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2 years ago

Does the point (4, 0) satisfy the equation y = x2?

1 answer:

8 0

Y=x^2
substitute the values into the original equation
0=4^2
0=16
No it does not satisfy the equation because 0 does not equal to 16

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The probability that a student has a Visa card (event V) is .73. The probability that a student has a MasterCard (event M) is .1

snow_lady [41]

We assumed in this answer that the question b is, Are the events V and M independent?

Answer:

(a). The probability that a student has either a Visa card or a MasterCard is $\\ P(V \cup M) = 0.88$ . (b). The events V and M are not independent.

Step-by-step explanation:

The key factor to solve these questions is to know that:

$\\ P(V \cup M) = P(V) + P(M) - P(V \cap M)$

We already know from the question the following probabilities:

$\\ P(V) = 0.73$

$\\ P(M) = 0.18$

The probability that a student has both cards is 0.03. It means that the events V AND M occur at the same time. So

$\\ P(V \cap M) = 0.03$

The probability that a student has either a Visa card or a MasterCard

We can interpret this probability as $\\ P(V \cup M)$ or the sum of both events; that is, the probability that one event occurs OR the other.

Thus, having all this information, we can conclude that

$\\ P(V \cup M) = P(V) + P(M) - P(V \cap M)$

$\\ P(V \cup M) = 0.73 + 0.18 - 0.03$

$\\ P(V \cup M) = 0.88$

Then, the probability that a student has either a Visa card or a MasterCard is $\\ P(V \cup M) = 0.88$ .

Are the events V and M independent?

A way to solve this question is by using the concept of conditional probabilities.

In Probability, two events are independent when we conclude that

$\\ P(A|B) = P(A)$ [1]

The general formula for a conditional probability or the probability that event A given (or assuming) the event B is as follows:

$\\ P(A|B) = \frac{P(A \cap B)}{P(B)}$

If we use the previous formula to find conditional probabilities of event M given event V or vice-versa, we can conclude that

$\\ P(M|V) = \frac{P(M \cap V)}{P(V)}$

$\\ P(M|V) = \frac{0.03}{0.73}$

$\\ P(M|V) \approx 0.041$

If M were independent from V (according to [1]), we have

$\\ P(M|V) = P(M) = 0.18$

Which is different from we obtained previously;

That is,

$\\ P(M|V) \approx 0.041$

So, the events V and M are not independent.

We can conclude the same if we calculate the probability

$\\ P(V|M)$ , as follows:

$\\ P(V|M) = \frac{P(V \cap M)}{P(M)}$

$\\ P(V|M) = \frac{0.03}{0.18}$

$\\ P(V|M) = 0.1666.....\approx 0.17$

Which is different from

$\\ P(V|M) = P(V) = 0.73$

In the case that both events were independent.

Notice that

$\\ P(V|M)*P(M) = P(M|V)*P(V) = P(V \cap M) = P(M \cap V)$

$\\ \frac{0.03}{0.18}*0.18 = \frac{0.03}{0.73}*0.73 = 0.03 = 0.03$

$\\ 0.03 = 0.03 = 0.03 = 0.03$

3 0

3 years ago

3567 +3569-1467+9087-3=

Stolb23 [73]

14753 is your answer

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3 years ago

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Surface area of a sphere with a diameter of 17.5 inches

statuscvo [17]

Answer:

Ans is behind

Step-by-step explanation:

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3 years ago

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This is a 30-60-90 triangle.

Triss [41]

Answer:

x = $\frac{2\sqrt{21}}{3}$

Step-by-step explanation:

For a 30-60-90 triangle:

Short leg = s
Long leg = s√3
Hypotenuse = 2s

Thus:

Short leg = √7/√3 = (√21)/3
Long leg = √7
Hypotenuse = x = 2(√21)/3

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2 years ago

What is the median of this set of data? 66, 51, 77, 68, 60, 75, 54, 80

labwork [276]

51, 54, 60, 66, 68, 75, 77, 80 The median is 67

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3 years ago

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