If x=4 what is the value of x to the power of 2-2 x

A student claims that the expression “9 times the sum of a number and 13" is translated to the algebraic expression 9n + 13. Is

FrozenT [24]

Answer:

No, the student is not correct. The correct expression is 9(n+13).

Step-by-step explanation:

"9 times the sum of a number and 13" means you must multiply 9 by the sum, not just by the number. 9n + 13 would be "the sum of 9 times a number and 13." Slightly different wording.

4 0

3 years ago

Read 2 more answers

Evaluate 11 to the 2nd power

lions [1.4K]

121)

11 times itself.

6 0

3 years ago

Choose the correct congruence statement

KengaRu [80]

Answer:

To write the congruence statement, you need to line up the corresponding parts in the triangles: \begin{align*}\angle R \cong \angle F, \angle S \cong \angle E,\end{align*} and \begin{align*}\angle T \cong \angle D\end{align*}. Therefore, the triangles are \begin{align*}\triangle RST \cong \triangle FED\end{align*}.

Step-by-step explanation:

3 0

3 years ago

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

4 years ago

Is the following conditional true? If p<2, then p≤2.

Romashka-Z-Leto [24]

If this is the only information you are given then no. This is because the < sign means that p is less than 2. The second sign cannot be true if the first one is because the second sign says that p is less than or equal to 2.

6 0

3 years ago