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Nina [5.8K]
2 years ago
13

What can you conclude about the setting, based on this passage? Buck is in the north, where sled dogs are in demand. Buck is exp

eriencing snow for the first time. Buck is curious about becoming a sled dog. Buck has been returned to his original owner.
Mathematics
2 answers:
Dovator [93]2 years ago
7 0

Answer:

We can conclude that the setting is stationed somewhere in the north where it's very cold, so cold that they can use sled dogs. We can also assume it's winter time because it normally snows during the winter.

So the setting is in the north where it is cold, during the winter time.

Nimfa-mama [501]2 years ago
3 0

Answer:

North during the winter time!

Step-by-step explanation:

We can conclude that Buck is in the north, in the winter time with snow!

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Y = ax² + bx + c is the equation of a parabola:
 y = 4x² -8x -6 is our equation.
The axis of symmetry in a parabola is x = -b/2a 
 x = -(-8)/(2.4) = - 8/8 and x = - 1 (axis of symmetry
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Ellie bought a pair of jeans on sale for 15% off. If the jeans were originally $35, how much did she save?
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Answer:5.25

Step-by-step explanation:

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For what value of a does 9 = (StartFraction 1 Over 27 EndFraction) Superscript a + 3?
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A

Step-by-step explanation:

9 = (1/27)^(a+3)

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3² = 3^(-3a-9)

2 = -3a-9

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DEF has vertices P(-3,8) Q(-6, -4) R (1,1). if you reflect pqr across the x axis what will be the coordinates of the vertices
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This should be correct! Hope I helped!

4 0
3 years ago
An urn contains 5 white and 10 black balls. A fair die is rolled and that number of balls is randomly chosen from the urn. What
galina1969 [7]

Answer:

Part A:

The probability that all of the balls selected are white:

P(A)=\frac{1}{6}(\frac{1}{3}+\frac{2}{21}+\frac{2}{91}+\frac{1}{273}+\frac{1}{3003}+0)\\      P(A)=\frac{5}{66}=0.075757576

Part B:

The conditional probability that the die landed on 3 if all the balls selected are white:

P(D_3|A)=\frac{\frac{2}{91}*\frac{1}{6}}{\frac{5}{66} } \\P(D_3|A)=\frac{22}{455}=0.0483516

Step-by-step explanation:

A is the event all balls are white.

D_i is the dice outcome.

Sine the die is fair:

P(D_i)=\frac{1}{6} for i∈{1,2,3,4,5,6}

In case of 10 black and 5 white balls:

P(A|D_1)=\frac{5_{C}_1}{15_{C}_1} =\frac{5}{15}=\frac{1}{3}

P(A|D_2)=\frac{5_{C}_2}{15_{C}_2} =\frac{10}{105}=\frac{2}{21}

P(A|D_3)=\frac{5_{C}_3}{15_{C}_3} =\frac{10}{455}=\frac{2}{91}

P(A|D_4)=\frac{5_{C}_4}{15_{C}_4} =\frac{5}{1365}=\frac{1}{273}

P(A|D_5)=\frac{5_{C}_5}{15_{C}_5} =\frac{1}{3003}=\frac{1}{3003}

P(A|D_6)=\frac{5_{C}_6}{15_{C}_6} =0

Part A:

The probability that all of the balls selected are white:

P(A)=\sum^6_{i=1} P(A|D_i)P(D_i)

P(A)=\frac{1}{6}(\frac{1}{3}+\frac{2}{21}+\frac{2}{91}+\frac{1}{273}+\frac{1}{3003}+0)\\      P(A)=\frac{5}{66}=0.075757576

Part B:

The conditional probability that the die landed on 3 if all the balls selected are white:

We have to find P(D_3|A)

The data required is calculated above:

P(D_3|A)=\frac{P(A|D_3)P(D_3)}{P(A)}\\ P(D_3|A)=\frac{\frac{2}{91}*\frac{1}{6}}{\frac{5}{66} } \\P(D_3|A)=\frac{22}{455}=0.0483516

7 0
3 years ago
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