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Veronika [31]
2 years ago
15

What is the perimeter of triangle ABC with coordinates of A(3,4), B(3,11), C(27,4)

Mathematics
1 answer:
BlackZzzverrR [31]2 years ago
5 0

Answer:

Is there supposed to be a picture or no?

Step-by-step explanation:

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Suppose that 10% of adults belong to health clubs, and 40% of these health club members go to the club at least twice a week. Wh
kozerog [31]

Answer:

4% of all adults go to a health club at least twice a week

Step-by-step explanation:

  • the proportion of adults who belong to health clubs is 10% that is 0.10
  • the proportion of these adults (health club members) go to the club at least twice a week is 40%, that is 0.40.

Thus, the proportion of all adults go to a health club at least twice a week is

0.10 × 0.40 = 0.04, that is 4%

3 0
3 years ago
QUESTION:
Elan Coil [88]

Answer:

b.

Step-by-step explanation:

6 0
3 years ago
K: Marcus has 8 pairs of jeans, 4 pairs of black
jasenka [17]

Answer:

19 bottoms to wear

Step-by-step explanation:

1. 8+4+2+5

2.8+4+2+5= 19 bottoms

3. if i did something wrong or didnt catch something tell me

4 0
3 years ago
Suppose William and Donald both drive the same car, and have the same
nataly862011 [7]

Answer: William

Step-by-step explanation: This is true because whichever has a better gas mileage pays less, so the bigger number per year has better mileage, so 15,000 would have better which is Donald, and William has 12,000 which is worse, so he would have to pay more.

5 0
3 years ago
Consider two independent tosses of a fair coin. Let A be the event that the first toss results in heads, let B be the event that
aliina [53]

Answer with Step-by-step explanation:

We are given that two independent tosses of a fair coin.

Sample space={HH,HT,TH,TT}

We have to find that A, B and C are pairwise independent.

According to question

A={HH,HT}

B={HH,TH}

C={TT,HH}

A\cap B={HH}

B\cap C={HH}

A\cap C={HH}

P(E)=\frac{number\;of\;favorable\;cases}{total\;number\;of\;cases}

Using the formula

Then, we get

Total number of cases=4

Number of favorable cases to event A=2

P(A)=\frac{2}{4}=\frac{1}{2}

Number of favorable cases to event B=2

Number of favorable cases to event C=2

P(B)=\frac{2}{4}=\frac{1}{2}

P(C)=\frac{2}{4}=\frac{1}{2}

If the two events A and B are independent then

P(A)\cdot P(B)=P(A\cap B)

P(A\cap)=\frac{1}{4}

P(B\cap C)=\frac{1}{4}

P(A\cap C)=\frac{1}{4}

P(A)\cdot P(B)=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}

P(B)\cdot P(C)=\frac{1}{4}

P(A)\cdot P(C)=\frac{1}{4}

P(A)\cdot P(B)=P(A\cap B)

Therefore, A and B are independent

P(B)\cdot P(C)=P(B\cap C)

Therefore, B and C are independent

P(A\cap C)=P(A)\cdot P(C)

Therefore, A and C are independent.

Hence, A, B and C are pairwise independent.

6 0
3 years ago
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