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QveST [7]
2 years ago
6

You need 200 mL of a 10% alcohol solution. On hand, you have a 25% alcohol mixture. How much of the 25%

Mathematics
1 answer:
Anastasy [175]2 years ago
4 0

so yull do math I'm guna do different number just plug these in for.your numbers 225/(375+x)=25

cross multiply land solve900-375 x=525 then cheak

225/(375+525) = .25

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16 - 2 = 14

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2 years ago
A researcher wishes to estimate the average blood alcohol concentration​ (BAC) for drivers involved in fatal accidents who are f
Mnenie [13.5K]

Answer:

A 90% confidence interval for the mean BAC in fatal crashes in which the driver had a positive BAC is [0.143, 0.177] .

Step-by-step explanation:

We are given that a researcher randomly selects records from 60 such drivers in 2009 and determines the sample mean BAC to be 0.16 g/dL with a standard deviation of 0.080 ​g/dL.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                               P.Q.  =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~   t_n_-_1

where, \bar X = sample mean BAC = 0.16 g/dL

            s = sample standard deviation = 0.080 ​g/dL

            n = sample of drivers = 60

            \mu = population mean BAC in fatal crashes

<em>Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation. </em>

So, a 90% confidence interval for the population mean, \mu is;

P(-1.672 < t_5_9 < 1.672) = 0.90  {As the critical value of t at 59 degrees of

                                              freedom are -1.672 & 1.672 with P = 5%}    P(-1.672 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 1.672) = 0.90

P( -1.672 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 1.672 \times {\frac{s}{\sqrt{n} } } ) = 0.90

P( \bar X-1.672 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+1.672 \times {\frac{s}{\sqrt{n} } } ) = 0.90

<u>90% confidence interval for</u> \mu = [ \bar X-1.672 \times {\frac{s}{\sqrt{n} } } , \bar X+1.672 \times {\frac{s}{\sqrt{n} } } ]

                                       = [ 0.16-1.672 \times {\frac{0.08}{\sqrt{60} } } , 0.16+1.672 \times {\frac{0.08}{\sqrt{60} } } ]

                                       = [0.143, 0.177]

Therefore, a 90% confidence interval for the mean BAC in fatal crashes in which the driver had a positive BAC is [0.143, 0.177] .

7 0
3 years ago
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