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4 years ago

Lee scored 15 goals out of 35 attempts. What fraction of her attempts were successful?

Mathematics

2 answers:

V125BC [204]4 years ago

8 0

15/35 of her attempts were successful.

matrenka [14]4 years ago

8 0

Answer:

3/7 simplified

Step-by-step explanation:

Hope this helps!!! :)

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Imagine you know the true length of a board to be 24 inches. measuring the board 10 times with instrument a, you find the measur

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The answer is simple it's:

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4 years ago

Suppose Cho deposits $200 into a bank account for 3 years. The annual interest rate is 5%. Find the simple interest earned.

Mars2501 [29]

Simple interest is given by the formula:
S.I=PRT/100
where:
P=principle
R=rate
T=time
From the information given:
P=$200, r=5%, t=3 years.
Plugging into the formula we obtain:
A=200×5/100×3
simplifying we obtain:
A=$30

Answer is: A] $30

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4 years ago

Using the triangle below, what is Cos A?

Stells [14]

Answer:

a\b

Step-by-step explanation:

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3 years ago

Read 2 more answers

Kellan’s lemonade recipe uses 4 scoops of lemonade powder for every 2.5 cups of water. Kellan uses 10 scoops of lemonade powder

3241004551 [841]

6.25 because 2.5+2.5= 5 and then half of 2.5 is 1.25 so 6.25

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3 years ago

The perimeter of a triangular field is 120m. Two of the sides are 21m and 40m.Calculate the largest angle of the field.

Mama L [17]

Answer:

the largest angle of the field is 149⁰

Step-by-step explanation:

Given;

perimeter of the triangular filed, P = 120 m

length of two known sides, a and b = 21 m and 40 m respectively

The length of the third side is calculated as follows;

a + b + c = P

21 m + 40 m + c = 120 m

61 m + c = 120 m

c = 120 m - 61 m

c = 59 m

↓ ↓

A → → → → → → → → → → → C

Consider ABC as the triangular field;

Angle A is calculated by applying cosine rule;

$a^2 = b^2 + c^2 - 2bc \ Cos A\\\\Cos \ A = \frac{b^2 + c^2 - a^2}{2bc} \\\\Cos \ A = \frac{40^2 + 59^2 - 21^2}{2 \times 40 \times 59} \\\\Cos \ A = 0.983\\\\A = Cos ^{-1} (0.983)\\\\A = 10.6 \ ^0$

Angle B is calculated as follows;

$Cos \ B = \frac{a^2 + c^2 - b^2}{2ac} \\\\Cos \ B = \frac{21^2 + 59^2 - 40^2}{2 \times 21 \times 59} \\\\Cos \ B = 0.937\\\\B= Cos ^{-1} (0.937)\\\\B = 20.5 \ ^0$

Angle C is calculated as follows;

$Cos \ C = \frac{a^2 + b^2 - c^2}{2ab} \\\\Cos \ C = \frac{21^2 + 40^2 - 59^2}{2 \times 21 \times 40} \\\\Cos \ C = -0.857\\\\C = Cos ^{-1} (-0.857)\\\\C = 149\ ^0$

Therefore, the largest angle of the field is 149⁰.

8 0

3 years ago