All categories

3 years ago

1)the mean of a, b, c, d, and e is 86. The mean of a, b, and c is 80 What is the mean of d and e?

Mathematics

1 answer:

Paraphin [41]3 years ago

8 0

Since a,b and c is 80 then d and e is 6

You might be interested in

Use matrix addition to solve this equation: B + 15 −7 4 0 1 2 = 1 2 12 4 0 2 b11 = b12 = b13 = b21 = 4 b22 = −1 b23 = 0

Arada [10]

Answer:

$b_{11}=-14,b_{12}=9,b_{13}=8,b_{21}=4,b_{22}=-1,b_{23}=0$

Step-by-step explanation:

The given matrix addition is

$B+\begin{bmatrix}15&-7&4\\ 0&1&2\end{bmatrix}=\begin{bmatrix}1&2&12\\ 4&0&2\end{bmatrix}$

We need to find the elements of matrix B.

Let $B=\begin{bmatrix}b_{11}&b_{12}&b_{13}\\ b_{21}&b_{22}&b_{23}\end{bmatrix}$

Substitute the value of matrix.

$\begin{bmatrix}b_{11}&b_{12}&b_{13}\\ b_{21}&b_{22}&b_{23}\end{bmatrix}+\begin{bmatrix}15&-7&4\\ 0&1&2\end{bmatrix}=\begin{bmatrix}1&2&12\\ 4&0&2\end{bmatrix}$

After addition of two matrix we get

$\begin{bmatrix}b_{11}+15&b_{12}-7&b_{13}+4\\ b_{21}+0&b_{22}+1&b_{23}+2\end{bmatrix}=\begin{bmatrix}1&2&12\\ 4&0&2\end{bmatrix}$

On equating both sides.

$b_{11}+15=1\Rightarrow b_{11}=-14$

$b_{12}-7=2\Rightarrow b_{12}=9$

$b_{13}+4=12\Rightarrow b_{13}=8$

$b_{21}+0=4\Rightarrow b_{21}=4$

$b_{22}+1=0\Rightarrow b_{22}=-1$

$b_{23}+2=2\Rightarrow b_{23}=0$

Therefore, the elements of matrix B are $b_{11}=-14,b_{12}=9,b_{13}=8,b_{21}=4,b_{22}=-1,b_{23}=0$ .

3 0

3 years ago

Each cube in the prisms below has a volume of 1 cubic unit. which prism has a volume of 60 cubic units

mr_godi [17]

For A, 5x4x3=60
for B 5x3x2=30
for C 5x4x1=20
for D 5x4x2=40
hence, A has a volume of 60cubic units!

7 0

3 years ago

Can someone help me pls

marta [7]

Answer:

237?

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Equation for dividing example 25 divided by 5

Dafna11 [192]

Answer: Example: 15 ÷ 5

Step-by-step explanation:

6 0

3 years ago

Find cot θ if csc θ = negative square root of thirty seven divided by six and tan θ > 0.

Ket [755]

You can use the trigonometric identity $1+\cot^2{x}=\csc^2{x}$ .

$1+\cot^2{\theta}=(\frac{-\sqrt{37}}{6} )^2 \\ 1+\cot^2{\theta}=\frac{37}{36} \\ \cot^2{\theta}=\frac{1}{36} \\
\cot{\theta}=\frac{1}{6}$

The requirement that $\tan{\theta}\ \textgreater \ 0$ eliminates -1/6 from being another solution.

6 0

3 years ago