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3 years ago

11

Can you find the first and third one please!

1 answer:

fenix001 [56]3 years ago

7 0

Answer:

The first ones area is 35 cm squared

The third ones area is 32 cm squared

Step-by-step explanation:

A=1/2H (B1+B2)

35=2.5(8+6)

32=2(7+9)

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Urgent! Need help solving

OLga [1]

Answer:

d 64

Step-by-step explanation:

f(-3)=1/2 ^-6=2^6=64

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3 years ago

Test each statement below to see if it is reversible. If so, write it as a true biconditional. If not, write not reversible

FinnZ [79.3K]

Statement: Complementary angles are two angles with measures that have a sum of 90.
Reverse: Two angles with measures that have a sum of 90 are complementary angles.

Here the reverse is true, therefore, the statement is true biconditional.

Statement: A rectangle is a four-sided figure with at least one right angle.
Reverse: A four sided figure with at least one right angle is a rectangle.

Here, the reverse is not true, therefore, the statement is not reversible.

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3 years ago

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Convert .2 to a fraction

shepuryov [24]

.2 converted to a fraction would be 1/5

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3 years ago

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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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3 years ago

How do you factor x^3-2x^2-35x?

natima [27]

X^3 -2x^2 -35x

=x(x^2-2x-35)

=x(x-7)(x+5)

5 0

3 years ago

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