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3 years ago

Which of the following properties of equality would you use to move the 34 to the other side of the equation 5h+34 = 17?

1 answer:

4 0

D. Subtraction property of equality, because since its +34 you have to change it to a subtraction sign and subtract it from both sides. So it should look something like this

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The weight of tigers follow a normal distribution with a mean of 220 kg and a SD of 30 kg. 1) If we randomly select a tiger, wha

mylen [45]

Answer:

0.89736

Step-by-step explanation:

We solve this question using z score formula

Z score = x - μ/σ

x = raw score

μ = population mean

σ = population standard deviation

Hence,

x = 258, μ = 220, σ = 30

Z = 258 - 220/30

=1.26667

Probability value from Z-Table:

P(x<258) = 0.89736

Therefore, the probability that his weights is less than 258 kg is 0.89736

6 0

3 years ago

Pls help me to solve this

zloy xaker [14]

Answer:

see below

Step-by-step explanation:

19-3 will be greater

We are multiplying 1/2 *(19-3) which is multiplying by a number less than 1 it will be less than 1

18-3 = 16

1/2 * (19-3)

Using PEMDAS

We do parentheses first

1/2 ( 16)

Then multiply

4 0

3 years ago

All vectors are in Rn. Check the true statements below:

Oduvanchick [21]

Answer:

A), B) and D) are true

Step-by-step explanation:

A) We can prove it as follows:

$Proy_{cv}y=\frac{(y\cdot cv)}{||cv||^2}cv=\frac{c(y\cdot v)}{c^2||v||^2}cv=\frac{(y\cdot v)}{||v||^2}v=Proy_{v}y$

B) When you compute the product Ax, the i-th component is the matrix of the i-th column of A with x, denote this by Ai x. Then, we have that $||Ax||=\sqrt{(A_1 x)^2+\cdots (A_n x)^2}$ . Now, the colums of A are orthonormal so we have that (Ai x)^2=x_i^2. Then $||Ax||=\sqrt{(x_1)^2+\cdots (x_n)^2}=||x||$ .

C) Consider $S=\{(0,2),(2,0)\}\subseteq \mathbb{R}^2$ . This set is orthogonal because $(0,2)\cdot(2,0)=0(2)+2(0)=0$ , but S is not orthonormal because the norm of (0,2) is 2≠1.

D) Let A be an orthogonal matrix in $\mathbb{R}^n$ . Then the columns of A form an orthonormal set. We have that $A^{-1}=A^t$ . To see this, note than the component $b_{ij}$ of the product $A^t A$ is the dot product of the i-th row of $A^t$ and the jth row of $A$ . But the i-th row of $A^t$ is equal to the i-th column of $A$ . If i≠j, this product is equal to 0 (orthogonality) and if i=j this product is equal to 1 (the columns are unit vectors), then $A^t A=I$

E) Consider S={e_1,0}. S is orthogonal but is not linearly independent, because 0∈S.

In fact, every orthogonal set in R^n without zero vectors is linearly independent. Take a orthogonal set $\{u_1,u_2\cdots u_p\}$ and suppose that there are coefficients a_i such that $a_1u_1+a_2u_2\cdots a_nu_n=0$ . For any i, take the dot product with u_i in both sides of the equation. All product are zero except u_i·u_i=||u_i||. Then $a_i||u_i||=0$ then $a_i=0$ .

5 0

4 years ago

PLEASE HELP WITHIN 5 MINUTES! Be as accurate as possible!

Nookie1986 [14]

Answer:

2,75in

4,25in

33in²

Step-by-step explanation:

shorter base is 2,75in

larger base is 4,25in

S=11+10,5+9+2,5=33in²

8 0

3 years ago

Segment CD has point E located on it such that CE:ED = 3:5. If the endpoints are located at C(5, -6) and D(11,18) then what are

wlad13 [49]

Answer:

E(29/4,3)

Step-by-step explanation:

Given that,

Segment CD has point E located on it such that CE:ED = 3:5

The coordinates of C and D are (5, -6) and (11,18) respectively.

We need to find the coordinates of E. Let the coordinates are (x,y). Using section formula to find it as follows :

$(x,y)=(\dfrac{m_1x_2+m_2x_1}{m_1+m_2},\dfrac{m_1y_2+m_2y_1}{m_1+m_2})\\\\(x,y)=(\dfrac{3(11)+5(5)}{3+5},\dfrac{3(18)+5(-6)}{3+5})\\\\=\dfrac{29}{4},3$

So, the coordinates of E are (29/4,3).

4 0

3 years ago