HOW do I solve these questions?

Verify the equation below with each of the values listed for t to find a solution. 4 = 2t +8

Varvara68 [4.7K]

The solution for t

4 = 2t + 8
4 + (-8) = 2t + 8 + (-8)
-4 = 2t
-4 / 2 = 2t / 2
-2 = t
t = -2

6 0

4 years ago

The members of the math club are buying t-shirts. The shirts will cost $5.00 each plus a one-time fee of $20.00 for the design o

marta [7]

Answer is the choice B

3 0

3 years ago

Read 2 more answers

1. Find the perimeter of the figure. Each unit is 1 centimeter

Oxana [17]

Answer:

20

Step-by-step explanation:

count the sides of the boxes for each side of the shape

8 0

3 years ago

Solve the matrix and prove that it is equal 0

Art [367]

Step-by-step explanation:

$\underline{ \underline{ \text{Given}}} :$

$\tt{ {A}^{T} = \begin{bmatrix} 2 & - 4 \\ 4 & 3 \\ \end{bmatrix}}$

$\underline{ \underline { \text{To \: Find}}} :$

$\sf{ {A}^{2} - 5A+ 22I= 0}$

$\underline{ \underline{ \text{Solution}}} :$

The new matrix obtained from a given matrix by interchanging it's rows and columns is called the transposition of matrix. It is denoted by $\sf{ {A}^{T}}$ . Again , Interchange it's rows and columns in order to find ' A '.

$\tt{A = \begin{bmatrix} 2 & 4 \\ - 4 & 3 \\ \end{bmatrix}}$

Now , LEFT HAND SIDE ( L.H.S )

$\tt{ {A}^{2} - 5A+ 22I}$

Here, I refers to identity matrix. A diagonal matrix in which all the elements of leading diagonal are 1 ( unit ) is called unit or identity matrix.

⟼ $\begin{bmatrix} 2 & 4 \\ - 4 & 3 \\ \end{bmatrix} \times \begin{bmatrix} 2 & 4 \\ - 4 & 3 \\ \end{bmatrix} - 5 \times \begin{bmatrix} 2 & 4 \\ - 4 & 3 \\ \end{bmatrix} + 22 \times \begin{bmatrix} 1 & 0 \\ 0 & 1\\ \end{bmatrix}$

⟼ $\begin{bmatrix} 2 \times 2 + 4 \times ( - 4)& 2 \times 4 + 4 \times 3 \\ - 4 \times 2 + 3 \times ( - 4) & - 4 \times 4 + 3 \times 3 \\ \end{bmatrix} - \begin{bmatrix} 10 & 20 \\ - 20& 15 \\ \end{bmatrix} + \begin{bmatrix} 22 & 0 \\ 0 & 22 \\ \end{bmatrix}$

⟼ $\begin{bmatrix} 4 + ( - 16) & 8 + 12 \\ - 8 + ( - 12) & - 16 + 9 \\ \end{bmatrix} - \begin{bmatrix} 10 & 20 \\ - 20 & 15 \\ \end{bmatrix} + \begin{bmatrix} 22 & 0 \\ 0 & 22 \\ \end{bmatrix}$

⟼ $\begin{bmatrix} - 12 & 20\\ - 20& - 7 \\ \end{bmatrix} - \begin{bmatrix} 10 & 20 \\ - 20 & 15 \\ \end{bmatrix} + \begin{bmatrix} 22 & 0 \\ 0 & 22 \\ \end{bmatrix}$

⟼ $\begin{bmatrix} - 22 & 0 \\ 0& - 22 \\ \end{bmatrix} + \begin{bmatrix} 22 & 0 \\ 0 & 22 \\ \end{bmatrix}$

⟼ $\begin{bmatrix} - 22 + 22 & 0 + 0 \\ 0 + 0 & - 22 + 22 \\ \end{bmatrix}$

⟼ $\begin{bmatrix} 0 & 0\\ 0 & 0 \\ \end{bmatrix}$

⟼ $\sf{0}$

RIGHT HAND SIDE ( R.H.S ) : 0

L.H.S = R.H.S [ Hence , proved ! ]

Hope I helped ! ♡

Have a wonderful day / night ! ツ

▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁

4 0

3 years ago

The data below are the temperatures on randomly chosen days during the summer and the number of employee absences at a local com

s2008m [1.1K]

Answer:

The critical region is t ≥ t(0.025, 7) = 2.365

Since the calculated value of t= 18.50249 falls in the critical region we reject the null hypothesis and conclude that there is sufficient reason to support the claim of a linear relationship between the two variables.

Step-by-step explanation:

We set up our hypotheses as

H0: β= 0 the two variable X and Y are not related

Ha: β ≠ 0. the two variables X and Y are related.

The significance level is set at α =0.05

The test statistic if, H0 is true, is t= b/s_b

Where Sb =S_yx/√(∑(X-X`)^2 )

Syx = √((∑(Y-Y`)^2 )/(n-2))

In the given question we have the estimated regression line as y= 0.449x - 30.27

X Y X2 Y2 XY

72 3 5184 9 216

85 7 7225 49 595

91 10 8281 100 910

90 10 8100 100 900

88 8 7744 64 704

98 15 9604 225 1470

75 4 5625 16 300

100 15 10000 225 1500

80 5 6400 25 400

∑779 77 68163 813 6995

Now finding the variances

∑(Y-Y`)^2 = ∑〖Y^2- a〗 ∑Y- b∑XY

= 813 – (- 30.27)77 - 0.449(6995)

= 813+2330.79 – 3140.755

= 3.035

∑(X-X`)^2 = ∑X^2 – (∑〖X)〗^2 /n

= 68163 – (779)2/9

= 736.22

Syx = √((∑(Y-Y`)^2 )/(n-2)) = √(3.035/7) = 0.65846 and

Sb =S_yx/√(∑(X-X`)^2 ) = (0.65846 )/27.13337 = 0.024267

t= b/s_b = 0.449/ 0.024267 = 18.50249

The critical region is t ≥ t(0.025, 7) = 2.365

Since the calculated value of t= 18.50249 falls in the critical region we reject the null hypothesis and conclude that there is sufficient reason to support the claim of a linear relationship between the two variables.

Download docx

6 0

3 years ago