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Valentin [98]
3 years ago
11

Harsh is part of a school cricket team, and this year he has scored and average of 46 runs. He has played 5 innings so far and h

is scores in 4 of them are 46,45, 39,55. What was his score on the last one?
Mathematics
2 answers:
Nataly [62]3 years ago
7 0

Answer: 45

Step-by-step explanation:

Harsh is part of a school cricket team, and he has scored an average of 46 runs in 5 innings. His scores in 4 of them are 46,45, 39, and 55. His fifth score is calculated as follows:

Since he made 5 innings and has an average of 46, the total scores will be:

= 46 × 5

= 230

Total of his first 4 scores will be:

= 46 + 45 + 39 + 55

= 185

5th score = 230 - 185

= 45

Mila [183]3 years ago
6 0

Answer: his score on the last one is 45.

Step-by-step explanation:

Let h represent the his score on the last inning.

Average number of runs = sum of runs/total number of innings played.

From the information given,

Average number of runs = 46

total number of innings played = 5

Sum of runs= 46 + 45 + 39 + 55 + h = 185 + h

Therefore,

46 = (185 + h)/5

Cross multiplying by 5, it becomes

46 × 5 = 185 + h

230 = 185 + h

h = 230 - 185

h = 45

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√34

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carlos walks 4 miles every night for exercise it takes him exactly 64 minutes to finish his walk Assuming he walks at a constant
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y = 1/16x

So let's check it...

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9.
A quadrilateral has 4 vertices. If you add two bases, its like adding two vertices. 4+2=6. A figure that has 6 vertices is a hexagon, so the answer is hexagon.

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3 years ago
Read 2 more answers
Quadrilateral PEST has vertices (-1, -5), (8, 2), (11, 13), and (2, 6), respectively. Classify the quadrilateral as a square, rh
Alchen [17]

Answer:

The figure PEST is a rhombus

Step-by-step explanation:

* Lets talk about the difference between all these shapes

- At first to prove the shape is a parallelogram you must have one

 of these conditions

# Each two opposite sides are parallel OR

# Each two opposite sides are equal in length OR

# Its two diagonals bisect each other

- After that to prove the parallelogram is:

* A rectangle you must have one of these conditions

# Two adjacent sides are perpendicular to each other OR

# Its two diagonals are equal in length

* A rhombus you must have one of these conditions

# Two adjacent sides are equal in length OR

# Its two diagonals perpendicular to each other OR

# Its diagonals bisect its vertices angles

* A square you must have two of these conditions

# Its diagonals are equal and perpendicular OR

# Two adjacent sides are equal and perpendicular

* Now lets solve the problem

∵ The vertices of the quadrilateral PEST are

   P (-1 , -5) , E (8 , 2) , S (11 , 13) , T (2 , 6)

- Lets find the slope from each two points using this rule :

 m = (y2 - y1)/(x2 - x1), where m is the slope and (x1 , y1) , (x2 , y2)

 are two points on the line

- Let (x1 , y1) is (-1 , -5) and (x2 , y2) is (8 , 2)

∴ m of PE = (2 - -5)/(8 - -1) = 7/9

- Let (x1 , y1) is (8 , 2) and (x2 , y2) is (11 , 13)

∴ m of ES = (13 - 2)/(11 - 8) = 11/3  

- Let (x1 , y1) is (11 , 13) and (x2 , y2) is (2 , 6)

∴ m of ST = (6 - 13)/(2 - 11) = -7/-9 = 7/9

- Let (x1 , y1) is (2 , 6) and (x2 , y2) is (-1 , -5)

∴ m of TP = (-5 - 6)/(-1 - 2) = -11/-3 = 11/3

∵ m PE = m ST = 7/9

∴ PE // ST ⇒ opposite sides

∵ m ES = m TP = 11/3

∴ ES // TP ⇒ opposite sides

- Each two opposite sides are parallel

∴ PEST is a parallelogram

- Lets check if the parallelogram can be rectangle or rhombus or

 square by one of the condition above

∵ If two line perpendicular , then the product of their slops = -1

- Lets check the slopes of two adjacent sides (PE an ES)

∵ m PE = 7/9

∵ m ES = 11/3

∵ m PE × m ES = 7/9 × 11/3 = 77/27 ≠ -1

∴ PE and ES are not perpendicular

∴ PEST not a rectangle or a square (the sides of the rectangle and

  the square are perpendicular to each other)

- Now lets check the length of two adjacent side by using the rule

 of distance between two points (x1 , y1) and (x2 , y2)

 d = √[(x2 - x1)² + (y2 - y1)²]

- Let (x1 , y1) is (-1 , -5) and (x2 , y2) is (8 , 2)

∴ PE = √[(8 - -1)² + (2 - -5)²] = √[9² + 7²] = √[81 + 49] = √130 units

- Let (x1 , y1) is (8 , 2) and (x2 , y2) is (11 , 13)

∴ ES = √[(11 - 8)² + (13 - 2)²] = √[3² + 11²] = √[9 + 121] = √130 units

∴ PE = ES ⇒ two adjacent sides in parallelogram

∴ The four sides are equal

* The figure PEST is a rhombus

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