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Anton [14]
3 years ago
6

A map has the scale 0.75 of an inch equals 220 miles. The distance from Springfield to Glenview is 6 inches. How many actual mil

es separate Springfield from Glenview?
Mathematics
1 answer:
DENIUS [597]3 years ago
8 0

Answer:

<em>1,760 actual miles separate Springfield from Glenview.</em>

Step-by-step explanation:

<u>Scaling</u>

Objects can be represented in a reduced or augmented size by using scaling which is basically multiplying or dividing real dimensions by a constant factor. We use scaling when representing geographic locations on a map.

The map has the scale:

0.75 inch => 220 miles

This gives us the scaling factor as 220/0.75. There is no need to divide.

We are given the distance from Springfield to Glenview is 6 inches on the map. The actual distance is

6*220/0.75 = 1,760

1,760 actual miles separate Springfield from Glenview.

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Pavel [41]

<em>(</em><em>t</em><em>he answer is on the picture. is this correct?</em><em>)</em> ^^'

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2 years ago
Several programs attempt to address the shortage of qualified teachers by placing uncertified instructors in schools with acute
ss7ja [257]

Answer:

We conclude that the mean scores with uncertified teachers is higher or equal as compared to certified teachers.

Step-by-step explanation:

We are given that reading scores of the students of certified teachers averaged 35.62 points with standard deviation 9.31. The scores of students instructed by uncertified teachers had mean 32.48 points with standard deviation 9.43 points on the same test.

There were 44 students in each group.

Let \mu_1 = <em><u>mean scores with uncertified teachers.</u></em>

\mu_2 = <em><u>mean scores with certified teachers.</u></em>

So, Null Hypothesis, H_0 : \mu_1\geq \mu_2     {means that the mean scores with uncertified teachers is higher or equal as compared to certified teachers}

Alternate Hypothesis, H_A : \mu_1     {means that the mean scores with uncertified teachers is lower as compared to certified teachers}

The test statistics that would be used here <u>Two-sample t test statistics</u> as we don't know about the population standard deviations;

                          T.S. =  \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }  ~ t__n__1-_n__2-2

where, \bar X_1 = sample mean scores of students instructed by uncertified teachers = 32.48 points

\bar X_2 = sample mean scores of students instructed by certified teachers = 35.62 points

s_1 = sample standard deviation of scores of students instructed by uncertified teachers = 9.43 points

s_2 = sample standard deviation of scores of students instructed by certified teachers = 9.31 points

n_1 = sample of students under uncertified teachers = 44

n_2 = sample of students under certified teachers = 44

Also,  s_p=\sqrt{\frac{(n_1-1)s_1^{2} +(n_2-1)s_2^{2} }{n_1+n_2-2} } = \sqrt{\frac{(44-1)\times 9.43^{2} +(44-1)\times 9.31^{2} }{44+44-2} } = 9.37

So, <u><em>the test statistics</em></u>  =  \frac{(32.48-35.62)-(0)}{9.37 \times \sqrt{\frac{1}{44} +\frac{1}{44} } }  ~ t_8_6

                                     =  -1.572

The value of t test statistics is -1.572.

Since, in the question we are not given the level of significance so we assume it to be 5%. <u>Now, at 5% significance level the t table gives critical values of -1.665 at 86 degree of freedom for left-tailed test.</u>

Since our test statistic is more than the critical values of t as -1.572 > -1.665, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis</u>.

Therefore, we conclude that the mean scores with uncertified teachers is higher or equal as compared to certified teachers.

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Step-by-step explanation:

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