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3 years ago

11

What average speed does he need to reach in his second round in order to attain the overall average speed of 60 mph required for

the test?

1 answer:

Dmitry_Shevchenko [17]3 years ago

3 0

10 think about it it’s 60s and it’s in the ten’s place because 10,20,30,40,50, and 60

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The age difference between the two brothers is 5. 7 years later, the age of the elder brother will be 3 times bigger than the ag

saw5 [17]

Answer:

<em><u>y = 4 and x = 9</u></em>

Step-by-step explanation:

I hope this is correct

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3 years ago

Write the population of lubbok in word form

Shtirlitz [24]

1651384 also known as "one million, six hundred fifty one thousand, three hundred eighty four"

6 0

3 years ago

3. Which rule best describes the last three numbers shown in the pattern? 10, 12, 9, 11

sergij07 [2.7K]

Answer:

it goes by 2 two times then subtracts 3 then again

Step-by-step explanation:

10,12,9,11,8,10,7

5 0

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

Let u = <3, -1>, v = <-6, -6>. Find 9u + 2v.

Gnom [1K]

9u means you're multiplying 9 into that vector, both components. Same with the 2v. 9*3 = 27 and 9*-1 = -9, so your new vector u is <27, -9>. Now let's do v. 2* -6 (twice) = -12, so your new v vector is <-12, -12>. Add those together now, first components of each and second components of each. 27 + (-12) = 15; -9+(-12)=-21. So the addition of those gives us a final vector with a displacement of <15, -21>

5 0

3 years ago