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lara [203]
3 years ago
13

The legs of a right triangle measure 6 centimeters and 8 centimeters. How long is the hypotenuse in centimeters?

Mathematics
2 answers:
klemol [59]3 years ago
8 0

Answer:

10cm ( A )

Step-by-step explanation:

This would use the Pythagorean Theorem. That means it would be -

a^2_b^2= c^2, in this case, you have 2 of your legs already so those two values would be a&b. you would stick those two into the equation and it should look like this:

6^2+8^2=c^2

then 36+64=c

and then 100=c

after that, you would take the square root of c to get

10!

zloy xaker [14]3 years ago
4 0

Answer:

10cm

Step-by-step explanation:

Hypotenuse^{2} = base^{2} + altitude^{2}

The hypotenuse can be taken as 'x'. As given in the question, the length of base is 8cm and the altitude is 6cm. So substituting these in the equation, we get:-

Hypotenuse^{2}  = 8^{2} + 6^{2}

Hypotenuse^{2} = 64 + 36

Hypotenuse^{2}  = 100

Hypotenuse = \sqrt{100}

Hypotenuse equals 10cm!

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3 years ago
A basketball coach keeps track of the points scored per game for all of the players . She gives a trophy to the player with the
Verdich [7]

Your Question was incomplete so I will solve a similar question that will help you to tackle this kind of questions.

Q)Two university female basketball players are being considered for an award to be given to the most consistent player with the highest scoring average for the season. The number of points scored per game for all games played during the season is given by the following table. ( Table Attached)

a) Calculate the population mean and population standard deviation of the number of points scored per game for each player.

(b) Using the results of part (a) which player should receive the award.

Answer:

<h3><u>Player B should receive the award.</u></h3><h3>Step-by-step explanation:</h3>

Points of player A:- 5,32,45,25,8,10,35,12,37,21

<u>First we will calculate the mean for player A</u>

∑x =  5 + 32 + 45 + 25 + 8 + 10 + 35 + 12 + 37 + 21

∑x = 230

N = 10

<u>Mean = µ= ∑x/N = 230/10 = 23</u>

<u>Now we will calculate the standard deviation for Player A</u>

x - µ = -18 , 9, 22, 2, -15, -13, 12, 11, 14, -2

(x - µ)^2 = 324, 81, 484, 4, 225, 169, 144, 121, 196, 4

∑(x - µ)^2 = 1752

σ = sqrt( ∑((x - µ)^2/N) )

σ = sqrt(1752/10)

<u>σ = 13.236</u>

<u>Calculating mean for player B:</u>

Points of player B = 18, 20, 22, 15, 35, 24, 29, 19, 25, 23

∑x =  18 + 20 + 22 + 15 + 35 + 24 + 29 + 19 + 25 + 23

N =10

µ = 230/10

<u>µ  = 23</u>

<u>Calculating standard deviation for player B:</u>

(x - µ ) = -5, -3, -1, -8, 12, 1, 6, -4, 2, 0

(x -µ )^2 = 25, 9, 1, 64, 144, 1, 36, 16, 4, 0

∑(x - µ)^2 = 300

σ = sqrt(300/10)

<u>σ = 5.48</u>

Since both Player A and Player B have same mean we will calculated the coefficient of variation for both the players.

<u>For player A:</u>

σa/µa *100 = 13.24/23 *100 = 57.56

<u>For Player B:</u>

σb/µb *100 = 5.48/23 * 100 = 23.82

<u>Since the coefficient of variation of Player B is less than Player A hence Player B should receive the award.</u>

3 0
3 years ago
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Can someone help me find the area of these? ​
andreev551 [17]

Answer:

1.283.5

2.34.56

3.92.96

4.39.9

5.34

6.18.7

Step-by-step explanation:

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The compound interest on a sum of money in
Yakvenalex [24]

Answer:

The difference between the principal and the compound interest in three years is Rs 17,994

Step-by-step explanation:

The compound interest is given according to the following formula;

C.I. = P \cdot \left ( 1 + \dfrac{r}{n} \right ) ^{n\cdot t} - P

The given amount of the compound after 2 years = Rs 5,460

The given amount of the compound after 4 years = Rs 12,066.60

Therefore, we have;

5,460 = P \cdot \left ( 1 + \dfrac{r}{100} \right ) ^{2} - P...(1)

12,066.60 = P \cdot \left ( 1 + \dfrac{r}{100} \right ) ^{4} - P ...(2)

Dividing equation (2) by (1), we have;

\dfrac{12,066.60}{5,460} = \dfrac{P \cdot \left ( \left ( 1 + \dfrac{r}{100} \right ) ^{4} - 1\right )}{P \cdot \left (\left ( 1 + \dfrac{r}{100} \right ) ^{2} -1 \right ) } =\dfrac{\left ( 1 + \dfrac{r}{100} \right ) ^{4} - 1}{\left ( 1 + \dfrac{r}{100} \right ) ^{2} -1  }

Let \  \left ( 1 + \dfrac{r}{100} \right ) ^{2} = x, we \ get;

\dfrac{12,066.60}{5,460} =\dfrac{\left ( 1 + \dfrac{r}{100} \right ) ^{4} - 1}{\left ( 1 + \dfrac{r}{100} \right ) ^{2} -1  } = \dfrac{x^2 - 1}{x - 1}

∴ 12,066.60 × (x - 1) = 5,460 × (x² - 1) = 5,460 × (x - 1) ×(x + 1)

∴ 12,066.60 × (x - 1)/(x - 1) = 5,460 × (x + 1)

12,066.60/5,460 = x + 1

x =  12,066.60/5,460 - 1 = 1.21 = 121/100

x = 121/100

\left ( 1 + \dfrac{r}{100} \right ) ^{2} = x = \dfrac{121}{100}

1 + \dfrac{r}{100}   =\sqrt{ \dfrac{121}{100}} = \dfrac{11}{10}

We get

\dfrac{12,066.60}{5,460} =\dfrac{221}{100}

\therefore \dfrac{12,066.60}{5,460} =\dfrac{221}{100} = \left ( 1 + \dfrac{r}{100} \right ) ^{2}

1 + \dfrac{r}{100} = \sqrt{ \dfrac{221}{100} } = \dfrac{\sqrt{221} }{10}

\dfrac{r}{100} = \dfrac{\sqrt{221} }{10} - 1

\dfrac{r}{100}   = \dfrac{11}{10} - 1 = \dfrac{1}{10} = 0.1

r = 100 × 0.1 = 10%

r = 10%

Therefore, we have;

5,460 = P \cdot \left ( 1 + \dfrac{r}{100} \right ) ^{2} - P = P \times \left ( 1 + 0.1\right ) ^{2} - P

5,460  = P \times \left ( 1 + 0.1\right ) ^{2} - P = P \times \left (\left ( 1 + 0.1\right ) ^{2} - 1\right) = P \times \dfrac{21}{100}

P = \dfrac{100}{21} \times 5,460 = 26,000

The principal = Rs. 26,000

The compound interest in 3 years is therefore;

CI_3 = 26,000 \times \left ( 1 + \dfrac{10}{100} \right ) ^{3} - 26,000= 8606

The difference, 'd', between the principal and the compound interest in three years, is given as follows;

d = P - CI₃

d = 26,600 - 8606 = 17994

The difference between the principal and the compound interest in three years, d = Rs 17,994.

5 0
3 years ago
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