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Inga [223]
2 years ago
12

Mahalia is making cloth napkins. She bought fabric that is 12 inches wide and 6 yards long. How much napkins can Mahalia make if

each napkin is 1 foot by 1 foot?
Mathematics
1 answer:
In-s [12.5K]2 years ago
8 0
She can make 72
Find the area of the fabric 12x6=72
Area of napkins 1x1 =1
divide fabric by napkins
72/1=72 napkins can be made
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In which quadrants is the ordinate positive?
Anni [7]

Answer:

I and II

Step-by-step explanation:

The ordinate is the y element in an ordered pair in a Cartesian coordinate system. The quadrants in which the ordinates are positive should be at quadrants I and II.

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Convert to the nearest tenth. <br> gal. = 319 qt.
lisov135 [29]
The answer is 79.75 converted to 80.0
Hope this helps!!
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I need help with finding the equation for #20 and # 22 for my brother we get the answer but it’s hard for us to write out how to
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Step-by-step explanation:

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2 years ago
A random sample of 35 undergraduate students who completed two years of college were asked to take a basic mathematics test. The
lisov135 [29]

Answer:

(75.1 -72.1) -1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= -1.96

(75.1 -72.1) +1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= 7.96

And the confidence interval would be given by:

-1.96 \leq \mu_1 -\mu_2 \leq 7.96

And since the confidence interval contains the value 0 we have enough evidence to conclude that we don't have significant differences between the two means at 10% of significance.

Step-by-step explanation:

For this case we have the following info given :

\bar X_1= 75.1 represent the sample mean for the scores of the undergraduate students

s_1 = 12.8 represent the standard deviation for the undergraduate students

n_1 =35 the sample size for the undergraduate

\bar X_2= 72.1 represent the sample mean for the scores of the high school students

s_2 = 14.6 represent the standard deviation for the high school students

n_2 =50 the sample size for the high school

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

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

The degrees of freedom are given by:

df=n_1 +n_2 -2= 35+50-2=83

The confidence level is 90% and the significance level is \alpha=0.1 and \alpha/2 =0.05 then the critical value would be:

t_{\alpha/2}= 1.99

And replacing the info we got:

(75.1 -72.1) -1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= -1.96

(75.1 -72.1) +1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= 7.96

And the confidence interval would be given by:

-1.96 \leq \mu_1 -\mu_2 \leq 7.96

And since the confidence interval contains the value 0 we have enough evidence to conclude that we don't have significant differences between the two means at 10% of significance.

8 0
1 year ago
Is 3/6 grater then 1/8
sashaice [31]
True

3/6 = 0,5

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