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3 years ago

If you roll the six sided dice six times, are you guaranteed to get a sum of 7 once? answer

Mathematics

1 answer:

BlackZzzverrR [31]3 years ago

5 0

Yes it can be guaranteed to get a sum of 7 once

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You are comparing two refrigerator models. Their annual electricity consumption is shown below. Assume that the cost of electric

Otrada [13]

Answer:

Savings = $13

Step-by-step explanation:

$0.12 = 1 kilowatt-hour

So,

For Top-Freezer Refridgerator:

$0.12 × 529 = 1 × 529 kilowatt-hour

$63.48 = 529 kilowatt-hour

For Side-by-Side Refridgerator:

$0.12 × 634 = 1 × 634 kilowatt-hour

$76.08 = 634 kilowatt-hour

Savings = Side-by-Side Refridgerator - Top-

Freezer Refridgerator

Savings = $( 76.08 -63.48 )

Savings = $12.6

Savings = $13

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3 years ago

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I will give brainiest to whoever answers correctly !!

Viktor [21]

Answer: 10k

Step-by-step explanation:

6 0

3 years ago

Help me pliss <br>which one is correct?

Anna71 [15]

D. (2,5) is the answer

4 0

2 years ago

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Solve the equation: only write the number is the box <br> 2( 5x+2)=54

Black_prince [1.1K]

Answer:

x = 5

Step-by-step explanation:

Step 1:

2 (5x + 2) = 54

Step 2:

10x + 4 = 54

Step 3:

10x = 50

Answer:

x = 5

Hope This Helps :)

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3 years ago

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Let e1= 1 0 and e2= 0 1 , y1= 4 5 , and y2= −2 7 , and let T: ℝ2→ℝ2 be a linear transformation that maps e1 into y1 and maps

Furkat [3]

Answer:

The image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is $\left[\begin{array}{c}24&-8\end{array}\right]$

Step-by-step explanation:

We know that $T:$ $IR^{2}$ → $IR^{2}$ is a linear transformation that maps $e_{1}$ into $y_{1}$ ⇒

$T(e_{1})=y_{1}$

And also maps $e_{2}$ into $y_{2}$ ⇒

$T(e_{2})=y_{2}$

We need to find the image of the vector $\left[\begin{array}{c}4&-4\end{array}\right]$

We know that exists a matrix A from $IR^{2x2}$ (because of how T was defined) such that :

$T(x)=Ax$ for all x ∈ $IR^{2}$

We can find the matrix A by applying T to a base of the domain ( $IR^{2}$ ).

Notice that we have that data :

$B_{IR^{2}}=$ { $e_{1},e_{2}$ }

Being $B_{IR^{2}}$ the cannonic base of $IR^{2}$

The following step is to put the images from the vectors of the base into the columns of the new matrix A :

$T(\left[\begin{array}{c}1&0\end{array}\right])=\left[\begin{array}{c}4&5\end{array}\right]$ (Data of the problem)

$T(\left[\begin{array}{c}0&1\end{array}\right])=\left[\begin{array}{c}-2&7\end{array}\right]$ (Data of the problem)

Writing the matrix A :

$A=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]$

Now with the matrix A we can find the image of $\left[\begin{array}{c}4&-4\\\end{array}\right]$ such as :

$T(x)=Ax$ ⇒

$T(\left[\begin{array}{c}4&-4\end{array}\right])=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]\left[\begin{array}{c}4&-4\end{array}\right]=\left[\begin{array}{c}24&-8\end{array}\right]$

We found out that the image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is the vector $\left[\begin{array}{c}24&-8\end{array}\right]$

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3 years ago