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Sidana [21]
3 years ago
7

Which geometric construction is shown here?

Mathematics
1 answer:
Ann [662]3 years ago
5 0

Answer:

B) construct the center of a given circle

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6x-14<-14 or 3x+10>13
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Step-by-step explanation:

not sure the question but I can solve the equations

6x-14<-14

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3x+10>13

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ray OJ bisects angle IOK, if angle IOJ= 2x-5 and measure of angle JOK=x+11, then find measure of angle IOJ
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Round 0.172 to the nearest hundredth.
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It is <span>Round 0.172 to the nearest hundredth.</span>
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What are the coordinates of the endpoints of the midsegment for DEF that is parallel to DE
Nutka1998 [239]

Answer:

\left(\dfrac{x_D+x_F}{2},\dfrac{y_D+y_F}{2}\right),  \left(\dfrac{x_E+x_F}{2},\dfrac{y_E+y_F}{2}\right).

Step-by-step explanation:

Let points D, E and F have coordinates (x_D,y_D),\ (x_E,y_E) and (x_F,y_F).

1. Midpoint M of segment DF has coordinates

\left(\dfrac{x_D+x_F}{2},\dfrac{y_D+y_F}{2}\right).

2. Midpoint N of segment EF has coordinates

\left(\dfrac{x_E+x_F}{2},\dfrac{y_E+y_F}{2}\right).

3. By the triangle midline theorem, midline MN is parallel to the side DE of the triangle DEF, then points M and N are endpoints of the midsegment for DEF that is parallel to DE.

6 0
3 years ago
34​% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and a
finlep [7]

Answer:

a) There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

b) There is a 71.62% probability that more than two students use credit cards because of the rewards program.

c) There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

Step-by-step explanation:

There are only two possible outcomes. Either the student use credit cards because of the rewards program, or they use for other reason. So, we can solve this problem by the binomial distribution.

Binomial probability

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{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 \pi is the probability of X happening.

In this problem, we have that:

10 student are sampled, so n = 10

34% of college students say they use credit cards because of the rewards program, so \pi = 0.34

(a) exactly​ two

This is P(X = 2).

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873

There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

(b) more than​ two

This is P(X > 2).

Either a value is larger than two, or it is smaller of equal. The sum of the decimal probabilities must be 1. So:

P(X \leq 2) + P(X > 2) = 1

P(X > 2) = 1 - P(X \leq 2)

In which

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)

So

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 0) = C_{10,0}.(0.34)^{0}.(0.66)^{10} = 0.0157

P(X = 1) = C_{10,1}.(0.34)^{1}.(0.66)^{9} = 0.0808

P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0157 + 0.0808 + 0.1873 = 0.2838

P(X > 2) = 1 - P(X \leq 2) = 1 - 0.2838 = 0.7162

There is a 71.62% probability that more than two students use credit cards because of the rewards program.

(c) between two and five inclusive

This is:

P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873

P(X = 3) = C_{10,3}.(0.34)^{3}.(0.66)^{7} = 0.2573

P(X = 4) = C_{10,4}.(0.34)^{4}.(0.66)^{6} = 0.2320

P(X = 5) = C_{10,5}.(0.34)^{5}.(0.66)^{5} = 0.1434

P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1873 + 0.2573 + 0.2320 + 0.1434 = 0.82

There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

6 0
3 years ago
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