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masha68 [24]
3 years ago
14

Equivalent ratios of 25:45

Mathematics
2 answers:
liberstina [14]3 years ago
4 0

Answer: one could be 5:9

Step-by-step explanation:just divide 25 and 45 by 5

Masja [62]3 years ago
4 0

Answer:

5:9

50:90

75:135

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Find f'(x) <br>f(x)=(1-2x^2)^3
Hunter-Best [27]

Answer:  f'(x) = 12x  + 48x³ - 48x⁵

<u>Step-by-step explanation:</u>

f(x) = (1 - 2x²)³

     = (1 - 2x²)(1 - 4x² + 4x⁴)

         1 - 4x² + 4x⁴

        <u>    -2x² + 8x⁴ - 8x⁶</u>

     =  1 - 6x² + 12x⁴ - 8x⁶

f'(x) = 0 -2(6)x²⁻¹ + 4(12)x⁴⁻¹ - 6(8)x⁶⁻¹

      =     12x  + 48x³ - 48x⁵

3 0
3 years ago
A student answers 160 of 200 questions on an exam correctly. What percent of the questions did he answer correctly?
Alexandra [31]

Dive correct answers by total questions:

160/200 = 0.80

Multiply by 100 to get percent:

0.80 x 100 = 80%

5 0
3 years ago
Read 2 more answers
How to use Pythagorean theorem to find the right triangle side length
MrRissso [65]

Answer:

You would use a^2 + b^2 = c^2

Step-by-step explanation:

a and b are the shorter side of the triangle whereas c is the hypotenuse or longest side of the triangle

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3 years ago
Which value is equivalent to sin 42?
mafiozo [28]

ANSWER-COS 48 DEGREE

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3 years ago
A student takes an exam containing 1414 multiple choice questions. The probability of choosing a correct answer by knowledgeable
Readme [11.4K]

Answer:

0.0082 = 0.82% probability that he will pass

Step-by-step explanation:

For each question, there are only two possible outcomes. Either the students guesses the correct answer, or he guesses the wrong answer. The probability of guessing the correct answer for a question is independent of other questions. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

In this problem we have that:

n = 14, p = 0.3.

If the student makes knowledgeable guesses, what is the probability that he will pass?

He needs to guess at least 9 answers correctly. So

P(X \geq 9) = P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 9) = C_{14,9}.(0.3)^{9}.(0.7)^{5} = 0.0066

P(X = 10) = C_{14,10}.(0.3)^{10}.(0.7)^{4} = 0.0014

P(X = 11) = C_{14,11}.(0.3)^{11}.(0.7)^{3} = 0.0002

P(X = 12) = C_{14,12}.(0.3)^{12}.(0.7)^{2} = 0.000024

P(X = 13) = C_{14,13}.(0.3)^{13}.(0.7)^{1} = 0.000002

P(X = 14) = C_{14,14}.(0.3)^{14}.(0.7)^{0} \cong 0

P(X \geq 9) = P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) = 0.0066 + 0.0014 + 0.0002 + 0.000024 + 0.000002 = 0.0082

0.0082 = 0.82% probability that he will pass

6 0
3 years ago
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