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Brrunno [24]
3 years ago
9

Find the product of 83 and 3.

Mathematics
1 answer:
egoroff_w [7]3 years ago
6 0

Hello!


Answer:

249


Step-by-step explanation:


83

 3  x

___

249


Hope this helps! ~Pooch ♥

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Consider a collection of envelopes consisting of 1 red envelope, 3 blue envelopes, 2 green envelopes, and 3 yellow envelopes if
nasty-shy [4]

Answer:

the probability that at least one envelope is a yellow envelope is 16/21

Step-by-step explanation:

The probability that at least one envelope is a yellow envelope is P(Y);

P(Y) = 1 - P(Y)'

P(Y)' is the probability that no envelope is a yellow envelope.

Given;

red envelope = 1

blue envelopes = 3

green envelopes = 2

yellow envelopes = 3

Total = 9

Number of non-yellow envelope = 9 -3 = 6

(6 envelope are not yellow)

P(Y)' = P1 × P2 × P3

Since there is no replacement;

P(Y)' = 6/9 × 5/8 × 4/7

P(Y)' = 5/21

From equation 1;

P(Y) = 1 - 5/21

P(Y) = 16/21

the probability that at least one envelope is a yellow envelope is 16/21.

8 0
3 years ago
The expression log8512 = 3 in exponential form is 38 = 512.
svet-max [94.6K]
\large\begin{array}{I} \mathtt{ log_8~512=3 } \end{array}

\large\begin{array}{I} \mathtt{ 3^8 = 512 } \end{array}
3 0
3 years ago
Each floor in a hotel has the same number of rooms. A member of the housekeeping staff cleaning every room on 5 floors cleaned 5
ch4aika [34]
There are 10 rooms per floor if there are 30 floors then there are 300 rooms
8 0
3 years ago
(2x + y) (2x + 2y) =
IceJOKER [234]

Answer: 4x^2 + 6xy + 2y^2

Step-by-step explanation:

(2x + y)(2x + 2y) =

2x*2x + 2x*2y + y*2x + y^2y =

4x^2 + 4xy + 2xy + 2y^2 =

4x^2 + 6xy + 2y^2

3 0
3 years ago
Read 2 more answers
Find the equation of the line that is perpendicular to the line y = (-1/3)x -1 and passes through the point (1, 5)?
Anit [1.1K]

bearing in mind that perpendicular lines have negative reciprocal slopes, so


\bf \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}~\hspace{10em}\stackrel{slope}{y=\stackrel{\downarrow }{-\cfrac{1}{3}}x-1} \\\\[-0.35em] ~\dotfill


\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{-\cfrac{1}{3}}\qquad \qquad \qquad \stackrel{reciprocal}{-\cfrac{3}{1}}\qquad \stackrel{negative~reciprocal}{+\cfrac{3}{1}\implies 3}}


so we're really looking for a line whose slope is 3 and runs through (1,5)


\bf (\stackrel{x_1}{1}~,~\stackrel{y_1}{5})~\hspace{10em} slope = m\implies 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-5=3(x-1) \\\\\\ y-5=3x-3\implies y=3x+2

4 0
3 years ago
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