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BigorU [14]
3 years ago
12

The radius of star A is 6.9203x10^5 km, which is 108.7 times the rafius of star B. what is the radius of the star B?

Mathematics
1 answer:
Kay [80]3 years ago
7 0

Answer:

6366.4213

Step-by-step explanation: 6.9203x10^5 is 692,030. I divided 692,030 by 108.7 to get the answer. Anybody can correct any error I may have.

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2/5 of 2/3
Vilka [71]

Answer:

4/15 is your answer my friend

Step-by-step explanation:

5 0
2 years ago
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The number of tickets sold for a concert during each hour ,x, of the first 10hours the tickets were in sale is given by g(x)=x^3
Mashutka [201]

The number in tickets sold A)107

8 0
3 years ago
Show that x = -2 is a solution of 3x² + 13x + 14 = 0​
Oliga [24]

Answer:

x= -2 is correct answer

Step-by-step explanation:

3(-2×-2)+13(-2)+14=0

3(4)+13(-2)+14=0

12+(-26)+14=0

12-26+14=0

12+14-26=0

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3 0
3 years ago
Two sides of a triangle are 8cm and 12cm. Which of the following CANNOT be the measure of the third side,
blondinia [14]

Answer:

4 cannot be the measure of the third side. This is because of the Triangle Inequality Theorem, which states that the sum of two sides of a triangle must be greater than the third side (A+B>C, A+C>B, B+C>A) In this example, if side C were 4, side C (4) plus side A (8) would be 12. Since side B is 12, and 12 cannot be greater than 12, 4 would not work.

Answer=8

Step-by-step explanation:

7 0
4 years ago
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Needdd hellpppppssssssss
Ratling [72]

Answer:

Choice number one:

\displaystyle \frac{5}{10}\cdot \frac{4}{9}.

Step-by-step explanation:

  • Let A be the event that the number on the first card is even.
  • Let B be the event that the number on the second card is even.

The question is asking for the possibility that event A and B happen at the same time. However, whether A occurs or not will influence the probability of B. In other words, A and B are not independent. The probability that both A and B occur needs to be found as the product of

  • the probability that event A occurs, and
  • the probability that event B occurs given that event A occurs.

5 out of the ten numbers are even. The probability that event A occurs is:

\displaystyle P(A) = \frac{5}{10}.

In case A occurs, there will only be four cards with even numbers out of the nine cards that are still in the bag. The conditional probability of getting a second card with an even number on it, given that the first card is even, will be:

\displaystyle P(B|A) = \frac{4}{9}.

The probability that both A and B occurs will be:

\displaystyle P(A \cap B) = P(B\cap A) =  P(A) \cdot P(B|A) = \frac{5}{10}\cdot \frac{4}{9}.

6 0
3 years ago
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