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3 years ago

How do I solve this problem

Mathematics

1 answer:

zysi [14]3 years ago

5 0

The following equations are solved in terms of x:
a) 2x - 3 = y
2x = y + 3
x = (1/2)y + 3/2

b) 2 + 2x - 1 = 4y
2x - 1 = 4y
2x = 4y + 1
x = 2y + 1/2

c) 2A + 4x + 3B = 10
4x = 10 - 2A - 3B
x = 5/2 - (1/2)A - (3/4)B

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Please help with this and explain how to do it, I'll give 15 points

MrRissso [65]

Answer:

Option (4).

Step-by-step explanation:

Outer diameter of the hollow metallic ball = 10 centimeters

Outer radius of this ball = $\frac{10}{2}$ = 5 cm

Volume of the outer ball $V_{1}$ = $\frac{4}{3}\pi (r)^3$

= $\frac{4}{3}\pi (5)^3$

Inner radius of the hollow metallic ball = (5 - 1) = 4 cm

Volume of the inner hollow ball $V_{2}$ = $\frac{4}{3}\pi (4)^3$

Volume of the metal used in the metallic ball = $V_{1}-V_{2}$

= $\frac{4}{3}\pi (5)^3-\frac{4}{3}\pi (4)^{3}$

= $\frac{4}{3}\pi [(5)^3-(4)^3]$

Therefore, expression given in option (4) will be used to measure the volume of the hollow metallic ball.

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3 years ago

Which of the following are types of insurance?

garik1379 [7]

Answer:

life insurance is a type n other are together called as non-life insurance.

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3 years ago

Read 2 more answers

4 is added to the numerator of the fraction 2/5 what number must be added to the denominator to make the fraction equivalent to

kondaur [170]

Answer: 7

Step-by-step explanation:

Hope this helps!

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2 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

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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:

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

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3 years ago