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3 years ago

Any solution to this?

Mathematics

2 answers:

Katyanochek1 [597]3 years ago

7 0

Answer: Turn on the lights

sukhopar [10]3 years ago

7 0

Answer:

I can't see the picture.

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So is the correct answer A, B, C, or D

yuradex [85]

Answer:

What do you mean I need a picture or text please to understand what you are talking about so i can help you

Step-by-step explanation:

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3 years ago

Read 2 more answers

P(3,-1) that has a slope of -2 on a graph

KonstantinChe [14]

Graph the points positive three and negativeone then we go down to over one

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4 years ago

Jens purchased a new ski jacked for $91. The Jacket was marked down 35%. What was the original price of the jacket?

White raven [17]

Answer:

1.00-.35=.65

.65×140.00=$91.00

3 0

3 years ago

Consider this scenario: The population of a city increased steadily over a ten-year span. The following ordered pairs show the p

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:

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

6 0

3 years ago

1/8 x 8<br>3/8 x 8<br>1/8 x 8<br>3/8 x 8

aivan3 [116]

Step-by-step explanation:

1/8 x 8 = 1

3/8 x 8 = 3

1/8 x 8 = 1

3/8 x 8 = 3

8 0

3 years ago