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2 years ago

8

Trevor gets 10 heads when flipping a weighted coin 18 times. What is the experimental probability that the next flip will come u

p heads?
Write your answer as a fraction or whole number.

1 answer:

oee [108]2 years ago

3 0

Answer:

5/9

i got it right

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a scaled drawing of a rectangular park is 5 inches wide and 7 inches long. the actual park is 210 yards long. what is the width

VMariaS [17]

Answer:

I think the width would be 150 yd

Step-by-step explanation:

just broke down the 210 yd we had already to inches found out how much it was scaled up and did the same on the other side

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2 years ago

Read 2 more answers

Filipo is building a rectangular sandbox . The length of the sandbox is 1 foot longer than twice the width of the sandbox .the p

Alla [95]

Answer:

Part A : D.) $2w + 2(2w+ 1) = 29$

Part B : Length of the sandbox is 10 feet.

Step-by-step explanation:

Given,

Perimeter = 29 ft

We need to find the equation for the perimeter and also the length of the sandbox.

Solution,

Let the width of the sandbox be 'w'.

Now as per question said;

The length of the sandbox is 1 foot longer than twice the width of the sandbox.

So we can say that;

Length = $2w+1$

Now we know that the perimeter is equal to the sum of twice of length and width.

framing in equation form, we get;

Perimeter = $2(2w+1)+2w$

we have given the perimeter, so on substituting the value, we get;

$29=2(2w+1)+2w$

Hence The equation used to find the width is $29=2(2w+1)+2w$ .

Now we solve for 'w'.

Applying distributive property, we get;

$4w+2+2w=29\\\\6w+2=29$

Subtracting both side by '2' we get

$6w+2-2=29-2\\\\6w=27$

Dividing both side by 6 we get;

$\frac{6w}{6}=\frac{27}{6}\\\\w=4.5\ ft$

Width of the sandbox = 4.5 ft

Length of the sandbox = $2w+1=2\times4.5 +1 = 9+1=10\ ft$

Hence Length of the sandbox is 10 feet.

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2 years ago

Find the value of x to the nearest tenth.

Alex73 [517]

64 - 9 = 55 \sqrt{55} = 7.4161984871

7.4161984871/2 ≈3.7

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2 years ago

a baseball diamond is a square that is 90' on each side. if a player throws the ball from 2nd base to home,how far will bar trav

9966 [12]

Well when you multiply it would most likely be 180

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2 years ago

At a sport clubhouse the coach wants to rope off a rectangular area that is adjacent to the building. He uses the length of the

Furkat [3]

Answer:

w < 8 meters ( w must be greater than 0 because length cannot be 0)

Step-by-step explanation:

One side with building is 23 meters, the other opposite side also will be 23 meters (with rope).

Let width (remaining 2 sides) be "w", he has AT MOST 39 meters of rope, so we can write:

Rope Needed = 23 + 2w < 39

Simplifying:

$23 + 2w < 39\\2w < 39 - 23\\2w < 16\\w < 8$

The range of possible values of w is $w$ meters (of course w has to be greater than 0)

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3 years ago