Is this table proportional? XY 1 3 3 4 12 5 15 7 21

I don't know how to do this. Can someone help me please?

Svetach [21]

1. y =220- x
150=220-70

5 0

3 years ago

What is 620,000 nearest 10 thousand

FinnZ [79.3K]

Here first we have to understand which place is 10 thousand.

6- hundred thousand

2- ten thousand

0- thousand

0- hundred

0- ten

0- one

so as we can see 2 is in ten thousands place.

Now rounding ten thousands place is done by looking up to thousands place.

If the digit at thousands place is 5 or greater than 5 then we change our ten thousands value to one digit higher else it remains same.

Here we have 0 at thousands place , which is less than 5 , so we keep the number as it is.

Answer: 620, 000

5 0

3 years ago

Determine if the ordered pair (6, 4) is a solution to the inequality

Ugo [173]

Answer:

$\Large \boxed{\mathrm{Option \ D}}$

Step-by-step explanation:

(6, 4)

x = 6 and y = 4

y > -1/2x + 7

Plug in the values to check if it is true.

4 > -1/2(6) + 7

4 > -3 + 7

4 > 4

This statement is false.

(6, 4) lies on the line.

8 0

3 years ago

Cards. A deck of 52 cards is shuffled and dealt. Find the probabilities of the following events: a) the tenth card is a queen; b

Veronika [31]

Answer:

0.0374
0.0010

Step-by-step explanation:

number of cards = 52

number of queen = 4

number of spades = 13

A) probability that the tenth card is a queen

drawn time (r) = 1

position of success(x) = 10th

p = 4/52

P( x,r,p) = $x-1Cr-1*p^(r) * q^(x-r)$

p(10,1,4/52) = 9C0(4/52)^1 * (48/52)^9 = 0.0374

B) probability the twentieth card is a spade

x = 20

r = 1

p = 13 / 52

P(20,1,26/52) = 19C0(26/52)^1 * (26/52)^19 = 0.0010

c) The last five cards been spades

p(last five cards been spades )

p(48..52, 5, 13/52 ) = 47...52C4(13/52)^5 * (39/52)^48..52 - 5

4 0

3 years ago

Question 3, help please!

White raven [17]

The answer is not defined.

Explanation:

The given matrix is $\left[\begin{array}{cc}{2} & {4} \\ {1} & {-6}\end{array}\right]+\left[\begin{array}{c}{1} \\ {0}\end{array}\right]$

The matrix $\left[\begin{array}{cc}{2} & {4} \\ {1} & {-6}\end{array}\right]$ has dimensions $2\times 2$

This means that the matrix has 2 rows and 2 columns.

Also, the matrix $\left[\begin{array}{l}{1} \\ {0}\end{array}\right]$ has dimensions $2\times1$

This means that the matrix has 2 rows and 1 column.

Since, the matrices can be added only if they have the same dimensions.

In other words, to add the matrices, the two matrices must have the same number of rows and same number of columns.

Since, the dimensions of the two matrices are not equal, the addition of these two matrices is not possible.

Hence, the addition of these two matrices is not defined.

4 0

3 years ago