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Jet001 [13]
3 years ago
12

A pharmacist needs 100 gallons of 50% alcohol solution. She has available a 30% solution and an 80% solution. How much of each s

hould she use?
Mathematics
2 answers:
Salsk061 [2.6K]3 years ago
8 0

Answer:

25

Step-by-step explanation:


Aneli [31]3 years ago
7 0
<h3>Answers: </h3><h3>Use <u>60 gallons</u> of the 30% solution</h3><h3>Mix with <u>40 gallons</u> of the 80% solution</h3>

===============================

Work Shown:

x = amount of the 30% solution (in gallons)

y = amount of the 80% solution (in gallons)

The two amounts must add to 100 as this is the total we want, so x+y = 100 which becomes y = 100-x after subtracting x from both sides

The expression 0.30*x represents the amount of pure alcohol from the first batch, while 0.80*y represents the amount of pure alcohol from the second batch. In total, we have 0.30*x+0.80*y gallons of pure alcohol. We want 50 gallons of pure alcohol (50% of 100 gallons is 50 gallons), therefore we end up with this equation: 0.30*x+0.80*y = 50

Let's use substitution to isolate the variables.

0.30*x+0.80*y = 50

0.30*x+0.80*(100-x) = 50 ... replace y with 100-x

0.30*x+0.80*(100)+0.80*(-x) = 50 ... distribute

0.30*x+80-0.80x = 50

-0.50*x+80 = 50

-0.50*x+80-80 = 50-80 ... subtract 80 from both sides

-0.50*x = -30

x = -30/(-0.50) .... divide both sides by -0.50

x = 60

If x = 60, then y is...

y = 100-x

y = 100-60

y = 40

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Given a normal distribution with mu equals 100μ=100 and sigma equals 10 commaσ=10, complete parts​ (a) through​ (d). loading...
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I believe the parts are:

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C.What is the probability that Upper X less than 85X<85 and Upper X greater than 125X>125?

D. 95​% of the values are between what two​ X-values (symmetrically distributed around the​ mean)?

 

<u>Solution:</u>

We use the equation for z score:

z = (X – μ) / σ

Then use the standard normal probability tables to locate for the value of P at indicated z score value.

 

A. Z = (95 – 100) / 10 = - 0.5

Using the tables, the probability at z = -0.5 using right tailed test is:

P = 0.6915

 

 

B. Z = (75 – 100) / 10 = - 2.5

Using the tables, the probability at z = -2.5 using left tailed test is:

P = 0.0062

 

 

C. Z = (85 – 100) / 10 = - 1.5

Using the tables, the probability at z = -2.5 using left tailed test is:

P = 0.0668

 

Z = (125 – 100) / 10 = 2.5

Using the tables, the probability at z = 2.5 using right tailed test is:

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So the probability that 85<X and X>125 is:

P(total) = 0.0668 + 0.0062

P(total) = 0.073

 

 

D. P(left) = 0.025, Z = -1.96

P(right) = 0.975, Z = 1.96

 

The X’s are calculated using the formula:

X = σz + μ

 

At Z = -1.96

X = 10 (-1.96) + 100 = 80.4

 

At Z = 1.96

X = 10 (1.96) + 100 = 119.6

 

<span>So 95% of the values are between 80.4 and 119.6 (80.4< X <119.6).</span>

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