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2 years ago

Find the value of X, I need help solving this.

Mathematics

2 answers:

kykrilka [37]2 years ago

7 0

Answer:

Step-by-step explanation:

add both expressions to equal 90 (bc of the right angle)- x+11+2x+1=90

when you solve, you get x=26

Veronika [31]2 years ago

6 0

Answer:

A) 26

Step-by-step explanation:

This is a right angle and the measure of a right angle is 90°

x+11+2x+1 = 90° add like terms

3x + 12 = 90° subtract 12 from both sides

3x = 78 divide by 3

x = 26

You might be interested in

Write an equation in slope intercept form of the line that passed through (7,2) and (2,12)

Sonbull [250]

Answer:

y= -2 x + 16

Step-by-step explanation:

bloop! hope it helps

6 0

3 years ago

Among a large group of patients recovering from shoulder injuries, it is found that 22% Visit both a physical therapist and a ch

EastWind [94]

Answer:

0.48

Step-by-step explanation:

First of all. let's recall probability formula needed for solving the problem.

We know that, If the events A and B are not mutually exclusive, the probability is:

<em>probability of event (A or B) = probability of event A + probability of event B – p(A and B). </em>

This formula will be used in calculations of this problem.

Let's consider that event A is patient visiting physical therapist; event B is patient visiting chiropractor.

P (A∩B) = 0.22 (visiting both physical therapist and chiropractor) ⇔ p(A and B)

P(B) = P(A) + 0.14 [ Probability patient visiting chiropractor is 0.14 more than probability visiting physical therapist]

Patients visiting none of these is 12%

Those who visit either therapist or chiropractor are amount to 1 - 0.12 = 0.88

Now we should use the formula mentioned in the beginning of the text:

0.88 = P(A) + P(B) - 0.22 ⇒ P(A) + P(B) = 1.1

If we replace P(B) by P(A) + 0.14 ⇒ 2P(A) + 0.14 = 1.1 ⇒ 2P(A) = 1.1 - 0.14 = 0.96

So, P(A) = 0.96/2 = 0.48 [Probability of patients visiting therapist]

7 0

2 years ago

What is the ones digit of a number if its cube ends in 7

Ostrovityanka [42]

Well, I have never seen a question posed this way, but let's check it out by trial and error.

1^3 = 3

2^3 = 8

3^3 = 27 Hey! There's one. And the ones digit ends in 3.

Let's try another number that ends in 3 and see if it works as well.

13^3 = 2197 Wow. It works again. I never noticed this before, so you taught me something new.

I will test one more.

33^3 = 35937 Bingo. I think we have a winner.

3 0

2 years ago

What is the equation for this line

pogonyaev

The equation of the line is $y=\frac{-1}{5} x-\frac{21}{5}$ .

Solution:

The points on the line are (–6, –3) and (4, –5).

$x_{1}=-6, y_{1}=-3, x_{2}=4, y_{2}=-5$

Slope of the line:

$$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$$m=\frac{-5-(-3)}{4-(-6)}$

$$m=\frac{-5+3}{4+6}$

$$m=\frac{-2}{10}$

$$m=\frac{-1}{5}$

Point-slope formula:

$y-y_{1}=m\left(x-x_{1}\right)$

$$y-(-3)=\frac{-1}{5}(x-(-6))$

$$y+3=\frac{-1}{5}(x+6)$

$$y+3=\frac{-1}{5} x-\frac{6}{5}$

Subtract 3 from both sides of the equation.

$$y=\frac{-1}{5} x-\frac{21}{5}$

The equation of the line is $y=\frac{-1}{5} x-\frac{21}{5}$ .

6 0

3 years ago

Conditional Distribution, Marginal Distribution, Joint Distribution. <br> What’s the difference?

lutik1710 [3]

Explanation:

Marginal distribution: This distribution gives the probability for each possible value of the Random variable ignoring other random variables. Basically, the values of other variables is not considered in the marginal distribution, they can be any value possible. For example, if you have two variables X and Y, the probability of X being equal to a value, lets say, 4, contemplates every possible scenario where X is equal to 4, independently of the value Y has taken. If you want the probability of a dice being a multiple of 3, you are interested that the dice is either 3 or 6, but you dont care if the dice is even or odd.

Conditional distribution: This distribution contrasts from the previous one in the sense that we are restricting the universe of events to specific condition for other variable, making a modification of our marginal results. If we know that throwing a dice will give us a result higher than 2, then to in order to calculate the probability of the dice being a multiple of 3 using that condition, we have two favourable cases (3 and 6) from 4 total possible results (3,4,5 and 6) discarding the impossible values (1 and 2) from this universe since they dont match the condition given (note that the restrictions given can also reduce the total of favourable cases).

The joint distribution calculates the probabilities for two different events (related to two different random variables) occuring simultaneously. If we want to calculate the joint probability of a dice being multiple of 3 and greater than 2 at the same time, our possible cases in this case are 3 and 6 from 6 possible results. We are not discarding 1 or 2 as possible results because we are not assuming, that the dice is greater than 2, that is another condition that we should met in the combination of events.

3 0

2 years ago