All categories

2 years ago

Draw a number line and show all numbers that are three units away from point 2

Mathematics

2 answers:

Arturiano [62]2 years ago

4 0

Numbers 3 units away from 2 are 5 if you move forward and -1 if you move backward.
2 + 3 =5
2 - 3 =-1

uysha [10]2 years ago

4 0

The numbers that are three units away from point 2 are -1 to the left and 5 to the right:

You might be interested in

Use substitution to determine if the solution is correct (PINK HIGHLIGHTED ONE!!!)

mart [117]

Answer:

It is not correct.

Step-by-step explanation:

If you plug 4 in for x, you get -2*4+5=13. This simplifies to: -8+5=13. This is an untrue statement since -8+5 is actually -3, not 13.

Hope this is helpful! :)

4 0

3 years ago

Read 2 more answers

In ΔSTU, m∠S=(11x+3)∘,∠T=(2x+16)∘, and m∠U=(x+7)∘. Find .

frozen [14]

Answer:

x = 11

Step-by-step explanation:

In ΔSTU ,

m∠S = (11x+3)° ; m∠T = (2x+16)° ; m∠U = (x+7)°

According to angle sum property of a triangle , sum of all the interior angles of the triangle is 180°.

So,

$11x + 3 + 2x + 16 + x + 7 = 180$

$= > 14x + 26 = 180$

$= > 14x = 180 - 26 = 154$

$= > x = \frac{154}{14} = 11$

6 0

2 years ago

Consider a discrete random variable x with pmf px (1) = c 3 ; px (2) = c 6 ; px (5) = c 3 and 0 otherwise, where c is a positive

PtichkaEL [24]

Looks like the PMF is supposed to be

$\mathbb P(X=x)=\begin{cases}\dfrac c3&\text{for }x\in\{1,5\}\\\\\dfrac c6&\text{for }x=2\\\\0&\text{otherwise}\end{cases}$

which is kinda weird, but it's not entirely clear what you meant...

Anyway, assuming the PMF above, for this to be a valid PMF, we need the probabilities of all events to sum to 1:

$\displaystyle\sum_{x\in\{1,2,5\}}\mathbb P(X=x)=\dfrac c3+\dfrac c6+\dfrac c3=\dfrac{5c}6=1\implies c=\dfrac65$

Next,

$\mathbb P(X>2)=\mathbb P(X=5)=\dfrac c3=\dfrac25$

$\mathbb E(X)=\displaystyle\sum_{x\in\{1,2,5\}}x\,\mathbb P(X=x)=\dfrac c3+\dfrac{2c}6+\dfrac{5c}3=\dfrac{7c}3=\dfrac{14}5$

$\mathbb V(X)=\mathbb E\bigg((X-\mathbb E(X))^2\bigg)=\mathbb E(X^2)-\mathbb E(X)^2$
$\mathbb E(X^2)=\displaystyle\sum_{x\in\{1,2,5\}}x^2\,\mathbb P(X=x)=\dfrac c3+\dfrac{4c}6+\dfrac{25c}3=\dfrac{28c}3=\dfrac{56}5$
$\implies\mathbb V(X)=\dfrac{56}5-\left(\dfrac{14}5\right)^2=\dfrac{84}{25}$

If $Y=X^2+1$ , then $X^2=Y-1\implies X=\sqrt{Y-1}$ , where we take the positive root because we know $X$ can only take on positive values, namely 1, 2, and 5. Correspondingly, we know that $Y$ can take on the values $1^2+1=2$ , $2^2+1=5$ , and $5^2+1=26$ . At these values of $Y$ , we would have the same probability as we did for the respective value of $X$ . That is,

$\mathbb P(Y=y)=\begin{cases}\dfrac c3&\text{for }y=2\\\\\dfrac c6&\text{for }y=5\\\\\dfrac c3&\text{for }y=26\\\\0&\text{otherwise}\end{cases}$

Part (5) is incomplete, so I'll stop here.

8 0

3 years ago

Square root of 1269 pl pl answer

Fiesta28 [93]

Square root of 1269 = 35.62

7 0

3 years ago

Read 2 more answers

In a school, 2/3 the students study a language.

xxTIMURxx [149]

Answer:

4:11