Use the multiplication property of equality to multiply both sides of the equation by 1/6 to get: (1/6)6x=(1/6)48.

If using the method of completing the square to solve the quadratic

In-s [12.5K]

Answer:

100

Step-by-step explanation:

20/2= 10 and 10 squared is 100 but you have to move the -6 to the opposite side

4 0

3 years ago

1/2 of the people in the clinic are men, 1/3 are elderly, how many people are not elderly or men?

Nostrana [21]

1/6 of the people are not elderly or men

<h3>How to determine the proportion?</h3>

The given parameters are:

Men = 1/2

Elderly = 1/3

The number of people that are not elderly or men are:

People = 1 - Men - Elderly

So, we have:

People = 1 - 1/2 - 1/3

Evaluate

People = 1/6

Hence, 1/6 of the people are not elderly or men

Read more about proportion at:

brainly.com/question/1781657

#SPJ1

7 0

2 years ago

Dylan delivers parcels.

vovikov84 [41]

Answer:

Dylan delivered 140 parcels on Wednesday.

Step-by-step explanation:

On Wednesday:

On Wednesday, he delivered x parcels.

Thursday:

10% more than Wednesday, so 100 + 10 = 110% of x = 1.1x

Friday:

50% pless than on Thursday, so 100 - 50 = 50% of 1.1x = 0.5*1.1*x.

THis is equals to 77. So

$0.5*1.1x = 77$

$x = \frac{77}{0.5*1.1}$

$x = 140$

Dylan delivered 140 parcels on Wednesday.

5 0

3 years ago

Identify the three similar right triangles in the given diagram.

irga5000 [103]

Answer:

B. ΔABD, ΔADC, ΔDBC

Step-by-step explanation

Step -1 In ΔABD and ΔADC (from figure).

∠DAB=∠CAD (common in both triangles) ,

∠DBA=∠CDA =90 degree, and

∠BDA=∠DCA (rest angle of the two triangles).

therefore ΔABD similar to ΔADC (by AAA similarity theorem).

Step -2 In ΔDBC and ΔADC (from figure).

∠DCB=∠ACD (common in both triangles) ,

∠DBC=∠ADC =90 degree, and

∠CDB=∠CAD (rest angle of the two triangles).

therefore ΔDBC similar to ΔADC (by AAA similarity theorem).

Step -3 In ΔABD and ΔDBC (from figure).

∠BDA=∠BCD (because , ∠ACD=ADB from stap-1 and ∠ACD=∠BCD from figure) ,

∠DBA=∠CBD =90 degree, and

∠BAC=∠BDC (rest angle of the two triangles).

therefore ΔABD similar to ΔDBC (by AAA similarity theorem).

In the above step- ΔABD similar to ΔADC, ΔDBC similar to ΔADC and ΔABD similar to ΔDBC.

Hence ΔABD, ΔADC, ΔDBC similar to each other in the given figure.

4 0

3 years ago

Test scores of the student in a school are normally distributed mean 85 standard deviation 3 points. What's the probability that

Mrrafil [7]

Answer:

The probability that a random selected student score is greater than 76 is $\\ P(x>76) = 0.99865$ .

Step-by-step explanation:

The Normally distributed data are described by the normal distribution. This distribution is determined by two parameters, the population mean $\\ \mu$ and the population standard deviation $\\ \sigma$ .

To determine probabilities for the normal distribution, we can use the standard normal distribution, whose parameters' values are $\\ \mu = 0$ and $\\ \sigma = 1$ . However, we need to "transform" the raw score, in this case x = 76, to a z-score. To achieve this we use the next formula:

$\\ z = \frac{x - \mu}{\sigma}$ [1]

And for the latter, we have all the required information to obtain z. With this, we obtain a value that represent the distance from the population mean in standard deviations units.

<h3>The probability that a randomly selected student score is greater than 76</h3>

To obtain this probability, we can proceed as follows:

First: obtain the z-score for the raw score x = 76.

We know that:

$\\ \mu = 85$

$\\ \sigma = 3$

$\\ x = 76$

From equation [1], we have:

$\\ z = \frac{76 - 85}{3}$

Then

$\\ z = \frac{-9}{3}$

$\\ z = -3$

Second: Interpretation of the previous result.

In this case, the value is three (3) standard deviations below the population mean. In other words, the standard value for x = 76 is z = -3. So, we need to find P(x>76) or P(x>-3).

With this value of $\\ z = -3$ , we can obtain this probability consulting the cumulative standard normal distribution, available in any Statistics book or on the internet.

Third: Determination of the probability P(x>76) or P(x>-3).

Most of the time, the values for the cumulative standard normal distribution are for positive values of z. Fortunately, since the normal distributions are symmetrical, we can find the probability of a negative z having into account that (for this case):

$\\ P(z>-3) = 1 - P(z>3) = P(z$

Then

Consulting a cumulative standard normal table, we have that the cumulative probability for a value below than three (3) standard deviations is:

$\\ P(z$

Thus, "the probability that a random selected student score is greater than 76" for this case (that is, $\\ \mu = 85$ and $\\ \sigma = 3$ ) is $\\ P(x>76) = P(z>-3) = P(z$ .

As a conclusion, more than 99.865% of the values of this distribution are above (greater than) x = 76.

We can see below a graph showing this probability.

As a complement note, we can also say that:

$\\ P(z3)$

Which is the case for the probability below z = -3 [P(z<-3)], a very low probability (and a very small area at the left of the distribution).

5 0

3 years ago