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Nadusha1986 [10]
3 years ago
10

PLEASE HELP MEEEEE THAnK YOU

Mathematics
1 answer:
laila [671]3 years ago
3 0
-6,2

That is da answer :)
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Use a proportion and the conversion to convert between metric and customary units.
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Answer:

(1 L = 33.8 oz)

Step-by-step explanation:

4 0
3 years ago
A bag contains 15 marbles. The probability of randomly selecting a green marble is 1/5. The probability of randomly
meriva

Answer:

2

Step-by-step explanation:

well if you get all of the fractions into percentage form they will be 20%for green 8% for blue so in turn it would mean 8% of 15 is 2

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Iron has a density of 7.9 g/cm³. Gold has a density of 19.3 g/cm³.
alexandr402 [8]
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3.86kg=3860g
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7 0
2 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
-3w=-w-6. How many solutions
GaryK [48]
There is one solution: w=3
6 0
3 years ago
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