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Mrrafil [7]
2 years ago
11

The ratio of boys to girls in Mr. Trent's

Mathematics
1 answer:
goldenfox [79]2 years ago
4 0

Answer:

12

Step-by-step explanation:

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Solve x<br> Log 10×=5.9<br> ×=
Alja [10]

Answer:

x=0.59

Step-by-step explanation:

because if u divide 5.9 by 10 it will give u 0.59

and isolate x

8 0
3 years ago
IM AWARDING POINTS AND BRAINLIEST ITS DUE ASAP
Lelechka [254]

Answer:

the answer would be c=81x^28-n I hope it helps!

4 0
2 years ago
Read 2 more answers
The probability that an event will occur is 95%. Which of these best describes the likelihood of the event occurring?
AfilCa [17]

Answer:

Likely

First option is correct.

Step-by-step explanation:

The probability for occurring the event = 95% = 0.95

Let us understand different types of events:-

  • Certain events: If the probability of occurring an even is 1, then it is a certain event.
  • Impossible events: If the probability of occurring an even is 0, then it is a impossible event.
  • Likely events: If the probability of occurring an even is close to 1, then it is a likely event.
  • Unlikely events: If the probability of occurring an even is close to 0, then it is a unlikely event.

Now, the probability for occurring the event = 95% = 0.95, which is close to 1.

Hence, we can conclude that the event is likely.

First option is correct.

4 0
3 years ago
Read 2 more answers
Which of the following is an example of a matched pairs design? A. A teacher calculates the average scores of students on a pair
Mashutka [201]

Answer:

B. A teacher compares the pre-test and post-test scores of students

Step-by-step explanation:

the answer is true, because it is a good example to compare the tests between students, we know that a matching pair design is a random model and is used when the experiment allows grouping subjects in pairs based on a variable and each pair will receive randomly a different handling, the answer A is not true because in the example all students are uniformly averaged and the variable is not correlated with subgroups, option c is incorrect because the variable was not randomized and generates classification bias and the option d is incorrect because the teacher compares a small sample as her class with a score of a total sample, but does not intervene on her students when selecting the corresponding group

5 0
3 years ago
) a, p and d are n×n matrices. check the true statements below:
BigorU [14]
A. False. Consider the identity matrix, which is diagonalizable (it's already diagonal) but all its eigenvalues are the same (1).

b. True. Suppose \mathbf P is the matrix of the eigenvectors of \mathbf A, and \mathbf D is the diagonal matrix of the eigenvalues of \mathbf A:


\mathbf P=\begin{bmatrix}\mathbf v_1&\cdots&\mathbf v_n\end{bmatrix}

\mathbf D=\begin{bmatrix}\lambda_1&&\\&\ddots&\\&&\lambda_n\end{bmatrix}

Then

\mathbf{AP}=\begin{bmatrix}\mathbf{Av}_1&\cdots&\mathbf{Av}_n\end{bmatrix}=\begin{bmatrix}\lambda_1\mathbf v_1&\cdots&\lambda_n\mathbf v_n\end{bmatrix}=\mathbf{PD}

In other words, the columns of \mathbf{AP} are \mathbf{Av}_i, which are identically \lambda_i\mathbf v_i, and these are the columns of \mathbf{PD}.

c. False. A counterexample is the matrix

\begin{bmatrix}1&1\\0&1\end{bmatrix}

which is nonsingular, but it has only one eigenvalue.

d. False. Consider the matrix

\begin{bmatrix}0&1\\0&0\end{bmatrix}

with eigenvalue \lambda=0 and eigenvector \begin{bmatrix}k&0\end{bmatrix}^\top, where k\in\mathbb R. But the matrix can't be diagonalized.
7 0
3 years ago
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