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3 years ago

Polygon ABCDE is congruent to polygon TVSRK. Angle EAB corresponds to _____

Mathematics

2 answers:

qaws [65]3 years ago

7 0

Answer d because it is the corresponding angle

ruslelena [56]3 years ago

3 0

The answer is d how do i know because i learned that before

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Two more than three times a number is twenty. What is that number?

spin [16.1K]

The number may be 6 because 6 • 3 = 18, and 2 more than 18 would be 20

3 0

2 years ago

At a school with 100? students, 33 were taking? Arabic, 32 ?Bulgarian, and 40 Chinese. 9 students take only? Arabic, 12 take onl

Stels [109]

Answer: The answer is 4 and 32.

Step-by-step explanation: Let "A", "B" and "C" represents the set of students who were taking Arabic, Bulgarian and Chinese respectively.

The, according to the given information, we have

$n(A)=33,~~n(B)=32,~~n(C)=40,~~n(A\cap B)=14.$

Let 'p' represents the number of students who take all the three languages, then

$n(A\cap B\cap C)=p.$

Also,

$n(A)-n(A\cap B)-n(A\cap C)+n(A\cap B\cap C)=9\\\\\Rightarrow n(A\cap B)+n(A\cap C)=24+p~~~~~~~~~~~~~~(a),\\\\n(A\cap B)+n(B\cap C)=20+p~~~~~~~~~~~~~~(b),\\\\n(B\cap C)+n(A\cap C)=20+p~~~~~~~~~~~~~~~(c).$

From here, we get after subtracting equation(c) from (b) that

$n(A\cap B)=n(A\cap C)=14.$

Therefore,

$p=14+14-24=4,$ and from equation (a), we find

$n(B\cap C)=24-14=10.$

Thus,

$n(A\cap B\cap C)=4$ and

$n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)\\\\\Rightarrow n(A\cap B\cap C)=33+32+40-9-12-20+4\\\\\Rightarrow n(A\cap B\cap C)=68.$

Thus, the number of students who take all the three languages is 4 and the number of students who take none of the languages is 100-68 = 32.

5 0

3 years ago

The distance between (0,-4) and (2,-5)

alex41 [277]

Step-by-step explanation:

d= $\sqrt{ (0-2)^2 + (-4-(-5))^2}$

d= $\sqrt{(-2)^2 + (1)^2}$

d= $\sqrt{4 + 1}$

d= $\sqrt{5}$

8 0

3 years ago

Are your finances, buying habits, medical records, and phone calls really private? A real concern for many adults is that comput

Andrej [43]

Answer:

a)There is a 4.88% probability that none is concerned that employers are monitoring phone calls.

b)There is a 7.89% probability that all are concerned that employers are monitoring phone calls.

c)There is a 37.23% probability that exactly two are concerned that employers are monitoring phone calls.

Step-by-step explanation:

The binomial probability is the probability of exactly x successes on n repeated trials in an experiment which has two possible outcomes (commonly called a binomial experiment).

It is given by the following formula:

$P = C_{n,x}.p^{n}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of a success.

In this problem, a success is being concerned that employers are monitoring phone calls.

53% of adults are concerned that employers are monitoring phone calls, so $p = 0.53$

(a) Out of four adults, none is concerned that employers are monitoring phone calls.

Four adults, so $n = 4$ .

Is the probability of 0 successes, so x = 0.

$P = C_{n,x}.p^{n}.(1-p)^{n-x}$

$P = C_{4,0}.(0.53)^{0}.(0.47)^{4}$

$P = 0.0488$

There is a 4.88% probability that none is concerned that employers are monitoring phone calls.

(b) Out of four adults, all are concerned that employers are monitoring phone calls.

Four adults, so $n = 4$ .

Is the probability of 4 successes, so x = 4.

$P = C_{n,x}.p^{n}.(1-p)^{n-x}$

$P = C_{4,0}.(0.53)^{4}.(0.47)^{0}$

$P = 0.0789$

There is a 7.89% probability that all are concerned that employers are monitoring phone calls.

(c) Out of four adults, exactly two are concerned that employers are monitoring phone calls.

Four adults, so $n = 4$ .

Is the probability of 4 successes, so x = 2.

$P = C_{n,x}.p^{n}.(1-p)^{n-x}$

$P = C_{4,2}.(0.53)^{2}.(0.47)^{2}$

$P = 0.3723$

There is a 37.23% probability that exactly two are concerned that employers are monitoring phone calls.

3 0

2 years ago

Fun fact

pav-90 [236]

Answer: wow

I'd want nothing more than to swim in a pool of saliva.

8 0

2 years ago