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3 years ago

Can someone help me with part B

Mathematics

1 answer:

leonid [27]3 years ago

4 0

B=5a-4
pick any a valve and replace it with a you will get b

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What is 12 over three equals four over one

Phoenix [80]

<h2><em>It's exactly what it says. 12/3 = 4. 4/1 = 4. Thus, 4 = 4. It's already solved.</em></h2>

4 0

3 years ago

Read 2 more answers

An umbrella sprinkler is positioned on a ceiling at a point whose x- coordinate is 0. Negative values of a indicate distances, i

vodka [1.7K]

Answer:

-1/4x² + 3

Step-by-step explanation:

Given that:

Path of sprinkler is modeled by the quadratic function :

w(x) = -1/4(x^2 -12 )

w(x) = height of the water, in meters, at position x

The height of ceiling as a constant or Coefficient = w(x)

Expanding w(x)

-1/4(x^2 -12 )

-1/4 * x² + 1/4 * 12

-1/4x² + 3

The height of ceiling is -1/4x² + 3

8 0

3 years ago

Mathematically verify the outlier(s) in the data set using the 1.5 rule.

statuscvo [17]

Given:

The data values are:

7, 8, 11, 13, 14, 14, 14, 15, 16, 16, 18, 19

To find:

The outliers of the given data set.

Solution:

We have,

7, 8, 11, 13, 14, 14, 14, 15, 16, 16, 18, 19

Divide the data set in two equal parts.

(7, 8, 11, 13, 14, 14), (14, 15, 16, 16, 18, 19)

Divide each parenthesis in two equal parts.

(7, 8, 11), (13, 14, 14), (14, 15, 16), (16, 18, 19)

Now,

$Q_1=\dfrac{11+13}{2}$

$Q_1=\dfrac{24}{2}$

$Q_1=12$

And

$Q_3=\dfrac{16+16}{2}$

$Q_3=\dfrac{32}{2}$

$Q_3=16$

The interquartile range is:

$IQR=Q_3-Q_1$

$IQR=16-12$

$IQR=4$

The data values lies outside the interval $[Q_1-1.5IQR,Q_3+1.5IQR]$ are known as outliers.

$[Q_1-1.5IQR,Q_3+1.5IQR]=[12-1.5(4),16+1.5(4)]$

$[Q_1-1.5IQR,Q_3+1.5IQR]=[12-6,16+6]$

$[Q_1-1.5IQR,Q_3+1.5IQR]=[6,22]$

All the data values lie in the interval [6,22]. So, there are no outliers.

Hence, the correct option is 4.

3 0

3 years ago

Ateacher uses 36 centimeters of tape to hang up 9 student projects. At that rate

Troyanec [42]

The teacher would need 40 centimeters of tape to hang 10 student projects.

Step-by-step explanation:

Given,

Tape used to hang projects = 36 centimeters

Number of projects hanged = 9

We will find unit rate;

9 projects = 36 centimeters

1 project = $\frac{36}{9}\ centimeters$

1 project = 4 centimeters

Therefore,

Tape needed to hang 10 projects = 4*10 = 40 centimeters

The teacher would need 40 centimeters of tape to hang 10 student projects.

Keywords: unit rate, multiplication

Learn more about unit rate at:

#LearnwithBrainly

3 0

3 years ago

If x/6=-3 then x= what?<br><br> Algebra 1

Gelneren [198K]

If x/6=-3 then x=-18

8 0

2 years ago