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2 years ago

What would the concentration be if we mixed 200g of 10% sugar solution and 300g of 20% sugar solution?

2 answers:

7 0

The new concentration is obtained by
$\frac{200 \times 10 + 300 \times 20}{200 + 300} = \frac{2000 + 6000}{500} = \frac{7000}{500} = 14\%$

stepan [7]2 years ago

3 0

200 g + 300 g = 500 g
200 * 0.1 + 300 * 0.2 = 500 * x
20 + 60 = 500 * x
80 = 500 * x
x = 80 : 500
x = 0.16
Answer: The concentration would be 16 % ( it must be closer to 20% than to 10% ).

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To make purple paint, Mya mixed 2/5 cup of red paint with 5/8 cup of blue paint. She wants to make a larger batch of the purple

Hitman42 [59]

She should use 5/8 cups i think

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2 years ago

If x is the midpoint of ZW, then ZX=XW

sasho [114]

Answer:

True.

Step-by-step explanation:

By definition of a mid point, it is located in the middle of two connected points. This means that the split is equal on both sides.

IF X is the midpoint of ZW, then ZX = XW.

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1 year ago

An investor believes that investing in domestic and international stocks will give a difference in the mean rate of return. They

arlik [135]

Answer:

So on this case the 90% confidence interval would be given by $-4.137 \leq \mu_1 -\mu_2 \leq 2.087$

For this case since the confidence interval for the difference of means contains the 0 we can conclude that we don't have significant differences at 10% of significance between the two means analyzed.

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".

$\bar X_1 =2.0233$ represent the sample mean 1

$\bar X_2 =3.048$ represent the sample mean 2

n1=15 represent the sample 1 size

n2=15 represent the sample 2 size

$s_1 =4.893387$ population sample deviation for sample 1

$s_2 =5.12399$ population sample deviation for sample 2

$\mu_1 -\mu_2$ parameter of interest at 0.1 of significance so the confidence would be 0.9 or 90%

We want to test:

H0: $\mu_1 = \mu_2$

H1: $\mu_1 \neq \mu_2$

And we can do this using the confidence interval for the difference of means.

Solution to the problem

The confidence interval for the difference of means is given by the following formula:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}$ (1)

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =2.0233-3.048=-1.0247$

The degrees of freedom are given by:

$df = n_1 +n_2 -2 = 15+15-2=28$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,28)".And we see that $t_{\alpha/2}=\pm 1.701$