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Stels [109]
3 years ago
13

Full working out for this question please

Mathematics
1 answer:
Darya [45]3 years ago
7 0

Answer:

C.

Step-by-step explanation:

according to the condition it is possible to calculate the values of ∠DEF; ∠CDE and ∠DCF:

1) m∠DEF=m∠BAE=104°;

2) m∠CDE=180-m∠BAE=180-104=76°;

3) m∠DCF=m∠BAE-m∠BCF=104-60=44°;

the required value of the angle EFC is:

4) m∠EFC=360- (m∠DEF+m∠CDE+m∠DCF)=360-104-76-44=136°

5) the correct answer is: <u>C. 136°.</u>

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A line l2 is perpendicular to the line x + 2y – 5 = 0 without computing the gradient of this
Stells [14]

Answer:

y=2x-2

Step-by-step explanation:

first, let's put the line into y=mx+b form (slope-intercept form), where m is the slope (or gradient) and b is the y intercept

add 5 to both sides

x+2y=5

subtract x from both sides

2y=-x+5

divide by 2

y=-1/2x+5/2

perpendicular lines have slopes (gradients) that are negative (one is positive, another one is negative) and reciprocal (they are essentially the same number, just "flipped")

to find the slope of I2:

since -1/2 is negative, that means the slope of I2 will be 2 (2/1 is the reciprocal of 1/2 (positive version of -1/2))

so here's our equation so far:

y=2x+b

now we need to find b

because line will pass through (3,4), we can use it to solve for b

substitute 4 as y and 3 as x

4=2(3)+b

multiply

4=6+b

-2=b

therefore the equation is y=2x-2

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3 0
3 years ago
In a certain Algebra 2 class of 30 students, 14 of them play basketball and 10 of them play baseball. There are 14 students who
Veseljchak [2.6K]

Answer:

In a certain Algebra 2 class of 30 students, 22 of them play basketball and 18 of them play baseball. There are 3 students who play neither sport. What is the probability that a student chosen randomly from the class plays both basketball and baseball?

I know how to calculate the probability of students play both basketball and baseball which is 1330 because 22+18+3=43 and 43−30 will give you the number of students plays both sports.

But how would you find the probability using the formula P(A∩B)=P(A)×p(B)?

Thank you for all of the help.

That formula only works if events A (play basketball) and B (play baseball) are independent, but they are not in this case, since out of the 18 players that play baseball, 13 play basketball, and hence P(A|B)=1318<2230=P(A) (in other words: one who plays basketball is less likely to play basketball as well in comparison to someone who does not play baseball, i.e. playing baseball and playing basketball are negatively (or inversely) correlated)

So: the two events are not independent, and so that formula doesn't work.

Fortunately, a formula that does work (always!) is:

P(A∪B)=P(A)+P(B)−P(A∩B)

Hence:

P(A∩B)=P(A)+P(B)−P(A∪B)=2230+1830−2730=1330

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3 years ago
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