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11 months ago

Can someone help me please!!

Mathematics

1 answer:

ratelena [41]11 months ago

8 0

The coordinates of E are (5.75, 0).

<h3>What is midpoint?</h3>

The midpoint is the middle point of a line segment. It is equidistant from both endpoints, and it is the centroid both of the segment and of the endpoints.

Given that, C is the midpoint of AB, D is the midpoint of CB and E is the midpoint of DB.

Formula for coordinates of midpoint,

(x, y) = [(x1+x2)/2, (y1+y2)/2]

Coordinates of C;

(x, y) = (-10+8)/2, (14-2)/2

(x, y) = (-1, 6)

Coordinates of D;

(x, y) = (-1+8)/2, (6-2)/2

(x, y) = (3.5, 2)

Coordinates of E;

(x, y) = (3.5+8)/2, (2-2)/2

(x, y) = (5.75, 0)

Hence, The coordinates of E are (5.75, 0).

For more references on midpoints, click;

brainly.com/question/28224145

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Refer to the accompanying data set and construct a % confidence interval estimate of the mean pulse rate of adult females; the

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Complete Question

The complete question is shown on the first uploaded image

Answer:

The 95% confidence interval of mean pulse rate for adult female is $71.98 < \mu < 79.88$

The 95% confidence interval of mean pulse rate for adult male is $62.89 < \mu < 70.57$

The correct option is C

Step-by-step explanation:

Generally the sample mean for Male pulse rate is mathematically represented as

$\= x_1 = \frac{\sum x_i }{n}$

= > $\= x_1 = \frac{81 + 74 + \cdots + 59 }{40 }$

= > $\= x_1 = 66.73$

Generally the standard deviation for male pulse rate is mathematically represented as

$\sigma_1 = \sqrt{\frac{\sum (x - \= x)^2 }{n-1 } }$

=> $\sigma_1 = \sqrt{\frac{\sum (81 - 66.73 )^2 + (74 - 66.73 )^2+ \cdot + (59 - 66.73 )^2 }{40-1 } }$

=> $\sigma_1 = 12.24$

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma_1 }{\sqrt{n} }$

=> $E = 1.96 * \frac{12.4 }{\sqrt{40} }$

=> $E =3.84$

Generally 95% confidence interval is mathematically represented as

$\= x_1 -E < \mu < \=x_1 +E$

=> $66.73 -3.84 < \mu < 66.73 +3.84$

=> $62.89 < \mu < 70.57$

Generally the sample mean for Female pulse rate is mathematically represented as

$\= x_2 = \frac{\sum x_i }{n}$

= > $\= x_2 = \frac{81 + 94 + \cdots + 73 }{40 }$

= > $\= x_2 = 75.93$

Generally the standard deviation for Female pulse rate is mathematically represented as

$\sigma_2 = \sqrt{\frac{\sum (x - \= x)^2 }{n-1 } }$

=> $\sigma_2 = \sqrt{\frac{\sum (81 - 66.73 )^2 + (94 - 66.73 )^2+ \cdot + (73 - 66.73 )^2 }{40 } }$

=> $\sigma_2 = 12.73$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma_2 }{\sqrt{n} }$

=> $E = 1.96 * \frac{12.73 }{\sqrt{40} }$

=> $E =3.95$

Generally 95% confidence interval is mathematically represented as

$\= x_2 -E < \mu < \=x_2 +E$

=> $75.93 -3.95 < \mu < 75.93 + 3.95$

=> $71.98 < \mu < 79.88$

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=============

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