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3 years ago

What is true about this graphs slope

Mathematics

1 answer:

Rom4ik [11]3 years ago

8 0

the graph has no slope because it is y=2.5

therefore the answer is

d) it has no slope

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What is 3/5 equal to

klio [65]

3/5 is equal to .6 and 60%

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4 years ago

Read 2 more answers

Find the values of the six trigonometric functions for angle θ.Give answers in simplest form.

natita [175]

Answer:

sin (theta) = $\frac{\sqrt{15} }{8}$

cos (theta) = $\frac{7}{8}$

tan (theta) = $\frac{\sqrt{15} }{7}$

cot (theta) = 7 square root 15/15

sec (theta) = $\frac{8}{7}$

csc (theta) = 8 square root 15/15

Step-by-step explanation:

7 0

3 years ago

Complete the table by writing the fraction as a decimal and as a percent

aniked [119]

Answer:

Decimal = 0.6

Percentage = 60%

Step-by-step explanation:

Let's simplify the fraction.

$\frac{120}{200} = \frac{60}{100} = 0.6$

Hence the Decimal is 0.6

To express 120/200 in percentage , we have to multiply the fraction with 100.

$\frac{120}{200} \times 100 = \frac{120}{2} = 60\%$

Hence the percentage value is 60%

7 0

3 years ago

Part A

alexgriva [62]

Answer:

6=2 , 12=4, 15=5, 24=8

Step-by-step explanation:

6 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