How do I solve this problem

saw5 [17]

Answer: 92 which is choice D

Look at the right side of values for the digit that comes up the most. In this case, it would be '2' in the second row. It shows up 3 times. Those leafs pair up with the stem of 9 to form the full value 92. So we have 92 show up three times which is the most frequent value.

The data set would expand out to

85, 88

92, 92, 92, 98

105, 106, 106, 109

113

126, 127

5 0

2 years ago

The ratio of boys to girls in math classes is 3.5. If there are 130 girls in these classes, how many boys must there be?

dsp73

Answer:

<h3><u>Required Answer</u><u>:</u><u>-</u></h3>

The ratio of boys and girls =3:5

Let

Number of boys=3x

Number of girls =5x

According to the question Numbers of girls=130

${:}\longrightarrow$ $\sf 5x=130$

Now interchange both sides and get x

${:}\longrightarrow$ $\sf x={\dfrac {130}{5}}$

${:}\longrightarrow$ $\sf x=26$

The number of boys=3x

=3×26=78

$\therefore$ Number of boys is 78.

4 0

2 years ago

A cube has six sides that are numbered 1 to 6. The sides numbered 1, 2, and 5 are green, and the sides numbered 3, 4,

Katarina [22]

Answer:

5/6

Step-by-step explanation:

The sides numbered 1,2,5 are green and 2,4,6 are even

We cannot count the 2 twice, so there are 5 unique possibilities of getting either a green or an even

There are 6 total possibilities when rolling the die 1,2,3,4,5,6

P (green or even) = green or even/ total

= 5/6

8 0

2 years ago

If you spin the spinner 7 times,what is the best prediction possible for the number of times it will land on pink?

Tems11 [23]

Depends how many times pink is on the spinner you put the number of pink spaces there is at the top out of seven for example a fraction 3/7

8 0

2 years ago

Assume {v1, . . . , vn} is a basis of a vector space V , and T : V ------> W is an isomorphism where W is another vector spac

Degger [83]

Answer:

Step-by-step explanation:

To prove that $w_1,\dots w_n$ form a basis for W, we must check that this set is a set of linearly independent vector and it generates the whole space W. We are given that T is an isomorphism. That is, T is injective and surjective. A linear transformation is injective if and only if it maps the zero of the domain vector space to the codomain's zero and that is the only vector that is mapped to 0. Also, a linear transformation is surjective if for every vector w in W there exists v in V such that T(v) =w

Recall that the set $w_1,\dots w_n$ is linearly independent if and only if the equation

$\lambda_1w_1+\dots \lambda_n w_n=0$ implies that

$\lambda_1 = \cdots = \lambda_n$ .

Recall that $w_i = T(v_i)$ for i=1,...,n. Consider $T^{-1}$ to be the inverse transformation of T. Consider the equation

$\lambda_1w_1+\dots \lambda_n w_n=0$

If we apply $T^{-1}$ to this equation, then, we get

$T^{-1}(\lambda_1w_1+\dots \lambda_n w_n) =T^{-1}(0) = 0$

Since T is linear, its inverse is also linear, hence

$T^{-1}(\lambda_1w_1+\dots \lambda_n w_n) = \lambda_1T^{-1}(w_1)+\dots + \lambda_nT^{-1}(w_n)=0$

which is equivalent to the equation

$\lambda_1v_1+\dots + \lambda_nv_n =0$

Since $v_1,\dots,v_n$ are linearly independt, this implies that $\lambda_1=\dots \lambda_n =0$ , so the set $\{w_1, \dots, w_n\}$ is linearly independent.

Now, we will prove that this set generates W. To do so, let w be a vector in W. We must prove that there exist $a_1, \dots a_n$ such that

$w = a_1w_1+\dots+a_nw_n$

Since T is surjective, there exists a vector v in V such that T(v) = w. Since $v_1,\dots, v_n$ is a basis of v, there exist $a_1,\dots a_n$ , such that

$a_1v_1+\dots a_nv_n=v$

Then, applying T on both sides, we have that

$T(a_1v_1+\dots a_nv_n)=a_1T(v_1)+\dots a_n T(v_n) = a_1w_1+\dots a_n w_n= T(v) =w$

which proves that $w_1,\dots w_n$ generate the whole space W. Hence, the set $\{w_1, \dots, w_n\}$ is a basis of W.

Consider the linear transformation $T:\mathbb{R}^2\to \mathbb{R}^2$ , given by T(x,y) = T(x,0). This transformations fails to be injective, since T(1,2) = T(1,3) = (1,0). Consider the base of $\mathbb{R}^2$ given by (1,0), (0,1). We have that T(1,0) = (1,0), T(0,1) = (0,0). This set is not linearly independent, and hence cannot be a base of $\mathbb{R}^2$

8 0

2 years ago