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Butoxors [25]
3 years ago
12

You are making a new type of deodorant from two different mixtures the first mixture is 30% smell proof and the second mixture i

s 80% smell proof to make a safe make sure it needs to be 50% smell proof how much of each makes you need if you want 1000 L total
Mathematics
1 answer:
Readme [11.4K]3 years ago
3 0

Answer:

600 L of 30% proof and 400 L of 80% proof is needed

Step-by-step explanation:

Let the amount of 30% smell proof be x while that of 80% smell proof is y

Then;

x + y = 1000 •••••(i)

Also;

30% of x + 80% of y = 50% of 1000

0.3x + 0.8y = 500 •••••(ii)

From i, x = 1000 - y

Put this into ii

0.3(1000-y) + 0.8y = 500

300 - 0.3y + 0.8y = 500

0.5y = 500-300

0.5y = 200

y = 200/0.5

y = 400 L

But x = 1000 - y

x = 1000 - 400 = 600 L

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Answer:

c) P(270≤x≤280)=0.572

d) P(x=280)=0.091

Step-by-step explanation:

The population of bearings have a proportion p=0.90 of satisfactory thickness.

The shipments will be treated as random samples, of size n=500, taken out of the population of bearings.

As the sample size is big, we will model the amount of satisfactory bearings per shipment as a normally distributed variable (if the sample was small, a binomial distirbution would be more precise and appropiate).

The mean of this distribution will be:

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The standard deviation will be:

\sigma_s=\sqrt{np(1-p)}=\sqrt{500*0.90*0.10}=\sqrt{45}=6.7

We can calculate the probability that a shipment is acceptable (at least 440 bearings meet the specification) calculating the z-score for X=440 and then the probability of this z-score:

z=(x-\mu_s)/\sigma_s=(440-450)/6.7=-10/6.7=-1.49\\\\P(z>-1.49)=0.932

Now, we have to create a new sampling distribution for the shipments. The size is n=300 and p=0.932.

The mean of this sampling distribution is:

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The standard deviation will be:

\sigma=\sqrt{np(1-p)}=\sqrt{300*0.932*0.068}=\sqrt{19}=4.36

c) The probability that between 270 and 280 out of 300 shipments are acceptable can be calculated with the z-score and using the continuity factor, as this is modeled as a continuos variable:

P(270\leq x\leq280)=P(269.5

d) The probability that 280 out of 300 shipments are acceptable can be calculated using again the continuity factor correction:

P(X=280)=P(279.5

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