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larisa [96]
3 years ago
9

HELP! The Last option is none of the above

Mathematics
1 answer:
IgorC [24]3 years ago
6 0

<em>The</em><em> </em><em>rel</em><em>ationship</em><em> </em><em>between</em><em> </em><em><</em><em>a</em><em> </em><em>and</em><em> </em><em><</em><em>B </em><em>is</em><em> </em><em>supplementary</em><em> </em><em>angles</em><em>.</em>

<em>both</em><em> </em><em><</em><em>a</em><em> </em><em>and</em><em> </em><em><</em><em>B </em><em>are</em><em> </em><em>in</em><em> </em><em>a</em><em> </em><em>straight</em><em> </em><em>line</em><em>.</em><em>.</em><em>so</em><em> </em><em>they</em><em> </em><em>are</em><em> </em><em>supplementary</em><em> </em><em>angles</em><em> </em>

<em>Supplementary</em><em> </em><em>angles</em><em> </em><em>are</em><em> </em><em>equal</em><em> </em><em>to</em><em> </em><em>1</em><em>8</em><em>0</em><em> </em><em> </em><em>degree</em><em>.</em>

<em>Hope</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>be</em><em> </em><em>helpful</em><em> </em><em>to</em><em> </em><em>you</em>

<em>Good</em><em> </em><em>luck</em><em> </em><em>on</em><em> </em><em>your</em><em> </em><em>assignment</em><em> </em><em>.</em><em>.</em><em>.</em>

<em>~</em><em>p</em><em>r</em><em>a</em><em>g</em><em>y</em><em>a</em>

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ivanzaharov [21]

Answer:

Option C is right

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Given that the  probability of drawing two aces from a standard deck is 0.0059

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Drawing ace in I draw has probability equal to 4/52, when we replace the I card again drawing age has probability equal to same 4/52

So if the two draws are defined as event A and event B,  the events are  independent

C. They are independent because, based on the probability, the first ace was replaced before drawing the second ace.

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Perform the indicated operation.<br> 1 1/6 + 7 3/6 =<br><br> I HAVE 5 MINUTES!!!
gizmo_the_mogwai [7]

Answer:

1 1/6 - 7 3/6 = -19/

3

= -6 1/

3

≅ -6.3333333

Step-by-step explanation:

Conversion a mixed number 1 1/

6

to a improper fraction: 1 1/6 = 1 1/

6

= 1 · 6 + 1/

6

= 6 + 1/

6

= 7/

6

To find new numerator:

a) Multiply the whole number 1 by the denominator 6. Whole number 1 equally 1 * 6/

6

= 6/

6

b) Add the answer from previous step 6 to the numerator 1. New numerator is 6 + 1 = 7

c) Write a previous answer (new numerator 7) over the denominator 6.

One and one sixth is seven sixths

Conversion a mixed number 7 3/

6

to a improper fraction: 7 3/6 = 7 3/

6

= 7 · 6 + 3/

6

= 42 + 3/

6

= 45/

6

To find new numerator:

a) Multiply the whole number 7 by the denominator 6. Whole number 7 equally 7 * 6/

6

= 42/

6

b) Add the answer from previous step 42 to the numerator 3. New numerator is 42 + 3 = 45

c) Write a previous answer (new numerator 45) over the denominator 6.

Seven and three sixths is forty-five sixths

Subtract: 7/

6

- 45/

6

= 7 - 45/

6

= -38/

6

= 2 · -19/

2 · 3

= -19/

3

For adding, subtracting, and comparing fractions, it is suitable to adjust both fractions to a common (equal, identical) denominator. The common denominator you can calculate as the least common multiple of both denominators - LCM(6, 6) = 6. In practice, it is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 6 × 6 = 36. In the next intermediate step, the fraction result cannot be further simplified by canceling.

In words - seven sixths minus forty-five sixths = minus nineteen thirds.

4 0
3 years ago
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