$F+G=\begin{bmatrix}{-1.8} & {-8.6} & {} \\ {2.85} & {-1.4} & {} \\ {-1.8} & {5.1} & {}\end{bmatrix}+\begin{bmatrix}{1.32} & {-1.9} & {} \\ {2.25} & {0.0} & {} \\ {-6.2} & {1.4} & {}\end{bmatrix}$

Then, add the elements that occupy the same position:

$H=\begin{bmatrix}{-1.8+1.32} & {-8.6+(-1.9)} & {} \\ {2.85+2.25} & {-1.4+0.0} & {} \\ {-1.8+(-6.2)} & {5.1+1.4} & {}\end{bmatrix}$

Solve

$H=\begin{bmatrix}{-0.48} & {-10.5} & {} \\ {5.1} & {-1.4} & {} \\ {-8} & {6.5} & {}\end{bmatrix}$

So, we find the element at address h31:

$H=\begin{bmatrix}{h11} & {h12} & {} \\ {h21} & {h22} & {} \\ {h31} & {h32} & {}\end{bmatrix}$

In this case, position h31 is - 8.0

8 0

1 year ago

A survey of 90 men found that an average amount spent on St. Patrick's day of $55 with a standard deviation of $18. A similar su

dem82 [27]

Answer:

The value of the test statistic is 4.70.

Step-by-step explanation:

The hypothesis for this test can be defined as follows:

H₀: Men do not spend more than women on St. Patrick's day, i.e. μ₁ = μ₂.

Hₐ: Men spend more than women on St. Patrick's day, i.e. μ₁ > μ₂.

The population standard deviations are not known.

So a t-distribution will be used to perform the test.

The test statistic for the test of difference between mean is:

$t=\frac{\bar x_{1}-\bar x_{2}}{\sqrt{\frac{s^{2}_{1}}{n_{1}}+\frac{s^{2}_{2}}{n_{2}}} }$

Given:

$\bar x_{1}=55\\s_{1}=18\\n_{1}=90\\\bar x_{1}=44\\s_{1}=16\\n_{1}=86$

Compute the value of the test statistic as follows:

$t=\frac{\bar x_{1}-\bar x_{2}}{\sqrt{\frac{s^{2}_{1}}{n_{1}}+\frac{s^{2}_{2}}{n_{2}}} }=\frac{55-44}{\sqrt{\frac{15^{2}}{90}+\frac{16^{2}}{86} }}=4.70$

Thus, the value of the test statistic is 4.70.

7 0

3 years ago

Add (13x – 4) and (–6x + 15)

kompoz [17]

(13x – 4) + (–6x + 15) =

13x - 4 - 6x + 15 =

7x + 11

4 0

3 years ago