Simplify square root of 306

to arrive at 2000 liters of 60% saline solution the attendant has to mix a 90% and a 40% saline solution. How many liters of eac

CaHeK987 [17]

Answer:

800 liters of 90% saline solution and 1200 liters of 40% saline solution should be used.

Step-by-step explanation:

Given:

At 2000 liters of 60% saline solution the attendant has to mix a 90% and a 40% saline solution.

Now, to find the number of liters of saline solution each should be used.

Let the liters of 90% saline solution mix be $x.$

And let the liters of 40% saline solution mix be $y.$

So, the total number of liters:

$x+y=2000.$

$y=2000-x\ \ \ ....(1)$

Now, the total percentage of saline solution:

$90\%\ of\ x+40\%\ of\ y=60\%\ of\ 2000$

$\frac{90}{100}\times x+\frac{40}{100}\times y=\frac{60}{100}\times 2000$

$0.9x+0.4y=1200$

Substituting the value of $y$ from equation (1) we get:

$0.9x+0.4(2000-x)=1200$

$0.9x+800-0.4x=1200$

$0.5x+800=1200$

Subtracting both sides by 800 we get:

$0.5x=400$

Dividing both sides by 0.5 we get:

$x=800.$

The liters of 90% saline solution mix = 800.

Now, substituting the value of $x$ in equation (1) to get the liters of 40% saline solution:

$y=2000-x$

$y=2000-800$

$y=1200.$

Thus, the liters of 40% saline solution = 1200.

Therefore, 800 liters of 90% saline solution and 1200 liters of 40% saline solution should be used.

8 0

4 years ago

45% of the 647 students at Royal Oak High attended the football game. How many students did not attend?

Lostsunrise [7]

First multiply
.45*647= 291.15
291.15 is the number of people that attended
subtract 291.15 from 647 and you get 355.85 wait.. but that's not possible so just round it so the answer to your question is 355 people did not attened

3 0

3 years ago

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In early 2012, the Pew Internet and American Life Project asked a random sample of U.S. adults, "Do you ever ... use Twitter or

8090 [49]

Answer:

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

And the confidence interval is given by:

(0.123, 0.177)

And for this case the interval contains the value 0.16, so then we can conclude at 5% of significance that the true proportion is not different from 0.16

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

And the confidence interval is given by:

(0.123, 0.177)

And for this case the interval contains the value 0.16, so then we can conclude at 5% of significance that the true proportion is not different from 0.16

3 0

3 years ago

5/8=n-2/10 9/x=10/x-7 p+9/p=3/10

lesya692 [45]

Download photo math for equations like this, it will also give you it step by step :)

4 0

3 years ago

Find the value of each variable. x = and y =

alex41 [277]

Answer:

x = 16

y = 16√3

Step-by-step explanation:

sin(30 degrees) = 1/2

x = 32/2 = 16

cos(30 degrees) = √3/2

y = 16√3

4 0

3 years ago

Read 2 more answers