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11111nata11111 [884]
3 years ago
12

What is the y value

Mathematics
2 answers:
rjkz [21]3 years ago
8 0

You can read this max y-value directly from the graph. It's 4. The vertex of this graph is at (2,4).


Dmitriy789 [7]3 years ago
3 0

It’s 4. The vertex if the graph is at (2,4). Hope this helps you out!

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What are the numerical measures of each angle in the diagram? 1 and 3 measure degrees. 2 and 4 measure degrees. (3x - 1)º (2x +
Tanya [424]

Answer:

Step-by-step explanation:

In the diagram shown, the measure of angle 1 is oppositely directed to angle 2 and oppositely directed angles are equal.

Hence <1 = <3

Given < 1 = 3x-1 and <3 = 2x+9

Hence 3x-1 =  2x+9

collect like terms

3x-2x = 9+1

x = 10°

Since <1 = 3x-1

on substituting x = 10

<1 = 3(10)-1

<1 = 30-1

<1 = 29°

<1+<2 = 180 (angle on a straight line)

29+<2 = 180

<2 = 180-29

<2 = 151°

Similarly, on substituting x = 10 into <3

<3 = 2x+9

<3 = 2(10)+9

<3 = 20+9

<3 = 29°

<3+<4 = 180 (angle on a straight line)

29+<4 = 180

<4 = 180-29

<4 = 151°

3 0
4 years ago
Read 2 more answers
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

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year 1

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put in 1591.35 in for x in year 3

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