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Andrei [34K]
2 years ago
3

Express this number in standard form. 2.456×10^​11

Mathematics
2 answers:
n200080 [17]2 years ago
6 0

Answer:

The standard form of the given number is 2.456\times 10^{11}=245600000000    

Step-by-step explanation:

Given : Number 2.456\times 10^{11}

To find : Express this number in standard form?

Solution :

We have given the number 2.456\times 10^{11} in scientific notation form.

The standard form is a way of writing a large or smaller number.

To convert scientific notation into standard form,

The exponent of the scientific notation is positive  11  so move the decimal point  11  places to the right.

The number is re-written as =2.456\times 100000000000

Remove the decimal also,

We get, =\frac{2456}{1000}\times 100000000000

=245600000000

Therefore, The standard form of the given number is 2.456\times 10^{11}=245600000000

Bad White [126]2 years ago
4 0
2.456 * 10^11.....since u have a positive exponent, u move ur decimal to the right

start at ur decimal, and count 11 spaces to ur right
245,600,000,000
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Dmitry_Shevchenko [17]

Mario's unit rate is given by 900 miles/hour.

Explanation:

We have been given that Mario walked at a rate of 2/3 miles every 10 minutes.

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\text{Speed}= \frac{\text{Distance}}{\text{Time}}

We have,

Distance = \frac{2}{3} \text{ miles}

Time = 10 minutes

On substituting these values in the above formula, we get

\text{Speed} = \frac{2/3}{10}\\\\ \text{Speed} =\frac{2}{10\cdot 3}\\\\ \text{Speed} =[tex]\frac{1}{15} \text{ miles/min}[/tex]

We have to find the unit rate in mile per hour, so we need to multiply it by 60

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2 years ago
Find the coefficient of variation for each of the two sets of data, then compare the variation. Round results to one decimal pla
svp [43]

Here is  the correct computation of the question given.

Find the coefficient of variation for each of the two sets of data, then compare the variation. Round results to one decimal place. Listed below are the systolic blood pressures (in mm Hg) for a sample of men aged 20-29 and for a sample of men aged 60-69.

Men aged 20-29:      117      122     129      118     131      123

Men aged 60-69:      130     153      141      125    164     139

Group of answer choices

a)

Men aged 20-29: 4.8%

Men aged 60-69: 10.6%

There is substantially more variation in blood pressures of the men aged 60-69.

b)

Men aged 20-29: 4.4%

Men aged 60-69: 8.3%

There is substantially more variation in blood pressures of the men aged 60-69.

c)

Men aged 20-29: 4.6%

Men aged 60-69: 10.2 %

There is substantially more variation in blood pressures of the men aged 60-69.

d)

Men aged 20-29: 7.6%

Men aged 60-69: 4.7%

There is more variation in blood pressures of the men aged 20-29.

Answer:

(c)

Men aged 20-29: 4.6%

Men aged 60-69: 10.2 %

There is substantially more variation in blood pressures of the men aged 60-69.

Step-by-step explanation:

From the given question:

The coefficient of variation can be determined by the relation:

coefficient \ of  \ variation = \dfrac{standard \ deviation}{mean}*100

We will need to determine the coefficient of variation both men age 20 - 29 and men age 60 -69

To start with;

The coefficient of men age 20 -29

Let's first find the mean and standard deviation before we can do that ;

SO .

Mean = \dfrac{\sum \limits^{n}_{i-1}x_i}{n}

Mean = \frac{117+122+129+118+131+123}{6}

Mean = \dfrac{740}{6}

Mean = 123.33

Standard deviation  = \sqrt{\dfrac{\sum (x_i- \bar x)^2}{(n-1)} }

Standard deviation =\sqrt{\dfrac{(117-123.33)^2+(122-123.33)^2+...+(123-123.33)^2}{(6-1)} }

Standard deviation  = \sqrt{\dfrac{161.3334}{5}}

Standard deviation = \sqrt{32.2667}

Standard deviation = 5.68

The coefficient \ of  \ variation = \dfrac{standard \ deviation}{mean}*100

coefficient \ of  \ variation = \dfrac{5.68}{123.33}*100

Coefficient of variation = 4.6% for men age 20 -29

For men age 60-69 now;

Mean = \dfrac{\sum \limits^{n}_{i-1}x_i}{n}

Mean = \frac{   130 +    153    +  141  +    125 +   164  +   139}{6}

Mean = \dfrac{852}{6}

Mean = 142

Standard deviation  = \sqrt{\dfrac{\sum (x_i- \bar x)^2}{(n-1)} }

Standard deviation =\sqrt{\dfrac{(130-142)^2+(153-142)^2+...+(139-142)^2}{(6-1)} }

Standard deviation  = \sqrt{\dfrac{1048}{5}}

Standard deviation = \sqrt{209.6}

Standard deviation = 14.48

The coefficient \ of  \ variation = \dfrac{standard \ deviation}{mean}*100

coefficient \ of  \ variation = \dfrac{14.48}{142}*100

Coefficient of variation = 10.2% for men age 60 - 69

Thus; Option C is correct.

Men aged 20-29: 4.6%

Men aged 60-69: 10.2 %

There is substantially more variation in blood pressures of the men aged 60-69.

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