Help will give crown Find the missing side length

He solution to the initial value problem x′′+2x′+26x=5cos(4t),x(0)=0,x′(0)=0 x″+2x′+26x=5cos⁡(4t),x(0)=0,x′(0)=0 is the sum of t

ddd [48]

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3 0

3 years ago

When results from a scholastic assessment test are sent to test-takers, the percentiles associated with their scores are also gi

hram777 [196]

Answer:

The correct option is (D).

Step-by-step explanation:

Percentiles are statistical measures that are used to interpret data. It represents the data value which is more that a specific percentage of the data set.

The <em>n</em>th percentile of a data set is the value that is more that <em>n</em>% of the data set.

⇒ It is provided that a test-taker's score was at the 94th percentile for their verbal grade.

This statement implies that the test taker scored a mark more than 94% of the other test-takers, i.e. he\she performed better than 94% of the other test-takers in the verbal grade.

⇒ Also the test-taker's score was at the 16th percentile for their quantitative grade.

This implies that the test taker scored a mark more than 16% of the other test-takers, i.e. he\she performed better than 16% of the other test-takers in the quantitative grade.

Thus, the correct option is (D).

8 0

3 years ago

a student takes two subjects A and B. Know that the probability of passing subjects A and B is 0.8 and 0.7 respectively. If you

aniked [119]

Answer:

0.64 = 64% probability that the student passes both subjects.

0.86 = 86% probability that the student passes at least one of the two subjects

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

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

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Passing subject A

Event B: Passing subject B

The probability of passing subject A is 0.8.

This means that $P(A) = 0.8$

If you have passed subject A, the probability of passing subject B is 0.8.

This means that $P(B|A) = 0.8$

Find the probability that the student passes both subjects?

This is $P(A \cap B)$ . So

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

$P(A \cap B) = P(B|A)P(A) = 0.8*0.8 = 0.64$

0.64 = 64% probability that the student passes both subjects.

Find the probability that the student passes at least one of the two subjects

This is:

$p = P(A) + P(B) - P(A \cap B)$

Considering $P(B) = 0.7$ , we have that:

$p = P(A) + P(B) - P(A \cap B) = 0.8 + 0.7 - 0.64 = 0.86$

0.86 = 86% probability that the student passes at least one of the two subjects

3 0

3 years ago

Divide 6x^4+2x^3-6x^2-14x-1 by 3x+1

kolezko [41]

Using synthetic division, we find the result of the division
(6x^4 +2x^3 -6x^2 -14x -1)/(3(x +1/3))
to be ...
(6x^3 -6x -12)/3 +3/(3x +1)

The result is ...
2x^3 -2x -4 +3/(3x+1)

3 0

3 years ago

PLS HELPPP MATHHH QUESTIONSSS REWARD BRAINLIEST AND STARS!!!!!!!!

CaHeK987 [17]

151 is a prime number, 616 prime factorization is 2^3 times 7 times 11, the gcf of 132 & 220 is 44, the prime factorization of 104 is 2^3 time 13, the the gcf for 333 & 441 is 9

8 0

3 years ago