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rewona [7]
3 years ago
10

the cooper family decided to hike to hillside lake,approximately 8⅝ miles away. after an hour tje lake was still 5⅓ miles away.

hpw far did the group so far?​
Mathematics
2 answers:
alisha [4.7K]3 years ago
7 0

Answer: 3 7/24

Step-by-step explanation:

Given: The Cooper family decided to hike to Hillside Lake.

The approximate distance to travel for Hillside Lake = 8 5/8

Since , After an hour the lake was still away. 5 1/3

Now, the distance hiked by group :

8 5/8 - 5 1/3 = 3 7/24

ollegr [7]3 years ago
7 0

Answer= 3\7-24 miles

Step-by-step explanation:

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-3x+4y=-10 in y=mx+b
belka [17]
Y = 3/4x -5/2

Hope this helps!
8 0
3 years ago
Darren has a wooden board that is Three-fifths of a meter long. He cuts the board into 3 equal parts. How long is each section o
Alik [6]

Answer:

Each section of the board is 1/3 or 0.334(0.333333333333) of a meter

Step-by-step explanation:

3/5 divided by 3 is the equation we have to do.

3/5 divided by 3 is also equal to 3/5 times the reciprocal of 3 or 1/3

3/5 * 1/3

3*1 = 3

5*3 = 15

3/15 which can simplify to 1/3. (divide numerator and denominator by 3.)

Each section of the board is 1/3 or 0.334(0.333333333333) of a meter

8 0
3 years ago
What is the best way to determine the graph of a situation?
elixir [45]

Answer:

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by drawing

3 0
3 years ago
The Rockford's have a square play area that is built around a center sandbox that is in the shape of a circle. The sides of the
stealth61 [152]
<span>The area of the square is d^2. The area of the circle sandbox is πr^2=πd^2/4. So the area of play area only is equal to square area minus circle area which is d^2-πd^2/4.</span>
5 0
4 years ago
(Anderson, 1.14) Assume that P(A) = 0.4 and P(B) = 0.7. Making no further assumptions on A and B, show that P(A ∩ B) satisfies 0
Goshia [24]

Answer with Step-by-step explanation:

We are given that

P(A)=0.4 and P(B)=0.7

We know that

P(A)+P(B)+P(A\cap B)=P(A\cup B)

We know that

Maximum value of P(A\cup B)=1 and minimum value of P(A\cup B)=0

0\leq P(A\cup B )\leq 1

0\leq P(A)+P(B)-P(A\cap B)\leq 1

0\leq 0.4+0.7-P(A\cap B)\leq 1

0\leq 1.1-P(A\cap B)\leq 1

0\leq 1.1-P(A\cap B)

P(A\cap B)\leq 1.1

It is not possible that P(A\cap B) is equal to 1.1

1.1-P(A\cap B)\leq 1

-P(A\cap B)\leq 1-1.1=-0.1

Multiply by (-1) on both sides

P(A\cap B)\geq 0.1

Again, P(A\cup B)\geq P(B)

0.4+0.7-P(A\cap B)\geq 0.7

1.1-P(A\cap B)\geq 0.7

-P(A\cap B)\geq -1.1+0.7=-0.4

Multiply by (-1) on both sides

P(A\cap B)\leq 0.4

Hence, 0.1\leq P(A\cap B)\leq 0.4

3 0
3 years ago
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