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1 year ago

The distance to the nearest exit doir is no more than 150 feet. Use d to represent the distance in feet to the nearest exit door

Mathematics

1 answer:

il63 [147K]1 year ago

5 0

Answer: $d\leq150$

Step-by-step explanation:

1. You have the following information given in the problem:

- The distance to the nearest exit door is no more than 150 feet.

- $d$ represents the distance to the nearest exit door, in feet.

2. Therefore, keeping the information above on mind, you know that the distance to the nearest exit door ( $d$ ) must be less than or equal to 150 feet, then, you can express this as following:

$d\leq150$

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Step-by-step explanation:

Circular orbit height ( from the center of the earth ) = 100 mile

angle between satellite and earth = 50°

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distance between surface of earth and satellite

x = cos ∅

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Which term of the AP. 8 , -4 , -16 , -28 ,......... is -880 ?

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Step-by-step explanation:

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Which set of ordered pairs represents a function?

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A random sample of 35 undergraduate students who completed two years of college were asked to take a basic mathematics test. The

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

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

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.

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1 year ago