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alukav5142 [94]

2 years ago

6

Which expression represents the algebraic phrase “twelve times the sum of a number and seven-tenths”? 12 y + StartFraction 7 ove

r 10 EndFraction 12 (y + StartFraction 7 over 10 EndFraction) StartFraction 7 over 10 EndFraction y + 12 y (StartFraction 7 over 10 EndFraction + 12)

2 answers:

kykrilka [37]2 years ago

7 0

Answer:

The answer is:

12 ( y+ $\frac{7}{10}$ )

this suits perfectly the description

Doss [256]2 years ago

6 0

Answer:

c

Step-by-step explanation:

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What is 5.316 - 1.942 btw (show ur work) :)

erastovalidia [21]

Answer:

3.374

Step-by-step explanation:

$\mathrm{Write\:the\:numbers\:one\:under\:the\:other,\:line\:up\:the\:decimal\:points.}$

$\mathrm{Add\:trailing\:zeroes\:so\:the\:numbers\:have\:the\:same\:length.}$

$\begin{matrix}\:\:&5&.&3&1&6\\ -&1&.&9&4&2\end{matrix}$

$\mathrm{Subtract\:each\:column\:of\:digits,\:starting\:from\:the\:right\:and\:working\:left}$

$\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:6-2=4\\$

$\frac{\begin{matrix}\:\:&5&.&3&1&\textbf{6}\\ -&1&.&9&4&\textbf{2}\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\:\:&\textbf{4}\end{matrix}}$

$\frac{\begin{matrix}\:\:&5&.&3&\textbf{1}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{\:\:}&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&5&.&\textbf{3}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&\textbf{4}&\:\:&10&\:\:&\:\:\\ \:\:&\textbf{\linethrough{5}}&.&3&1&6\\ -&\textbf{1}&.&9&4&2\end{matrix}}{\begin{matrix}\:\:&\textbf{\:\:}&\:\:&\:\:&\:\:&4\end{matrix}}\\$

$\frac{\begin{matrix}\:\:&4&\:\:&\textbf{13}&\:\:&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{3}}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&4&\:\:&\textbf{12}&10&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{13}}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&4&\:\:&12&\textbf{11}&\:\:\\ \:\:&\linethrough{5}&.&\linethrough{13}&\textbf{\linethrough{1}}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{\:\:}&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&4&\:\:&12&\textbf{11}&\:\:\\ \:\:&\linethrough{5}&.&\linethrough{13}&\textbf{\linethrough{1}}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{7}&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&4&\:\:&\textbf{12}&11&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{13}}&\linethrough{1}&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{3}&7&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&4&\textbf{\:\:}&12&11&\:\:\\ \:\:&\linethrough{5}&\textbf{.}&\linethrough{13}&\linethrough{1}&6\\ -&1&\textbf{.}&9&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\textbf{.}&3&7&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&\textbf{4}&\:\:&12&11&\:\:\\ \:\:&\textbf{\linethrough{5}}&.&\linethrough{13}&\linethrough{1}&6\\ -&\textbf{1}&.&9&4&2\end{matrix}}{\begin{matrix}\:\:&\textbf{3}&.&3&7&4\end{matrix}}$

=3.374

6 0

1 year ago

X + 2y = -4 in graph

MariettaO [177]

Equation Answer:

y = -½x - 2

Graph:

3 0

2 years ago

A factory worker productivity is normally distributed. one worker produces an average of 75 units per day with a standard deviat

Angelina_Jolie [31]

Let Xi be the random variable representing the number of units the ﬁrst worker produces in day i.
Deﬁne X = X1 + X2 + X3 + X4 + X5 as the random variable representing the number of units the
ﬁrst worker produces during the entire week. It is easy to prove that X is normally distributed with mean µx = 5·75 = 375 and standard deviation σx = 20√5.
Similarly, deﬁne random variables Y1, Y2,...,Y5 representing the number of units produces by
the second worker during each of the ﬁve days and deﬁne Y = Y1 + Y2 + Y3 + Y4 + Y5. Again, Y is normally distributed with mean µy = 5·65 = 325 and standard deviation σy = 25√5. Of course, we assume that X and Y are independent. The problem asks for P(X > Y ) or in other words for P(X −Y > 0). It is a quite surprising fact that the random variable U = X−Y , the diﬀerence between X and Y , is also normally distributed with mean µU = µx−µy = 375−325 = 50 and standard deviation σU, where σ2 U = σ2 x+σ2 y = 400·5+625·5 = 1025·5 = 5125. It follows that σU = √5125. A reference to the above fact can be found online at http://mathworld.wolfram.com/NormalDiﬀerenceDistribution.html.
Now everything reduces to ﬁnding P(U > 0) P(U > 0) = P(U −50 √5125 > − 50 √5125)≈ P(Z > −0.69843) ≈ 0.757546 .

5 0

3 years ago

Determine the area of the square ABCD with AB=3cm<br>

Svetach [21]

Side=3cm

$\\ \tt\longmapsto Area=(side)^2$

$\\ \tt\longmapsto Area=3^2$

$\\ \tt\longmapsto Area=9cm^2$

5 0

2 years ago

Read 2 more answers

Find the total area of the kitchen floor that is to be tiled as shown below.

schepotkina [342]

Answer:

58

Step-by-step explanation:

There is two rectangles in this problem.

For the first one 5×10=50

And the second one is 2×4=8

So the final answer will be 58

Hope this helpedn

7 0

3 years ago

Read 2 more answers