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3 years ago

Tanner made his aunt a garden. The width is 9 4/5 ft and the length is 5 3/4 ft. What is the area of

Mathematics

1 answer:

ratelena [41]3 years ago

3 0

Answer:

56.35 ft^2, or 56 7/20 ft^2

Step-by-step explanation:

One way of doing this problem quickly is to convert both measurements into improper fractions before multiplying them:

9 4/5 = 49/5 and 5 3/4 = 23/4.

Then the area of the garden is 49/5 times 23/4, or 1127/20 ft^2, or

56.35 ft^2, or 56 7/20 ft^2.

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Solve the system by graphing. Y=2x-9 y=-2x-1

sattari [20]

Answer:

(2,-5)

Step-by-step explanation:

See attachment

One can also solve this by calculation:

y=2x-9

y=-2x-1

Rearrange either equation to find x. I'll use the first:

y=2x-9

2x = y+9

x = (y+9)/2

Now use this value of x in the second equation:

y = -2x-1

y =-2((y+9)/2)-1

y = (-2y-18)/2)-1

y = -y -9 - 1

2y = -10

y = -5

Now use -5 for y in the rearranged equation:

y = -2x-1

-5 = -2x-1

-2x = -4

x = 2

Solution is (2,-5)

But the question wants a graph solution, which is also fun when you use DESMOS.

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2 years ago

Anyone can help me with problem 3?

Fudgin [204]

Answer: 1 and 3

It's because they both divide down lower than 1

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3 years ago

Read 2 more answers

The quotient of two times a number k and -2.6 is 12

a_sh-v [17]

-15.6 Is the answer. You do inverse

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4 years ago

Martin is making squares by arranging 26 sticks. how many squares can he make? Write and solve the number fact you used to find

Alexxx [7]

He could make 6 squares, because each square has 4 sides. And 24 divided by 4 is 6. The number fact that i used is 24 divided by 4

6 0

3 years ago

Isle Royale, an island in Lake Superior, has provided an important study site of wolves and their prey. Of special interest is t

STatiana [176]

Answer:

Part a) Probability that a moose in that age group is killed by a wolf

P (0.5 year) = 0.361

P (1 - 5) = 0.159

P (6 - 10) = 0.267

P (11 - 15) = 0.203

P (16 - 20) = 0.010

Part b)

Expected age of a moose killed by a wold

μ = 5.61 years

Stantard deviation of the ages

σ = 4.97 years

Explanation:

1) Start by arranging the table to interpret the information:

Age of Moose in years Number killed by woolves

Calf (0.5) 107

1 - 5 47

6 - 10 79

11 - 15 60

16 - 20 3

You can now verify the total number of moose deaths identified as wolf kills: 107 + 47 + 79 + 60 + 3 = 296, such as stated in the first part of the question.

2) First question: a) For each age, group, compute the probability that a moos in that age group is killed by a wolf.

i) Formula:

Probability = number of positive outcomes / total number of events.

ii) Probability that a moose in an age group is killed by a wolf = number of moose killed by a wolf in that age group / total number of moose deaths identified as wolf kills.

iii) P (0.5 year) = 107 / 296 = 0.361

iv) P (1 - 5) = 47 / 296 = 0.159

v) P (6 - 10) = 79 / 296 = 0.267

vi) P (11 - 15) = 60 / 296 = 0.203

vii) P (16 - 20) = 3 / 296 = 0.010

3) Second question: b) Consider all ages in a class equal to the class midpoint. Find the expected age of a moose killed by a wolf and the standard deviation of the ages.

i) Class midpoint

0.5 0.5

1 - 5 (1 + 5) / 2 = 3

6 - 10 (6 + 10) / 2 = 8

11 - 15 (11 + 15) / 2 = 13

16 - 20 (16 + 20) = 18

ii) Expected age of a moose killed by a wolf = mean of the distribution = μ

μ = Sum of the products of each probability times its age (mid point)

μ = 0.5 ( 0.361) + 3 ( 0.159) + 8 ( 0.267) + 13 ( 0.203) + 18 ( 0.010) = 5.61 years

μ = 5.61 years ← answer

iii) Stantard deviation of the ages = σ

σ = square root of the variace

variance = s = sum of the products of the squares of the differences between the mean and the class midpoint time the probability.

s = (0.5 - 5.61)² (0.361) + (3 - 5.61)² ( 0.159) + (8 - 5.61)² ( 0.267) + (13 - 5.61)² ( 0.203) + (18 - 5.61)² ( 0.010) = 24.65

σ = √ (24.65) = 4.97 years ← answer

7 0

3 years ago