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Angelina_Jolie [31]
3 years ago
7

In 1996, only 193 people owned cellphones in metropolis. Each year, cell phone use has continuously grown by 58%. How many cellp

hone users were there in 2008? (Show your work)
Mathematics
2 answers:
zubka84 [21]3 years ago
6 0
A = 193e^(0.58*12)
A = 203351 people
lutik1710 [3]3 years ago
4 0
For this case we have an equation of the form:
 y = A * (b) ^ x

 Where,
 A: initial amount
 b: growth rate
 x: time in years
 Substituting values we have:
 y = 193 * (1.58) ^ x

 For the year 2008 we have:
 x = 2008 - 1996

 x = 12

 y = 193 * (1.58) ^ {12}

 y = 46713.49403
 Round to the whole whole integer:
 y = 46714

 Answer:
 there were 46714 cellphone users in 2008
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jarptica [38.1K]

Answer:

870.497 ÷ 0.823=1.057 so to the nearest whole number it would be 1

Step-by-step explanation:

cause 0-4 numbers are rounded down

5 0
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Write 508/1000 as a decimal
natka813 [3]

Answer:

0.508

Hope this helps!

4 0
3 years ago
<img src="https://tex.z-dn.net/?f=1%20%5Cfrac%7B3%7D%7B4%7D%20" id="TexFormula1" title="1 \frac{3}{4} " alt="1 \frac{3}{4} " ali
LenaWriter [7]

Answer:

3  \frac{1}{16}

Step-by-step explanation:

first lets change the question into proper fraction

1  \frac{3}{4}  =  \frac{7}{4}

now lets find the square of the number

{ \frac{7}{4} }^{2}  =  \frac{49}{16}

The question says the answer must be a mixed number so lets change it

\frac{49}{16}  = 3  \frac{1}{16}

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3 0
3 years ago
Often, frequency distributions are reported using unequal class widths because the frequencies of some groups would otherwise be
RUDIKE [14]

Answer:

The sample standard deviation is 13.22°F

Step-by-step explanation:

Given - Often, frequency distributions are reported using unequal

            class widths because the frequencies of some groups would

            otherwise be small or very large. Consider the following​ data,

            which represent the daytime household temperature the

            thermostat is set to when someone is home for a random sample

            of households. Determine the class​ midpoint, if​ necessary, for

             each class and approximate the mean and standard deviation

             temperature.

Temp                    Frequency                        Class Midpoint

61-64                         34                                         63

65-67                         68                                         66.5

68-69                         196                                        69

70                              191                                         70.5

71-72                         122                                         72

73-76                          81                                           75

77-80                          52                                           79

To find - The sample standard deviation is _____degrees°F.

Proof -

Temp         Frequency(f)            Midpoint(m)           m×f        ( m - 70.73 )²×f

61-64                34                            63                   2142          2031.6

65-67                68                            66.5                4522           1216.7

68-69               196                            69                  13524          586.6

70                     191                           70.5               13465.5        10.1

71-72                122                           72                   8784             196.8

73-76                  81                            75                   6075             2224.4

77-80                  52                            79                    4108            3556.4

                   ∑f = 744                                    ∑m×f = 52620.5     ∑ = 9822.6

So, Mean = \frac{52620.5}{744} = 70.73

Sample standard deviation = \frac{9822.6}{744 - 1} =  \frac{9822.6}{743} = 13.22

∴ we get

The sample standard deviation is 13.22°F

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3 years ago
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aleksley [76]
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