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nexus9112 [7]
3 years ago
15

which list shows the numbers in order from least to greatest? a. 5.4 × 104, 5.4 × 103, 4.5 × 104 b. 5.4 × 103, 5.4 × 104, 4.5 ×

104 c. 5.4 × 103, 4.5 × 104, 5.4 × 104 d. 4.5 × 104, 5.4 × 103, 5.4 × 104
Mathematics
1 answer:
Umnica [9.8K]3 years ago
4 0

Answer with explanation:

Option A

 1.\rightarrow 5.4 \times 10^4=5.4 \times 10000\\\\=54000

2.\rightarrow 5.4 \times 10^3=5.4 \times 1000\\\\=5400

3.\rightarrow 4.5 \times 10^4=4.5 \times 10000\\\\=45000

Option B

 1.\rightarrow 5.4 \times 10^3=5.4 \times 1000\\\\=5400

2.\rightarrow 5.4 \times 10^4=5.4 \times 10000\\\\=54000

3.\rightarrow 4.5 \times 10^4=4.5 \times 10000\\\\=45000

Option C

 1.\rightarrow 5.4 \times 10^3=5.4 \times 1000\\\\=5400

2.\rightarrow 4.5 \times 10^4=4.5 \times 10000\\\\=45000

3.\rightarrow 5.4 \times 10^4=5.4 \times 10000\\\\=54000

Option D

 1.\rightarrow 4.5 \times 10^4=4.5 \times 10000\\\\=45000

2.\rightarrow 5.4 \times 10^3=5.4 \times 1000\\\\=5400

3.\rightarrow 5.4 \times 10^4=5.4 \times 10000\\\\=54000

⇒Option C ,shows number and exponents from least to greatest.

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After 10 months the simple interest of a 6,000 investment was $769. What is the interest rate
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Answer:

Interest rate ( <em>r </em>) = 15.38% or 0.1538

Step-by-step explanation:

Given the principal (P) = $6,000, and the simple interest of $769 earned after 10 months:

We can determine the interest rate (<em>r</em> ) by creating its formula algebraically, then use the equation to solve for its value.  

Using the simple interest formula,  

<em>I </em>=<em> P </em>× <em>r </em>×<em> t </em>

where:

I = interest amount = $769

P = present value of the principal = $6,000

r = interest rate

t = time  = 10/12 (10 months out of 12 months) = 0.83333

I = P × r × t

Divide both sides by P × t:

I / ( P × t ) = P × t × r / ( P × t )

I / ( P × t ) = r

Therefore, the formula for the interest rate is:<em> </em><em>r</em> = <em>I</em> / ( <em>P</em> × <em>t</em> )

Substitute the given values into the formula to solve for r:

r = I / ( P × t )

r = $769 / ($6,000 × 0.83333)

r = $769/ 4,999.99999

r = 0.1538 or 15.38%

I'd greatly appreciate it if you could please mark my answers as the Brainliest, if you find this solution helpful :)

8 0
2 years ago
After eating at the restaurant, Tom, Mary, and
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Answer:

$126

Step-by-step explanation:

since they divided it evenly therefore everyone paid $42 you just times that by the amount of people. there is tom, mary and sara therefore 3. 42*3=126

8 0
2 years ago
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Whats the answer<br> i dont know this
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Answer:

9.2

Step-by-step explanation:

a^2 + b^2 = c^2 (pythogoras theorem)

7^2 + 6^2 = 85

Square root of 85 = 9.2 ( nearest tenth)

I think this is right if not srry :(

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2 years ago
Suppose Yakov is the only seller in the market for bottled water and Rajiv is the only seller
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Answer:

\huge \underline \mathtt \colorbox{cyan}{didn't get 'chu}

Step-by-step explanation:

Incomplete question

7 0
3 years ago
The director of training for an equipment manufacturing company is interested in determining whether different training methods
Free_Kalibri [48]

Answer:

The null hypothesis is that there is no difference between the mean time it takes an online trained employee or a team-based trained employee to assemble the given part

Step-by-step explanation:

The information the director of the equipment manufacturing company is interested in determining is weather the productivity of assembly line employees is affected by the method used in their training

The total number newly hired employees in the sample = 42

The number of newly hired employees that receive training online = 21

The number of newly hired employees that receive training in a team = 21

The given data of the result of the time it takes an employee to assemble a part is presented as follows;

On-Line 19.4, 16.7, 20.7, 19.3, 21.8, 16.8, 14.1, 17.7, 16.1, 19.8, 16.8, 19.3, 14.7, 16.0, 16.5, 17.7, 16.2, 17.4, 16.4, 16.8, 18.5

Team; 22.4, 13.8, 18.7, 18.0, 19.3, 20.8, 15.6, 17.1, 18.0, 28.2, 21.7, 20.8, 30.7, 24.7, 23.7, 17.4, 23.2, 20.1, 12.3, 15.2, 16.0

The mean time for the of the on-line trained employee, \overline x_1 = 17.55714

The standard deviation of the time for the of the on-line trained employee, s₁ = 1.93328

The mean time for the of the team based trained employee, \overline x_2 = 19.89048

The standard deviation of the time for the of the team based trained employee, s₂ = 4.5667

The null hypothesis, H₀:  \overline x_1 = \overline x_2

The alternative hypothesis, Hₐ:  \overline x_1 ≠ \overline x_2

t=\dfrac{(\bar{x}_{1}-\bar{x}_{2})}{\sqrt{\dfrac{s_{1}^{2} }{n_{1}}-\dfrac{s _{2}^{2}}{n_{2}}}}

Therefore, we have;

t=\dfrac{(19.89048 - 17.55714)}{\sqrt{\dfrac{4.57667^{2}}{21} - \dfrac{1.93328^{2} }{21}}} \approx 2.5776

The degrees of freedom, df = n₁ - 1 = 21 - 1 = 20

At 95% confidence level, we have α = 1 - 0.95 = 0.05, and t = 2.086

Therefore, given that the test statistic is larger than the critical 't' value, we reject the null hypothesis. There is sufficient statistical evidence to show that there is a difference between the mean time of assembly value for the on-line trained and team-based employee

7 0
2 years ago
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