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sammy [17]
3 years ago
9

Two-thirds of students in a class are girls. If one-half of the girls wear glasses, what fractional part of students are girls w

ho wear glasses?
(A) 1/8
(B) 1/6
(C) 1/5
(D) 1/4
(E) 1/3
Mathematics
1 answer:
Fantom [35]3 years ago
5 0

Let's assume the number of students in the class be x.

Given that, Two-thirds of students in a class are girls.

So, number of girls =\frac{2}{3} x

Now ne-half of the girls wear glasses. So, one-half of \frac{2}{3} x is

= \frac{1}{2} *\frac{2}{3} x

=\frac{1}{3} x Cancel out 2 from both numerator and denominator.

So, one third of the students are girls who wear glasses.

So, the correct choice is E.

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vova2212 [387]

Answer:

Regression function: y=1986.406+0.0059x

The function predicts that population will reach 14,000 in year 2068.

Step-by-step explanation:

We have to determine a function y=b_0+b_1x_1 by applying linear regression. The data we have is 5 pair of points which relates population to year.

According to the simple regression model (one independent variable), if we minimize the error between the model (the linear function) and the points given, the parameters are:

b_0=\bar{y}+b_1\bar{x}\\\\b_1=\frac{\sum\limits^5_{i=1} {(x_i-\bar x)(y_i-\bar y)}}{\sum\limits^5_{i=1} {(x_i-\bar x)^2}}

We start calculating the average of x and y

\bar x=\frac{2500+2650+3000+3500+4200}{5}=\frac{15850}{5}=3170\\\\ \bar y=\frac{2001+2002+2004+2007+2011}{5}=\frac{10025}{5}=2005

The sample covariance can be calculated as

\sum\limits^5_{i=1} {(x_i-\bar x)(y_i-\bar y)}=(2500-3170)(2001-2005)+(2650-3170)(2002-2005)+(3000-3170)(2004-2005)+(3500-3170)(2007-2005)+(4200-3170)(2011-2005)\\\\\sum\limits^5_{i=1} {(x_i-\bar x)(y_i-\bar y)}=2680+1560+170+660+6180\\\\ \sum\limits^5_{i=1} {(x_i-\bar x)(y_i-\bar y)}=11250

The variance of x can be calculated as

\sum\limits^5_{i=1} {(x_i-\bar x)^2}=(2500-3170)^2+(2650-3170)^2+(3000-3170)^2+(3500-3170)^2+(4200-3170)^2\\\\\sum\limits^5_{i=1} {(x_i-\bar x)^2}=448900+270400+28900+108900+1060900\\\\\sum\limits^5_{i=1} {(x_i-\bar x)^2}=1918000

Now we can calculate the parameters of the regression model

b_1=\frac{\sum\limits^5_{i=1} {(x_i-\bar x)(y_i-\bar y)}}{\sum\limits^5_{i=1} {(x_i-\bar x)^2}}=\frac{11250}{1918000}=0.005865485  \\\\ b_0=\bar{y}+b_1\bar{x}=2005-0.005865485*3170=1986.406413

The function then become:

y=1986.406+0.0059x

With this linear equation we can predict when the population will reach 14,000:

y=1986.406+0.0059(14,000)=1986.406+82.117=2068.523

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