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RoseWind [281]
3 years ago
14

Arrange the following in decreasing order. 3.8 m^2 380 cm? 0.038 km2 3,800 mm Which of the following shows the answers in decrea

sing order? O A. 3,800 mm², 380 cm², 0.038 km², 3.8 m² B. 380 cm², 3.8 m², 0.038 km², 3,800 mm? OC. 0.038 km², 3.8 m², 380 cm², 3,800 mm? OD. 3,800 mm², 380 cm², 3.8 m², 0.038 km? Click to select your answer. Save for Later O Type here to search t e
Mathematics
1 answer:
Snezhnost [94]3 years ago
6 0

Answer:

C. 0.038 km² > 3.8 m² > 380 cm² > 3800 mm²  

Step-by-step explanation:

Convert every measurement to the same unit, say, square metres.

(a) 3.8 m²

No conversion necessary

(b) 380 cm²

\text{Area} = \text{380 cm}^{2} \times \left (\dfrac{\text{1 m}}{\text{100 cm}} \right )^{2} = \textbf{0.038 m}^{2}

(c) 0.038 km²

\text{Area} = \text{0.038 km}^{2} \times \left (\dfrac{\text{1000 m}}{\text{1 km}} \right )^{2} = \textbf{38 000 m}^{2}

(d) 3800 mm²

\text{Area} = \text{3800 mm}^{2} \times \left (\dfrac{\text{1 m}}{\text{1000 mm}} \right )^{2} = \textbf{0.0038 m}^{2}

In decreasing order, 0.038 km² > 3.8 m² > 380 cm² > 3800 mm²

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3 years ago
NEED HELP ASAP! 40 POINTS!
vladimir1956 [14]

Same height when both equations are equal

-16x^2 +  74x + 9 = -16x^2 + 82x

82x - 74x = 9

8x = 9

x = 9/8

x = 1.125

x = 1.13 ---->rounded to nearest hundredths of seconds

answer

1.13 seconds


3 0
3 years ago
Read 2 more answers
A Ferris Wheel 22.0m in diameter rotates once every 12.5s. What is the ratio of a persons apperenet weight to her real weight (a
AnnZ [28]
  <span>Acceleration of a passenger is centripetal acceleration, since the Ferris wheel is assumed at uniform speed: 
a = omega^2*r 

omega and r in terms of given data: 
omega = 2*Pi/T 
r = d/2 

Thus: 
a = 2*Pi^2*d/T^2 

What forces cause this acceleration for the passenger, at either top or bottom? 

At top (acceleration is downward): 
Weight (m*g): downward 
Normal force (Ntop): upward 

Thus Newton's 2nd law reads: 
m*g - Ntop = m*a 

At top (acceleration is upward): 
Weight (m*g): downward 
Normal force (Nbottom): upward 

Thus Newton's 2nd law reads: 
Nbottom - m*g = m*a 

Solve for normal forces in both cases. Normal force is apparent weight, the weight that the passenger thinks is her weight when measuring by any method in the gondola reference frame: 
Ntop = m*(g - a) 
Nbottom = m*(g + a) 


Substitute a: 
Ntop = m*(g - 2*Pi^2*d/T^2) 
Nbottom = m*(g + 2*Pi^2*d/T^2) 

We are interested in the ratio of weight (gondola reference frame weight to weight when on the ground): 
Ntop/(m*g) = m*(g - 2*Pi^2*d/T^2)/(m*g) 
Nbottom/(m*g) = m*(g + 2*Pi^2*d/T^2)/(m*g) 

Simplify: 
Ntop/(m*g) = 1 - 2*Pi^2*d/(g*T^2) 
Nbottom/(m*g) = 1 + 2*Pi^2*d/(g*T^2) 

Data: 
d:=22 m; T:=12.5 sec; g:=9.8 N/kg; 

Results: 
Ntop/(m*g) = 71.64%...she feels "light" 
Nbottom/(m*g) = 128.4%...she feels "heavy"</span>
7 0
3 years ago
Read 2 more answers
A survey was conducted to determine the average age at which college seniors hope to retire in a simple random sample of 101 sen
tatyana61 [14]

Answer:

96% confidence interval for desired retirement age of all college students is [54.30 , 55.70].

Step-by-step explanation:

We are given that a survey was conducted to determine the average age at which college seniors hope to retire in a simple random sample of 101 seniors, 55 was the  average desired retirement age, with a standard deviation of 3.4 years.

Firstly, the Pivotal quantity for 96% confidence interval for the population mean is given by;

                         P.Q. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample average desired retirement age = 55 years

            \sigma = sample standard deviation = 3.4 years

            n = sample of seniors = 101

            \mu = true mean retirement age of all college students

<em>Here for constructing 96% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.</em>

<u>So, 96% confidence interval for the population mean, </u>\mu<u> is ;</u>

P(-2.114 < t_1_0_0 < 2.114) = 0.96  {As the critical value of t at 100 degree

                                               of freedom are -2.114 & 2.114 with P = 2%}  

P(-2.114 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 2.114) = 0.96

P( -2.114 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 2.114 \times {\frac{s}{\sqrt{n} } } ) = 0.96

P( \bar X-2.114 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+2.114 \times {\frac{s}{\sqrt{n} } } ) = 0.96

<u>96% confidence interval for</u> \mu = [ \bar X-2.114 \times {\frac{s}{\sqrt{n} } } , \bar X+2.114 \times {\frac{s}{\sqrt{n} } } ]

                                           = [ 55-2.114 \times {\frac{3.4}{\sqrt{101} } } , 55+2.114 \times {\frac{3.4}{\sqrt{101} } } ]

                                           = [54.30 , 55.70]

Therefore, 96% confidence interval for desired retirement age of all college students is [54.30 , 55.70].

7 0
3 years ago
What is 0.4 divided by 5.12 in long division
Svetradugi [14.3K]
Start by mult. both 0.4 and 5.12 by 100.  What is 40 divided by 512?

        ___.078_______________
512 /  40.000
           3584
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             4096
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