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Anit [1.1K]
3 years ago
11

The perimeter of a rectangular garden is 36 1/2 feet. One side of the garden is 9 feet long. What is the area, in square feet, o

f the garden?
Group of answer choices

81 1/4

83 1/4

85 1/2

85 9/16
Mathematics
1 answer:
olganol [36]3 years ago
3 0

Answer:

83 1/4 sq ft

Step-by-step explanation:

let x = length or width of two identical unknown sides

2(9) + 2x = 36 1/2

18 + 2x = 36 1/2

2x = 18 1/2

x = 37/2 ÷ 2 = 37/4 or 9 1/4

therefore, multiply 9 by 9 1/4

Area is:  9/1 x 37/4 = 333/4 = 83 1/4

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Answer:

The 1st graph

Step-by-step explanation:

20 = 2t + 12

2t = 8

t = 4

At most she can afford 4 toppings which means she can have 4 toppings to less: t ≤ 4. This is represented in the 1st graph.

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3 years ago
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Find the value of x
romanna [79]

they are 2 similar triangles, one is smaller than the other. they are in the ratio of 2:1,

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3 years ago
−2(x+1/4)+1=5 PLEASE HELP
PilotLPTM [1.2K]

Answer: x=−9/4

Step-by-step explanation:

Let's solve your equation step-by-step.

−2(x+14)+1=5

Step 1: Simplify both sides of the equation.

−2(x+14)+1=5(−2)(x)+(−2)(14)+1=5(Distribute)−2x+

−1

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−2x+

1

2

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−2x+

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Step 2: Subtract 1/2 from both sides.

−2x+

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1

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1

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Step 3: Divide both sides by -2.

−2x

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aleksklad [387]
The answer should be B! Hope this helped :)
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3 years ago
The prior probabilities for events A1 and A2 are P(A1) = 0.20 and P(A2) = 0.80. It is also known that P(A1 ∩ A2) = 0. Suppose P(
Umnica [9.8K]

Answer:

(a) A_1 and A_2 are indeed mutually-exclusive.

(b) \displaystyle P(A_1\; \cap \; B) = \frac{1}{20}, whereas \displaystyle P(A_2\; \cap \; B) = \frac{1}{25}.

(c) \displaystyle P(B) = \frac{9}{100}.

(d) \displaystyle P(A_1 \; |\; B) \approx \frac{5}{9}, whereas P(A_1 \; |\; B) = \displaystyle \frac{4}{9}

Step-by-step explanation:

<h3>(a)</h3>

P(A_1 \; \cap \; A_2) = 0 means that it is impossible for events A_1 and A_2 to happen at the same time. Therefore, event A_1 and A_2 are mutually-exclusive.

<h3>(b)</h3>

By the definition of conditional probability:

\displaystyle P(B \; | \; A_1) = \frac{P(B \; \cap \; A_1)}{P(B)} = \frac{P(A_1 \; \cap \; B)}{P(B)}.

Rearrange to obtain:

\displaystyle P(A_1 \; \cap \; B) = P(B \; |\; A_1) \cdot  P(A_1) = 0.25 \times 0.20 = \frac{1}{20}.

Similarly:

\displaystyle P(A_2 \; \cap \; B) = P(B \; |\; A_2) \cdot  P(A_2) = 0.80 \times 0.05 = \frac{1}{25}.

<h3>(c)</h3>

Note that:

\begin{aligned}P(A_1 \; \cup \; A_2) &= P(A_1) + P(A_2) - P(A_1 \; \cap \; A_2) = 0.20 + 0.80 = 1\end{aligned}.

In other words, A_1 and A_2 are collectively-exhaustive. Since A_1 and A_2 are collectively-exhaustive and mutually-exclusive at the same time:

\displaystyle P(B) = P(B \; \cap \; A_1) + P(B \; \cap \; A_2) = \frac{1}{20} + \frac{1}{25} = \frac{9}{100}.

<h3>(d)</h3>

By Bayes' Theorem:

\begin{aligned} P(A_1 \; |\; B) &= \frac{P(B \; | \; A_1) \cdot P(A_1)}{P(B)} \\ &= \frac{0.25 \times 0.20}{9/100} = \frac{0.05 \times 100}{9} = \frac{5}{9}\end{aligned}.

Similarly:

\begin{aligned} P(A_2 \; |\; B) &= \frac{P(B \; | \; A_2) \cdot P(A_2)}{P(B)} \\ &= \frac{0.05 \times 0.80}{9/100} = \frac{0.04 \times 100}{9} = \frac{4}{9}\end{aligned}.

6 0
3 years ago
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