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alexira [117]
3 years ago
8

Halle had 15 pounds of sand to give to 6 of her friends for an art project. How many pounds of sand did each friend receive if H

alle divided it evenly? Leave answer as a reduced mixed number with a space between the whole number and the fractional part.
Mathematics
1 answer:
zubka84 [21]3 years ago
7 0
Each friend received 2 1/2 pounds of sand
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Christopher and Edward were painting wagon wheels for the school play. The pictures show how much of the wheels they painted. Ho
Marina CMI [18]

Using fractions, it is found that \frac{43}{45} of a wheel still needs to be painted.

  • On the first wheel, 4 of the 9 equal parts have been painted, that is, \frac{4}{9} of the wheel has been painted, hence \frac{5}{9} has not been painted.
  • On the second wheel, 6 of the 10 equal parts have been painted, that is, \frac{6}{10} = \frac{3}{5} of the wheel has been painted, hence \frac{2}{5} has not been painted.

The total fraction that still has to be painted is:

\frac{5}{9} + \frac{2}{5} = \frac{25 + 18}{45} = \frac{43}{45}

\frac{43}{45} of a wheel still needs to be painted.

To learn more about fractions, you can take a look at brainly.com/question/14387610

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3 years ago
Consider the accompanying data on breaking load (kg/25mm width) for various fabrics in both an unabraded condition and an abrade
WITCHER [35]

Answer:

The critical value for a one-tailed test with 7 degrees of freedom and level of significance α=0.01 is tc=2.998.

The test statistic (t=1.729) is below the critical value and falls within the acceptance region, so it is failed to reject the null hypothesis.

At a significance level of 0.01, there is not enough evidence to support the claim that the true difference is significantly bigger than 0.

Step-by-step explanation:

We have a matched-pair t-test for the difference.

We have the null and alternative hypothesis written as:

H_0: \mu_d=0\\\\H_a: \mu_d>0

We have n=8 pairs of data. We calculate the difference as:

d_1=U_1-A_1=36.4-28.5=7.9

Then, with this procedure we get the sample for d:

d=[7.9,\, 35,\, 5.5,\, 4.2,\, 6.7,\, -3.7,\, -0.9,\, 3.3]

The sample mean and standard deviation are:

M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{8}(7.9+35+5.5+. . .+3.3)\\\\\\M=\dfrac{58}{8}\\\\\\M=7.25\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{7}((7.9-7.25)^2+(35-7.25)^2+(5.5-7.25)^2+. . . +(3.3-7.25)^2)}\\\\\\s=\sqrt{\dfrac{985.08}{7}}\\\\\\s=\sqrt{140.73}=11.86\\\\\\

Now, we can perform the one-tailed hypothesis test.

The significance level is 0.01.

The sample has a size n=8.

The sample mean is M=7.25.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=11.86.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{11.86}{\sqrt{8}}=4.1931

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{7.25-0}{4.1931}=\dfrac{7.25}{4.1931}=1.729

The degrees of freedom for this sample size are:

df=n-1=8-1=7

This test is a right-tailed test, with 7 degrees of freedom and t=1.729, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=P(t>1.729)=0.0637

As the P-value (0.0637) is bigger than the significance level (0.01), the effect is not significant.

The null hypothesis failed to be rejected.

At a significance level of 0.01, there is not enough evidence to support the claim that the true difference is significantly bigger than 0.

If we use the critical value approach, the critical value for a one-tailed test with 7 degrees of freedom and level of significance α=0.01 is tc=2.998. The test statistic is below the critical value and falls within the acceptance region, so it is failed to reject the null hypothesis.

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They won 77 games and lost 20
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Answer:2

Step-by-step explanation:

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