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zalisa [80]
3 years ago
8

What can be concluded about the spread of the histogram? A histogram titled Maria's Monthly Jogging has miles run on the x-axis

and number of runs on the y-axis. 1 to 1.99 miles is 2 runs; 2 to 2.99 miles is 5 runs; 3 to 3.99 miles is 5 runs; 4 to 4.99 miles is 3 runs; 5 to 5.99 miles is 4 runs; 6 to 6.99 is 7 runs; 7 to 7.99 miles is 4 runs. The left side of the histogram is the mirror image of the right side. The histogram is not symmetrical. The histogram is evenly distributed. The left side of the histogram has a cluster.
Physics
2 answers:
Aleonysh [2.5K]3 years ago
8 0

Answer:

I think it is A

Explanation:

l

Ne4ueva [31]3 years ago
5 0

Answer:

the answer is B the historgram is not symetrical

Explanation:

just took the test and got it right!

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A battery of voltage V delivers power P to a resistor of resistance R connected to it. By what factor will the power delivered t
Anettt [7]

Answer:

Explanation:

Power P = V² / R

a ) The resistance is changed to 2.90R

Power will become 1 / 2.9 times .

b )The voltage of the battery is now 2.90V, but the resistance is R

P = (2.9V)² / R

= 8.41 x V² / R

So power becomes 8.41 times

c )The resistance is 2.90R and voltage is 2.90V

Power P = (2.9V)² / 2.9 R

= 2.9 V²/R

So power becomes 2.9 times

d ) The resistance is 2.90R and the voltage is V/2.90

Power P = ( V/2.90)² x 1 / 2.90R

1 / ( 2.9 )³ x V² / R

= 1 / 24.389 x V² / R

So power becomes  1 / 24.389 times .

4 0
3 years ago
A marching band consists of rows of musicians walking in straight, even lines. When a marching band performs in an event, such a
Svetllana [295]

Answer:

a. 4v

Explanation:

Alf moves with speed v

Alf travel during the same amount of time that is Δt = (1/4)s

v = (1/4)s /  Δt = s / 4 Δt

s / Δt  = 4 v

Beth travels a distance s during time Δt,

speed of Beth = s / Δt = 4 v .

7 0
2 years ago
Gravitational potential energy is a form of potential energy
Gre4nikov [31]
True because it has "falling" ability
5 0
3 years ago
A tree falls in a forest. How many years must pass before the 14C activity in 1.03 g of the tree's carbon drops to 1.02 decay pe
Illusion [34]

Answer:

t = 5.59x10⁴ y

Explanation:

To calculate the time for the ¹⁴C drops to 1.02 decays/h, we need to use the next equation:

A_{t} = A_{0}\cdot e^{- \lambda t}    (1)

<em>where A_{t}: is the number of decays with time, A₀: is the initial activity, λ: is the decay constant and t: is the time.</em>

To find A₀ we can use the following equation:  

A_{0} = N_{0} \lambda   (2)

<em>where N₀: is the initial number of particles of ¹⁴C in the 1.03g of the trees carbon </em>

From equation (2), the N₀ of the ¹⁴C in the trees carbon can be calculated as follows:        

N_{0} = \frac{m_{T} \cdot N_{A} \cdot abundance}{m_{^{12}C}}

<em>where m_{T}: is the tree's carbon mass, N_{A}: is the Avogadro's number and m_{^{12}C}: is the ¹²C mass.  </em>

N_{0} = \frac{1.03g \cdot 6.022\cdot 10^{23} \cdot 1.3\cdot 10^{-12}}{12} = 6.72 \cdot 10^{10} atoms ^{14}C    

Similarly, from equation (2) λ is:

\lambda = \frac{Ln(2)}{t_{1/2}}

<em>where t 1/2: is the half-life of ¹⁴C= 5700 years </em>

\lambda = \frac{Ln(2)}{5700y} = 1.22 \cdot 10^{-4} y^{-1}

So, the initial activity A₀ is:  

A_{0} = 6.72 \cdot 10^{10} \cdot 1.22 \cdot 10^{-4} = 8.20 \cdot 10^{6} decays/y    

Finally, we can calculate the time from equation (1):

t = - \frac{Ln(A_{t}/A_{0})}{\lambda} = - \frac {Ln(\frac{1.02decays \cdot 24h \cdot 365d}{1h\cdot 1d \cdot 1y \cdot 8.20 \cdot 10^{6} decays/y})}{1.22 \cdot 10^{-4} y^{-1}} = 5.59 \cdot 10^{4} y              

I hope it helps you!

4 0
3 years ago
State the Newton 2nd law of motion and also prove that F= ma<br>​
saw5 [17]

Explanation:

F = ma is the formula of Newton's Second Law of Motion. Newton's Second Law of Motion is defined as Force is equal to the rate of change of momentum. For a constant mass, force equals mass times acceleration.

...

3 0
3 years ago
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