2 * 10* 10) * x  = 108                           =>  36 * 3 = 108. Divide by 36

1/(200) *x = 108/36 = 3                             => multiply both sides by 200

x = 3 * 200 = 600

Thank you,

MrB

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MATH HELP!!!!!<br><br> If y = 15 when x = 2, what is the value of x when y = 7?
Ede4ka [16]

<u>Direct Variation:</u>  \frac{y}{x} = k

\frac{15}{2} = k

when y = 7:      \frac{7}{x} = \frac{15}{2}

14 = 15x

\frac{14}{15} = x

<u>Inverse variation:</u>   y*x = k

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3 0
3 years ago
Find the​ (a) mean,​ (b) median,​ (c) mode, and​ (d) midrange for the data and then​ (e) answer the given question. Listed below
mafiozo [28]

Answer:

a) \bar X = 369.62

b) Median=175

c) Mode =450

With a frequency of 4

d) MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5

<u>e)</u>s = 621.76

And we can find the limits without any outliers using two deviations from the mean and we got:

\bar X+2\sigma = 369.62 +2*621.76 = 1361

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case

Step-by-step explanation:

We have the following data set given:

49 70 70 70 75 75 85 95 100 125 150 150 175 184 225 225 275 350 400 450 450 450 450 1500 3000

Part a

The mean can be calculated with this formula:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

Replacing we got:

\bar X = 369.62

Part b

Since the sample size is n =25 we can calculate the median from the dataset ordered on increasing way. And for this case the median would be the value in the 13th position and we got:

Median=175

Part c

The mode is the most repeated value in the sample and for this case is:

Mode =450

With a frequency of 4

Part d

The midrange for this case is defined as:

MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5

Part e

For this case we can calculate the deviation given by:

s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

And replacing we got:

s = 621.76

And we can find the limits without any outliers using two deviations from the mean and we got:

\bar X+2\sigma = 369.62 +2*621.76 = 1361

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case

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