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marusya05 [52]
1 year ago
11

14.7 ml and a mass of 22.8 g

Mathematics
1 answer:
aliya0001 [1]1 year ago
5 0

The density of the liquid is 1.55 g/mL.

Given the following data:

Mass of liquid = 22.8 grams

Volume of liquid = 14.7 mL

To find the density (unknown) of the liquid:

Density is mass all over the volume of a substance or an object.  Thus, density is mass per unit volume of an object.

Mathematically, the density of a substance is given by the formula;

Substituting the values into the formula, we have;

Density = 1.55 g/mL

Therefore, the density of the liquid is 1.55 g/mL.

Read more: brainly.com/question/18320053

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A student takes an exam containing 1414 multiple choice questions. The probability of choosing a correct answer by knowledgeable
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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

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a total of $9500 is invested, part at 10% simple interest and part at 9%. if the total annual return from the two investments is
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a total of $9500 is invested, part at 10% simple interest and part at 9%

if the total annual return from the two investments is $914.00

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Let x amount  is invested at 10% simple interest

Interest for first part =  10 % times x = 0.10x

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Interest for remaining amount = 9% (9500 -x) = 0.09(9500-x)

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0.10x + 0.09(9500-x) = 914

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0.01x = 59 ( divide both sides by 0.01)

So x = 5900

$5900 is invested in 10% simple interest

9500 - 5900 = $3600 is invested in 9% simple interest


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