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Lerok [7]
3 years ago
13

A scientist mixes water (containing no salt) with a solution that contains 35% salt. She wants to obtain 105 ounces of a mixture

that is 15% salt. How many ounces of water and how many ounces of the 35% salt solution should she use?
Mathematics
1 answer:
Naya [18.7K]3 years ago
7 0

Answer:

60 oz of water to add

45 oz of 35% salt solution should be used

Step-by-step explanation:

Let x = amt no. ounces in the first solution

and  105 - x = amt no. of ounces in the second solution

Ist solution contains 0% salt

2nd solution contains .35(105 - x) salt

.15(105= amt of salt in mix

So.

amt of salt in 1st solution + amt of salt in 2nd solution = amt of salt in the mix

0 + .35(105 - x) = .15(105)

Let's get rid of the fractions by multipying thru the equation by 100

35(105 - x) = 15(105)

3675 - 35x = 1575

-35x = -2100

x = 60 oz of water to add

105 - 60 = 45 oz of 35% salt solution should be used

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The probability that the next toss will be heads is 1/8.

<h3>What is probability?</h3>

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Using the probability formula, one can determine the likelihood of an event by dividing the favorable number of possibilities by the total number of options. Since the favorable number of outcomes can never be greater than the entire number of outcomes, the probability of an event happening can range from 0 to 1.

Probability of getting two tails and next heads in three tosses is,

=1/2*1/2*1/2

=1/8

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4 0
2 years ago
Suppose that bugs are present in 1% of all computer programs. A computer de-bugging program detects an actual bug with probabili
lawyer [7]

Answer:

(i) The probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.

(ii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.

(iii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.

Step-by-step explanation:

Denote the events as follows:

<em>B</em> = bugs are present in a computer program.

<em>D</em> = a de-bugging program detects the bug.

The information provided is:

P(B) =0.01\\P(D|B)=0.99\\P(D|B^{c})=0.02

(i)

The probability that there is a bug in the program given that the de-bugging program has detected the bug is, P (B | D).

The Bayes' theorem states that the conditional probability of an event <em>E </em>given that another event <em>X</em> has already occurred is:

P(E|X)=\frac{P(X|E)P(E)}{P(X|E)P(E)+P(X|E^{c})P(E^{c})}

Use the Bayes' theorem to compute the value of P (B | D) as follows:

P(B|D)=\frac{P(D|B)P(B)}{P(D|B)P(B)+P(D|B^{c})P(B^{c})}=\frac{(0.99\times 0.01)}{(0.99\times 0.01)+(0.02\times (1-0.01))}=0.3333

Thus, the probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.

(ii)

The probability that a bug is actually present given that the de-bugging program claims that bug is present is:

P (B|D) = 0.3333

Now it is provided that two tests are performed on the program A.

Both the test are independent of each other.

The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is:

P (Bugs are actually present | Detects on both test) = P (B|D) × P (B|D)

                                                                                     =0.3333\times 0.3333\\=0.11108889\\\approx 0.1111

Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.

(iii)

Now it is provided that three tests are performed on the program A.

All the three tests are independent of each other.

The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is:

P (Bugs are actually present | Detects on all 3 test)

= P (B|D) × P (B|D) × P (B|D)

=0.3333\times 0.3333\times 0.3333\\=0.037025927037\\\approx 0.037

Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.

4 0
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Line a is parallel to line b, m∠1=2x +44, and m∠5x+38. Find the value of x.
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Answer:

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Step-by-step explanation:

You will first add all the students the mentioned which gives you 13 then since they ask what is the probability it the teacher would pick girls. You know there's 10 girls and 3 boys so it would be 10/13.                                  

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Goryan [66]

Answer: 37.5%


Explanation:


Percentage is defined as parts of an amount  in 100 parts of a basis. This must be stated mathamatically for better understanding and for calculations. It is the ratio:

             % = (parts / base) × 100.


In this case you need to express the amount of butter as a part in 100 parts of all purpose flour, which, using the above definition, is:


        % butter = (0.75 pounds of butter / 2 pound of flour) × 100

         % butter = 0.75 × 100 / 2 = 75 / 2 = 37.5


The answer is expressed as 37.5%.

4 0
3 years ago
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