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Inessa05 [86]
2 years ago
13

What weight of dry substance is in 150g of a 3% substance solution? What weight of an 8% solution can we have with the same weig

ht of dry substance?
Mathematics
1 answer:
Leto [7]2 years ago
4 0
<span>4.5 g 56.25 g Since the only type of measurement mentioned in this question is weight or mass, I'll assume that the percentage concentration is % m/m (mass/mass). For that type of concentration measurement, simply multiple the percentage by the total mass to get the mass of the desired substance. So 150 g * 3% = 150 g * 0.03 = 4.5g For the amount of 8% solution with the same amount of dry substance, there's 2 ways of calculating the mass of solution. First, use the ratio of percentages, multiplied by the mass of the original solution to get the desired amount of new solution: 3/8 * 150 g = 56.35 g Or calculate it from scratch, like 4.5/X = 8/100 450/X = 8 450 = 8X 56.25 = X In both cases, the result is that you desire 56.25 grams of 8% solution.</span>
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Four boxes are shown in order from the largest volume to the smallest volume. For each box, all of the dimensions are 3.5cm less
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LESSON 1 SESSION 1
denpristay [2]

Answer:

  • <em>Between which two tens does it fall?</em><em> </em><u>Between 25 and 26 tens</u>

<em><u /></em>

  • <em>Between which two hundreds does it fall?</em> <u>Between 2 and 3 hundreds</u>

Explanation:

The place-value chart is:

Hundreds         Tens      Ones

       2                   5             3

<em><u /></em>

<em><u>a)  Between which two tens does it fall? </u></em>

Using the place values you can write 253 = 25 × 10 + 3, i.e. 25 tens and 3 ones.

From that you can write:

  • 250 < 253 < 260
  • 250 = 25 × 10 = 25 tens
  • 260 = 26 × 10 = 26 tens

Then, you conclude that 253 is between 25 and 26 tens.

<u><em>b) Between which two hundreds does it fall?</em></u>

Using the same reasoning:

  • 253 = 2 × 100 + 5 × 10 + 3 = 253

  • 200 < 253 < 300
  • 200 = 2 hundreds
  • 300 = 3 hundreds

Conclusion: 253 is between 2 hundreds and 3 hundreds.

3 0
2 years ago
Two samples each of size 20 are taken from independent populations assumed to be normally distributed with equal variances. The
Harlamova29_29 [7]

Answer:

t=\frac{(43.5 -40.1)-(0)}{3.678\sqrt{\frac{1}{20}+\frac{1}{20}}}=2.923

The degrees of freedom are

df=20+20-2=38

And the p value is given by:

p_v =2*P(t_{38}>2.923) =0.0058

Since the p value for this cae is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true means for this case are significantly different

Step-by-step explanation:

When we have two independent samples from two normal distributions with equal variances we are assuming that  

\sigma^2_1 =\sigma^2_2 =\sigma^2

And the statistic is given by this formula:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

Where t follows a t distribution with n_1+n_2 -2 degrees of freedom and the pooled variance S^2_p is given by this formula:

S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}

The system of hypothesis on this case are:

Null hypothesis: \mu_1 = \mu_2

Alternative hypothesis: \mu_1 \neq \mu_2

We have the following data given:

n_1 =20 represent the sample size for group 1

n_2 =20 represent the sample size for group 2

\bar X_1 =43.5 represent the sample mean for the group 1

\bar X_2 =40.1 represent the sample mean for the group 2

s_1=4.1 represent the sample standard deviation for group 1

s_2=3.2 represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

\S^2_p =\frac{(20-1)(4.1)^2 +(20 -1)(3.2)^2}{20 +20 -2}=13.525

And the deviation would be just the square root of the variance:

S_p=3.678

The statistic is givne by:

t=\frac{(43.5 -40.1)-(0)}{3.678\sqrt{\frac{1}{20}+\frac{1}{20}}}=2.923

The degrees of freedom are

df=20+20-2=38

And the p value is given by:

p_v =2*P(t_{38}>2.923) =0.0058

Since the p value for this cae is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true means for this case are significantly different

6 0
2 years ago
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