If x ml pure alcohol is added to 400 ml of a 15% solution and after adding its strength becomes 32%, then the value of x is

1 answer:

6 0

We calculate as follows:

15 mL pure ethanol / 100 mL solution (400 mL) = 60 mL pure ethanol

32% = 32 mL pure ethanol/100 mL solution = 60 mL pure ethanol + x mL pure ethanol / (400 + x)

x = 100 mL pure ethanol was added to the solution ------> OPTION D

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PIT_PIT [208]

Answer:

Step-by-step explanation:

A. S=strawberry, C=chocolate

S+C=14 equation 1

2s+3c=30 equation 2

B. -2(s+c)=-2(14) multiply equation 1 by -2 to use elimination method

-2s+-2c=-28 modified Equation 1

2s+ 3c=30. Equation 2

C=2 add above two equations.

solve for s

s+c=14

s+2=14

s=12

Used elimination method since it was easier to add the two equations.

check answer by inputting s,c values in either equation.

2s+3c=30

2(12)+3(2)=30

24+6=30

30=30

7 0

3 years ago

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LuckyWell [14K]

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3 years ago

Final numeric grades in a certain course are normally distributed with a mean of 72.3 and a standard deviation of 6.4. The profe

Dimas [21]

Answer:

The minumum numeric grade you have to earn to obtain an A is 81.29.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 72.3, \sigma = 6.4$

The professor curves the grades so that the top 8% of students will receive an A. What is the minumum numeric grade you have to earn to obtain an A?

The minimum numeric value is the value of X when Z has a pvalue of 1-0.08 = 0.92. So it is X when Z = 1.405.

$Z = \frac{X - \mu}{\sigma}$

$1.405 = \frac{X - 72.3}{6.4}$