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3 years ago

17. The total bank loan for Sarah's new car is $15,265. The bank

Mathematics

2 answers:

Andru [333]3 years ago

5 0

Answer:

Hi there!

Your formula is:

15265- 295.8x

Step-by-step explanation:

The 295.8x represents the total months & cost per month. Since paying off a bank loan makes the loan decrease ,you subtract

Hope this helps

mezya [45]3 years ago

5 0

Answer: F(x)= 295.80x-15,265

Reasoning: If you are withdrawing $295.80 per month you would need to multiply the amount taken by the amount of months and then subtract that by 15,265

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Two number cubes whose sides are numbered 1 through 6 are rolled on a table. The two numbers showing are added. If you repeat th

ankoles [38]

Answer:

50 times.

Step-by-step explanation:

When the 2 number cubes are rolled once:

The possible outcomes for a total of 7 are:

1 and 6

6 and 1

2 and 5

5 and 2

3 and 4

4 and 3

= 6 outcomes.

The total of all possible outcomes = 6 * 6 = 36.

So Prob( total of 7 in one throw of teh 2 cubes) = 6/36 = 1/6.

For 300 throws you would expect the total of 7 to occur

1/6 * 300 = 50.

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Alina [70]

The ratio would be 1:4, I would start with dividing each number by 2 then working from there. With diving by two you would end up with 3:12 which you then can divide by 3 since 12 is divisible by 3. This would get you 1:4.

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3 years ago

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Snezhnost [94]

Answer:

Prime.

Step-by-step explanation:

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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

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