Ask question

Ask question

All categories

3 years ago

15

Suppose you spin the pointer on this spinner. (circle with 4 outcomes from 1 to 4). What is the probability the pointer does not

land on 3?

1 answer:

sveticcg [70]3 years ago

8 0

The probability of getting 1,2,or 4 is .75 or 75%

You might be interested in

Can anyone help me ASAP!

TiliK225 [7]

Answer:

17.3 in

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Suppose that we were going to conduct a study to determine if there is an decrease in risk of getting lung cancer based on wheth

nalin [4]

Answer: Option C

Step-by-step explanation:

In group(1), the risk is= 5/2000 x 100

= 0.25

In group(2), the risk is= 5/1000 x 100

= 0.5

The sample relative risk is= 0.25/0.5

= 0.5

So, option C is correct.

7 0

3 years ago

System of Equations <br> Need help<br> Trig

Yanka [14]

Looks like the system is

$\begin{cases}-2x+y+6z=1\\3x+2y+5z=16\\7x+3y-4z=11\end{cases}$

We can eliminate $y$ by taking

$(3x+2y+5z)-2(-2x+y+6z)=16-2(1)$

$\implies3x+2y+5z+4x-2y-12z=16-2$

$\implies7x-7z=14$

$\implies x-z=2$

so that $z=x-2$ , and

$(7x+3y-4z)-3(-2x+y+6z)=11-3(1)$

$\implies7x+3y-4z+6x-3y-18z=11-3$

$\implies13x-22z=8$

Substitute $z=x-2$ into this last equation and solve for $x$ :

$13x-22(x-2)=8$

$\implies13x-22x+44=8$

$\implies-9x=-36$

$\implies x=4$

Then

$z=x-2$

$\implies z=4-2$

$\implies z=2$

Plug these values into any one of the original equation to solve for $y$ :

$-2x+y+6z=1$

$\implies-2(4)+y+6(2)=1$

$\implies-8+y+12=1$

$\implies y=-3$

Hence the solution is x = 4, y = -3, and z = 2.

6 0

3 years ago

Helpppp lol pleaseeeeee

makkiz [27]

Answer: i thinks it's D. none of the above because they are all angles?

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

The square of the sum of two consecutive positive even integers is 4048 more than the sum of the squares of these two numbers. F

Burka [1]

Answer:

The numbers are 44 and 46

Step-by-step explanation:

Let

x, x+2 ----> the two consecutive positive even integers

we know that

$(x+x+2)^{2} =4,048+x^{2} +(x+2)^{2} \\ \\(2x+2)^{2} =4,048+x^{2} +x^{2}+4x+4\\ \\4x^{2}+8x+4=2x^{2}+4x+4,052\\ \\2x^{2} +4x-4,048=0$

Solve the quadratic equation using a graphing calculator

The solution is x=44

see the attached figure

$x+2=44+2=46$

therefore

The numbers are 44 and 46

4 0

3 years ago