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3 years ago

jessica had some pebbles her mother gave her 36 more pebbles now she has 77 pebbles how many pebbles did jessica start with?

Mathematics

2 answers:

uysha [10]3 years ago

5 0

Answer: 41

Step-by-step explanation:

all you have to do is subtract 36 from 77

OlgaM077 [116]3 years ago

4 0

Answer:

77-36= 41

41+36=77

She had 41 pebbles to start with

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a monument that is 169.4 ft tall is built on a site that is 67.3 Ft below sea level how many feet above sea level is the top of

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Answer:

102.1 ft

Explanation:

We can represent the situation as follows:

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So, the top of the monument is 102.1 ft above sea level.

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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:

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

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3 years ago