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vredina [299]
3 years ago
15

5000 people went to vote. Candidate Smith claimed 52% of the votes. Candidate Adams claimed 2681 votes. What percentage does can

didate Adams claim.
Mathematics
1 answer:
boyakko [2]3 years ago
3 0

The percentage of votes claimed by Adam is 53.62 %

<em><u>Solution:</u></em>

Given that, 5000 people went to vote

Candidate Smith claimed 52% of the votes. Candidate Adams claimed 2681 votes

To find: Percentage claimed by Adam

From given,

Total number of votes = 5000

Votes claimed by Adam = 2681

<em><u>The formula used is:</u></em>

Percentage\ claimed\ by\ Adam = \frac{\text{votes claimed by adam}}{\text{Total number of votes}} \times 100

<em><u>Substituting the values we get,</u></em>

Percentage\ claimed\ by\ Adam = \frac{2681}{5000} \times 100\\\\Percentage\ claimed\ by\ Adam = 0.5362 \times 100\\\\Percentage\ claimed\ by\ Adam = 53.62

Thus percentage of votes claimed by Adam is 53.62 %

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x subtracted by negative 8 is equal to 14.

Step-by-step explanation:

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Poppy is 5 years older than Cairo. Their ages in years added up to 25. How old is Poppy and Cairo?​
postnew [5]

Answer:

10 and 15

Step-by-step explanation:

let Cairo's age be x then Poppy's age is x + 5 , then

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3 years ago
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wo point charges lie on the $x$ axis. A charge of +6.24 $\mu C$ is at the origin, and a charge of -9.55 $\mu C$ is at $x$ = 12.0
Andreyy89

Answer:

E_n=34,467,075.42\ N/C

Step-by-step explanation:

<u>Electric Field</u>

The electric field produced by a point charge Q at a distance d is given by

\displaystyle E=K\cdot \frac{Q}{d^2}

Where

K = 9\cdot 10^9\ Nw.m^2/c^2

The net electric field is the vector addition of the individual electric fields produced by each charge. The direction is given by the rule: If the charge is positive, the electric field points outward, if negative, it points inward.

Let's calculate the electric fields of each charge at the given point. The first charge q_1=+6.24\mu C=6.24\cdot 10^{-6}C is at the origin. We'll calculate its electric field at the point x=-3.85 cm. The distance between the charge and the point is d=3.85 cm = 0.0385 m, and the electric field points to the left:

\displaystyle E_1=9\cdot 10^9\cdot \frac{6.24\cdot 10^{-6}}{0.0385^2}

E_1=37,888,345.42\ N/C

Similarly, for q_2=-9.55\mu C=-9.55\cdot 10^{-6}C, the distance to the point is 12 cm + 3.85 cm = 15.85 cm = 0.1585 m. The electric field points to the right:

\displaystyle E_2=9\cdot 10^9\cdot \frac{9.55\cdot 10^{-6}}{0.1585^2}

E_2=3,421,270\ N/C

Since E1 and E2 are opposite, the net field is the subtraction of both

E_n=37,888,345.42\ N/C-3,421,270\ N/C

\boxed{E_n=34,467,075.42\ N/C}

6 0
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Answer:

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Step-by-step explanation:

Hello!

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