Mathematics, 20.10.2020 02:01, paco38

Use the normal distribution of SAT Critical reading scores for which the mean is 510 and the standard deviation is 105. Assume the variable x is normally distributed (a) What percent of the SAT verbal scores are less than 6252
(D) 1000 SAT verbal scores are randomly selected, about how many would you expect to be greater than 550?
(a) Approximately % of the SAT verbal scores are less than 625.
(Round to two decimal places as needed.)

Answers: 2

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answered: Guest

check the picture below.

well, the angle at the vertex d is clearly not a right-angle, let's check for the one at the vertex f which does look like, however if it's indeed, that means ef is perpendicular to df, in which case their slopes will be negative reciprocal of each other.

$\bf e(\stackrel{x_1}{-2}~,~\stackrel{y_1}{2})\qquad f(\stackrel{x_2}{0}~,~\stackrel{y_2}{0}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{0-2}{)}\implies \implies \cfrac{-2}{0+2}\implies \cfrac{-2}{2}\implies -1 \\\\[-0.35em] ~\dotfill$

$\bf d(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-1})\qquad f(\stackrel{x_2}{0}~,~\stackrel{y_2}{0}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{)}{)}\implies \cfrac{0+1}{0+2}\implies \cfrac{1}{2} \\\\[-0.35em] ~\dotfill$

$\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope~of~ef}{-1\implies \cfrac{-1}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{1}{-1}}\qquad \stackrel{negative~reciprocal}{+\cfrac{1}{1}\implies 1}}$

so the negative reciprocal of the slope of ef is 1, however the slope of df is not 1, is 1/2, so nope, they're not perpendicular lines.

answered: Guest

well my understanding of this question is they want you to use the y=14.1x -2.3 and put the coordinates they gave you onto the line segment of y=14.1x -2.3

step-by-step explanation:

answered: Guest

x= 1

step-by-step explanation: convert 5\frac{11}{12}5

to improper fraction. use this rule: a \frac{b}{c}=\frac{ac+b}{c}a

ac+b

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{5\times 12+11}{12}3.25+0.5x+1

5×12+11

2 simplify 5\times 125×12 to 6060

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{60+11}{12}3.25+0.5x+1

60+11

3 simplify 60+1160+11 to 7171

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{71}{12}3.25+0.5x+1

4 simplify \frac{5}{6}x

x to \frac{5x}{6}

3.25+0.5x+1\frac{1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+1

5 convert 1\frac{1}{3}1

to improper fraction. use this rule: a \frac{b}{c}=\frac{ac+b}{c}a

ac+b

3.25+0.5x+\frac{1\times 3+1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

1×3+1

6 simplify 1\times 31×3 to 33

3.25+0.5x+\frac{3+1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

3+1

7 simplify 3+13+1 to 44

3.25+0.5x+\frac{4}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

8 simplify 3.25+0.5x+\frac{4}{3}+\frac{5x}{6}3.25+0.5x+

to \frac{13.75}{3}+\frac{4x}{3}

13.75

\frac{13.75}{3}+\frac{4x}{3}=\frac{71}{12}

13.75

9 simplify \frac{13.75}{3}

13.75

to 4.5833334.583333

4.583333+\frac{4x}{3}=\frac{71}{12}4.583333+

10 subtract 4.5833334.583333 from both sides

\frac{4x}{3}=\frac{71}{12}-4.583333

−4.583333

11 simplify \frac{71}{12}-4.583333

−4.583333 to \frac{4}{3}

\frac{4x}{3}=\frac{4}{3}

12 multiply both sides by 33

4x=\frac{4}{3}\times 34x=

×3

13 simplify \frac{4}{3}\times 3

×3 to \frac{12}{3}

4x=\frac{12}{3}4x=

14 simplify \frac{12}{3}

to 44

4x=44x=4

15 divide both sides by 44

x=1

answered: Guest

i will with my problems and all the other things that need to be on by the way

step-by-step explanation:

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Use the normal distribution of SAT Critical reading scores for which the mean is 510 and the standar...