What would the cords be

HELPPPPPPP PLSSSSSSSSS

Rainbow [258]

Answer:

c

Step-by-step explanation:

c. j = a + 5

such that 7 = 2 +5

8 0

2 years ago

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

2 years ago

A right triangle has two equal sides. The hypotenuse is 10 cm. What is the perimeter of the triangle?

Scorpion4ik [409]

Answer:

Isosceles Right Triangle Example

Step-by-step explanation:

Find the area and perimeter of an isosceles right triangle whose hypotenuse side is 10 cm. Therefore, the length of the congruent legs is 5√2 cm. Therefore, the perimeter of an isosceles right triangle is 24.14 cm.

Hope this answer helps ^^

4 0

3 years ago

You weigh six packages and find the weights to be 34, 24, 74, 29, 69, and 64 ounces. If you include a package that weighs 154 ou

saw5 [17]

Answer:

C. the mean increases more

Step-by-step explanation:

the mean is more affected by outliers because it is an average of all the numbers.

7 0

3 years ago

Read 2 more answers

Three populations have proportions 0.1, 0.3, and 0.5. We select random samples of the size n from these populations. Only two of

IRINA_888 [86]

Answer:

(1) A Normal approximation to binomial can be applied for population 1, if n = 100.

(2) A Normal approximation to binomial can be applied for population 2, if n = 100, 50 and 40.

(3) A Normal approximation to binomial can be applied for population 2, if n = 100, 50, 40 and 20.

Step-by-step explanation:

Consider a random variable X following a Binomial distribution with parameters n and p.

If the sample selected is too large and the probability of success is close to 0.50 a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

np ≥ 10
n(1 - p) ≥ 10

The three populations has the following proportions:

p₁ = 0.10

p₂ = 0.30

p₃ = 0.50

(1)

Check the Normal approximation conditions for population 1, for all the provided n as follows:

$n_{a}p_{1}=10\times 0.10=1$

Thus, a Normal approximation to binomial can be applied for population 1, if n = 100.

(2)

Check the Normal approximation conditions for population 2, for all the provided n as follows:

$n_{a}p_{1}=10\times 0.30=310\\\\n_{c}p_{1}=50\times 0.30=15>10\\\\n_{d}p_{1}=40\times 0.10=12>10\\\\n_{e}p_{1}=20\times 0.10=6$

Thus, a Normal approximation to binomial can be applied for population 2, if n = 100, 50 and 40.

(3)

Check the Normal approximation conditions for population 3, for all the provided n as follows:

$n_{a}p_{1}=10\times 0.50=510\\\\n_{c}p_{1}=50\times 0.50=25>10\\\\n_{d}p_{1}=40\times 0.50=20>10\\\\n_{e}p_{1}=20\times 0.10=10=10$

Thus, a Normal approximation to binomial can be applied for population 2, if n = 100, 50, 40 and 20.

8 0

2 years ago