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leva [86]
2 years ago
5

A jar contains only one blue tile, one green tile, one red tile, and one yellow tile. if a tile is chosen at random and then rep

laced in the jar, and then a second tile is chosen at random, how many of the possible outcomes contain at least one red tile?
Mathematics
2 answers:
MArishka [77]2 years ago
7 0
Its 7 just had the test
Maurinko [17]2 years ago
6 0

Answer:  

Possible outcomes that contain at least one red tile = 7

Step-by-step explanation:

Number of blue tile = 1

Number of green tile = 1

Number of red tile = 1

Number of yellow tile = 1

Total number of tiles = 4

Possible number of outcomes in choosing one tile = 4

Possible number of outcomes in selecting the second tile = 4

⇒ Possible number of outcomes in selecting both the tiles = 4 × 4 = 16

We need to find the find the possible outcomes that contain atleast one red ⇒ One tile must be atleast of red color.

Now, Possible outcomes of selecting first tile if not red = 3

Possible outcomes of selecting second tile if not red = 3

⇒ Possible outcomes of selecting both the tiles if not red = 3 × 3 = 9

Hence, Possible outcomes that contain at least one red tile = 16 - 9 = 7

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A plumber charges a ​$70 flat rate for all home repairs. His hourly rate is​ $15 per half hour. How much would a 2 hour and 26 m
Gnesinka [82]

The plumber's 2 hour and 26 min plumbing bill​ cost is $73

What is the plumber's per minute rate?

The plumber's per minute rate is the per half-hour rate divided by 30 minutes since the plumber charges $15 for every 30-minute job

per-minute rate=$15/30

per-minute rate=$0.50

Now, let us convert 2hours and 26 minutes to minutes, in other words, every 1 hour is 60 minutes, the two hours equal 120 minutes plus the 26 minutes gives a total of 146 minutes

Total minutes=120 min+26 min

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Total plumbing bill cost for 146 minutes is $0.50, the per-minute rate multiplied by the number of minutes the plumber needs to work in this case

2 hour and 26 min plumbing bill​ cost=146*$0.50

2 hour and 26 min plumbing bill​ cost=$73

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4 0
2 years ago
The top one I need help with
brilliants [131]
Same thing as what you did on the bottom. Find numbers with both 7 as the base and numbers that add to 14 on the top. Possibilities:
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3 years ago
Help please algebra 8th grade need to finish ASAP!!<br><br> A.27<br> B.33<br> C.60<br> D.93
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7 0
3 years ago
A circle has the order pairs (-1, 2) (0, 1) (-2, -1) what is the equation . Show your work.
olga55 [171]
We know that:

(x-a)^2+(y-b)^2=r^2

is an equation of a circle.

When we substitute x and y (from the pairs we have), we'll get a system of equations:

\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}

and all we have to do is solve it for a, b and r.

There will be:

\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}\\\\\\&#10;\begin{cases}1+2a+a^2+4-4b+b^2=r^2\\a^2+1-2b+b^2=r^2\\4+4a+a^2+1+2b+b^2=r^2\end{cases}\\\\\\&#10;\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\\\\\&#10;

From equations (II) and (III) we have:

\begin{cases}a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\--------------(-)\\\\a^2+b^2-2b+1-a^2-b^2-4a-2b-5=r^2-r^2\\\\-4a-4b-4=0\qquad|:(-4)\\\\\boxed{-a-b-1=0}

and from (I) and (II):

\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\end{cases}\\--------------(-)\\\\a^2+b^2+2a-4b+5-a^2-b^2+2b-1=r^2-r^2\\\\2a-2b+4=0\qquad|:2\\\\\boxed{a-b+2=0}

Now we can easly calculate a and b:

\begin{cases}-a-b-1=0\\a-b+2=0\end{cases}\\--------(+)\\\\-a-b-1+a-b+2=0+0\\\\-2b+1=0\\\\-2b=-1\qquad|:(-2)\\\\\boxed{b=\frac{1}{2}}\\\\\\\\a-b+2=0\\\\\\a-\dfrac{1}{2}+2=0\\\\\\a+\dfrac{3}{2}=0\\\\\\\boxed{a=-\frac{3}{2}}

Finally we calculate r^2:

a^2+b^2-2b+1=r^2\\\\\\\left(-\dfrac{3}{2}\right)^2+\left(\dfrac{1}{2}\right)^2-2\cdot\dfrac{1}{2}+1=r^2\\\\\\\dfrac{9}{4}+\dfrac{1}{4}-1+1=r^2\\\\\\\dfrac{10}{4}=r^2\\\\\\\boxed{r^2=\frac{5}{2}}

And the equation of the circle is:

(x-a)^2+(y-b)^2=r^2\\\\\\\left(x-\left(-\dfrac{3}{2}\right)\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}\\\\\\\boxed{\left(x+\dfrac{3}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}}
7 0
3 years ago
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