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goldenfox [79]
3 years ago
9

One container is filled with a mixture that is 30%acid a second container is filled with a mixture that is 50%acid the second co

ntainet is 50%larger than the first and that twocontainers are are emptied into a third what percent of acid is the third container
Mathematics
1 answer:
julsineya [31]3 years ago
4 0

Answer:

The amount of acid in third container is =42%

Step-by-step explanation:

Given , one container is filled with a mixture that is 30% acid a second container filled with a mixture that is 50% acid and the second container 50% larger than the first .

Let, the volume of first container is = x

Then , the volume of second container = (x+ x of 50%)

                                                                 = x + 0.5 x

                                                                 = 1.5 x

Therefore the amount of acid in first container =x \times \frac{30}{100} = 0.3 x

The amount of acid in second container =1.5x \times \frac{50}{100}  = 0.75x

Total amount of acid= 0.3x + 0.75x = 1.05 x

Total amount  of solution = x+1.5x = 2.5x

The amount of acid in third container is = \frac{1.05x}{2.5x} \times 100% %

                                                                     = 42%

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Find the surface area of the part of the sphere x2+y2+z2=81 that lies above the cone z=x2+y2−−−−−−√
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4 0
3 years ago
A health psychologist knew that corporate executives in general have an average score of 2.55 with a standard deviation of 0.5 o
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Answer:

Kindly check explanation

Step-by-step explanation:

Given the following :

Population mean (μ) = 2.55

Population standard deviation (σ) = 0.5

Sample size (n) = 30

Sample mean (x) = 2.76

α = 0.05

STEP 1:

Stress score in general executive (s1)

Stress score in exercising executive (s2)

Null : s1 = s2

Alternative : s1 < s2

STEP 2:

Shape of distribution = normal

Population mean (μ) = 2.55

Population standard deviation (σ) = 0.5

Sample size (n) = 30

Sample mean (x) = 2.76

α = 0.05

Decision rule :

α = 0.05 which corresponds to a t score (t0) ;

df = n - 1 = 30 - 1= 30 at 0.05 = 1.699

If :

(Test statistic (t) > t0) ; reject the Null

(right tailed test)

Test statistic (t) :

(x - μ) / (σ/√n)

(2.76 - 2.55) / (0.5/√30)

0.21 / 0.0913

= 2.30

t > t0

2.30 > 1.699

t is more extreme than t0

Hence, reject the null at α = 0.05

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Which of the following graphs would represent the solution set for y &gt; 4/5 x – 1?
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It's this graph. You didn't give me any options in the question.

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