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3 years ago

If LN = 6x – 5, LM = x + 7, and MN = 3x + 20, find MN.

Mathematics

1 answer:

nikklg [1K]3 years ago

4 0

Answer:

LM = LN + MN [ assuming N is a point between L and M ]

$(x + 7) = (6x - 5) + (3x + 20) \\ x + 7 = 9x + 15 \\ 8x = - 8 \\ { \underline{x = - 1}}$

then find MN:

$MN = 3x + 20 \\ MN = 3( - 1) + 20 = - 3 + 20\\ { \boxed{ \boxed{MN = 17}}}$

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Taking a test i don’t know anything about please help

PilotLPTM [1.2K]

Answer:

See below for answers and explanations

Step-by-step explanation:

Problem 4:

The line of regression is $\hat y = a+bx$ where:

$a=\overline y-b \overline x$

$b=\frac{r*s_y}{s_x}$

We are given that $\overline y=75$ , $s_y=8$ , $\overline x=280$ , $s_x=30$ , and $r=0.60$ , therefore our slope, $b$ , is:

$b=\frac{r*s_y}{s_x}$

$b=\frac{(0.60)(8)}{30}$

$b=0.16$

Therefore, the slope of the regression line is 0.16, which can be used along with the values of $\overline y$ and $\overline x$ to find the constant $a$ :

$a=\overline y-b \overline x$

$a=75-(0.16)(280)$

$a=30.2$

This means our final regression line is $\hat y = 30.2 + 0.16x$

Problem 5:

The slope, $b=0.16$ , means that for every 1 point earned for a student's pre-exam total, their final exam score will increase by 0.16 points for each point they earned on the pre-exam.

Problem 6:

The y-intercept (or constant), $a=30.2$ , means that if a student's pre-exam total were 0, then they would expect to get a 30.2 on the final exam.

Problem 7:

$R^{2}=(0.60)^{2}=0.36=36\%$

Problem 8:

The fraction of variability (aka. coefficient of determination), $R^2$ , means that a certain proportion (or percentage) of the variance in the response variable can be explained by the explanatory variable. In context, this means that 36% of the variance in final exam scores can be explained by the pre-exam scores.