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Nadya [2.5K]
3 years ago
10

Josh likes to play an outdoor beanbag toss game in his spare time. On any toss, he has about a 30% chance of getting a bag into

the hole. As a challenge one day, he decides to keep tossing beanbags until he makes it. Starting on line 5 of the random number table, how many times does Josh have to toss the beanbag in order to make it in the hole if we let success = 0–2 and failure = 3–9?
Mathematics
1 answer:
MArishka [77]3 years ago
5 0

Answer:

B

Step-by-step explanation:

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Please help me i need to find the area of shaded region please help me​
Montano1993 [528]

Answer:

Part 1) A=60\ ft^2

Part 2) A=80\ cm^2

Part 3) A=96\ m^2

Part 4) A=144\ cm^2

Part 5) A=13\ m^2

Part 6) A=(49\pi -33)\ in^2

Step-by-step explanation:

Part 1) we know that

The shaded region is equal to the area of the complete rectangle minus the area of the interior rectangle

The area of rectangle is equal to

A=bh

where

b is the base of rectangle

h is the height of rectangle

so

A=(12)(7)-(8)(3)\\A=84-24\\A=60\ ft^2

Part 2) we know that

The shaded region is equal to the area of the complete rectangle minus the area of the interior square

The area of square is equal to

A=b^2

where

b is the length side of the square

so

A=(12)(8)-(4^2)\\A=96-16\\A=80\ cm^2

Part 3) we know that

The area of the shaded region is equal to the area of four rectangles plus the area of one square

so

A=4(4)(5)+(4^2)\\A=80+16\\A=96\ m^2

Part 4) we know that

The shaded region is equal to the area of the complete square minus the area of the interior square

so

A=(15^2)-(9^2)\\A=225-81\\A=144\ cm^2

Part 5) we know that

The area of the shaded region is equal to the area of triangle minus the area of rectangle

The area of triangle is equal to

A=\frac{1}{2}(b)(h)

where

b is the base of triangle

h is the height of triangle

so

A=\frac{1}{2}(6)(7)-(4)(2)\\A=21-8\\A=13\ m^2

Part 6) we know that

The area of the shaded region is equal to the area of the circle minus the area of rectangle

The area of the circle is equal to

A=\pi r^{2}

where

r is the radius of the circle

so

A=\pi (7^2)-(3)(11)\\A=(49\pi -33)\ in^2

6 0
3 years ago
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