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3 years ago

9

If M/DFC= 40 and M CD= 55 the BG= 25 80 135

2 answers:

timofeeve [1]3 years ago

8 0

Answer:

BG = 25°

Step-by-step explanation:

Given

∠DFC=40°

arc CD=55°

The inner angle of an arc is calculated by the semi-sum of the arcs comprising it and its opposite.

Mathematically, it is represented as follows;

∠DFC=½[CD+BG]

Open bracket

∠DFC=½ CD+½ BG]l

Make arc BG the subject of formula

½ BG = ∠DFC - ½CD

Multiply both sides by 2

BG = 2∠DFC - arc CD

By substituton,

BG = 2 * 40 - 55

BG =80°-55°

BG =25°

Hence the measurement of Arc BG is 25°

Y_Kistochka [10]3 years ago

6 0

Answer:

25

Step-by-step explanation:

GC and BD are 2 intersecting chords thus the measure of the angle formed is

∠DFC = 0.5( m arc CD + m arc BG ), that is

0.5(55 + BG) = 40 ( multiply both sides by 2 )

55 + BG = 80 ( subtract 55 from both sides )

BG = 25

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Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl; then a seco

kari74 [83]

Answer:

C. $\frac{1}{18}$

Step-by-step explanation:

Given: Six cards numbered from $1$ to $6$ are placed in an empty bowl. First one card is drawn and then put back into the bowl then a second card is drawn.

To Find: If the cards are drawn at random and if the sum of the numbers on the cards is $8$ , what is the probability that one of the two cards drawn is numbered $5$ .

Solution:

Sample space for sum of cards when two cards are drawn at random is $\{(1,1),(1,2),(1,3)......(6,6)\}$

total number of possible cases $=36$

Sample space when sum of cards is $8$ is $\{(3,5),(5,3),(6,2),(2,6),(4,4)\}$

Total number of possible cases $=5$

Sample space when one of the cards is $5$ is $\{(5,3),(3,5)\}$

Total number of possible cases $=2$

Let A be the event that sum of cards is $8$

$p(\text{A})$ $=\frac{\text{total cases when sum of cards is 8}}{\text{all possible cases}}$

$p(\text{A})=\frac{5}{36}$

Let B be the event when one of the two cards is $5$

probability than one of two cards is $5$ when sum of cards is $8$

$p(\frac{\text{B}}{\text{A}})=\frac{\text{total case when one of the number is 5}}{\text{total case when sum is 8}}$

$p(\frac{\text{B}}{\text{A}})=\frac{2}{5}$

Now,

probability that sum of cards $8$ is and one of cards is $5$

$p(\text{A and B}=p(\text{A})\times p(\frac{\text{B}}{\text{A}})$

$p(\text{A and B})=\frac{5}{36}\times\frac{2}{5}$

$p(\text{A and B})=\frac{1}{18}$

if sum of cards is $8$ then probability that one of the cards is $8$ is $\frac{1}{18}$ , option C is correct.

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