I NEED HELPPPP!!!!!!! PLEASEEEEE!!!

Mathematics

1 answer:

Soloha48 [4]2 years ago

5 0

for number one, the mean would be 45 minutes

for number 9, he wants to use the number of minutes that occurred most often as that is the definition of mode

please mark branliest it’d be appreciated ♡

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If you roll a dice 600 times, on how many occasions would you expect to land on a 6

egoroff_w [7]

Answer:

100

Step-by-step explanation:

A dice has 6 sides and, theoretically, each side has the same probability of being rolled. So, if a dice was rolled a number of times, it would land on each side the same number of times. This means that 6 would be rolled 1/6 of the time. Since 1/6 of 600 is 100, that is how many times 6 would be rolled.

5 0

2 years ago

Read 2 more answers

5 Exam-style ABCD is a kite.

Mnenie [13.5K]

let's recall that in a Kite the diagonals meet each other at 90° angles, Check the picture below, so we're looking for the equation of a line that's perpendicular to BD and that passes through (-1 , 3).

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of BD

$y = \stackrel{\stackrel{m}{\downarrow }}{3}x-1\impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}$

$\stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{3\implies \cfrac{3}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{3}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{3}}}$

so we're really looking for the equation of a line whose slope is -1/3 and passes through point A

$(\stackrel{x_1}{-1}~,~\stackrel{y_1}{3})\qquad \qquad \stackrel{slope}{m}\implies -\cfrac{1}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{3}=\stackrel{m}{-\cfrac{1}{3}}[x-\stackrel{x_1}{(-1)}]\implies y-3=-\cfrac{1}{3}(x+1) \\\\\\ y-3=-\cfrac{1}{3}x-\cfrac{1}{3}\implies y=-\cfrac{1}{3}x-\cfrac{1}{3}+3\implies y=-\cfrac{1}{3}x+\cfrac{8}{3}$