All categories

3 years ago

Laura is graphing an absolute-value function in the form

Mathematics

1 answer:

Ilya [14]3 years ago

7 0

Answer:

The equation of the function is;

y = |x-3| -3

Step-by-step explanation:

Here, we want to give the absolute-value function

For the minimum value to be -3, it means that k is negative

since the domain is the set of all real numbers, then we do not have any undefined value for the range y

So, the equation is ;

y = |x-3| - 3

You might be interested in

Find the value of t in the equation t+5+3t=1

Sedaia [141]

$t+5+3t=1\\t+3t+5=1\\4t+5=1\\4t+5-5=1-5\\4t=1-5\\4t =-4\\\\\boxed{\bf{t=-1}}$

6 0

3 years ago

Read 2 more answers

Help needed with both 9 & 10

fomenos

To find 9 you have to do Diameter x pi and when you do that you get 7.85

6 0

3 years ago

4) A garden is in the shape of a square with a perimeter of

Leni [432]

Answer:

It costs $154.88 to build the fences.

Step-by-step explanation:

Perimeter of square = $4 \times Side$

We are given that A garden is in the shape of a square with a perimeter of 56 feet.

So, $4 \times side = 56$

Side = $\frac{56}{4}=14 feet$

So, Side of square park = 14 feet

One fence is around the perimeter of the garden, whereas the second fence is 2 feet from the first fence on the outside.

Outer length = 14+2+2=18 feet

So, Perimeter of outer fence = $4 \times side = 4 \times 18 = 72 feet$

Cost of fencing 1 foot = $1.21

Cost of fencing 56 feet = $56 \times 1.21=67.76$

Cost of fencing 72 feet = $72 \times 1.21=87.12$

Total cost = 67.76+87.12=154.88

Hence It costs $154.88 to build the fences.

7 0

3 years ago

I don’t know how to solve number 8?

dlinn [17]

2C^2 + 6 - 5C + 4
=2C^2 -5C + 10

3 0

3 years ago

Read 2 more answers

If you are a dog lover, having your dog with you may reduce your stress level. Does having a friend with you reduce stress? To e

Pepsi [2]

The outlier of a dataset is a data element that is relatively far from the remaining data elements

99 is an outlier of pet group
See attachment for the parallel box plots

(a) Prove that 99 is an outlier for Pet

We have:

Pets: 58 64 65 68 69 69 69 70 70 72 76 79 85 86 99

$n = 15$

The quartiles positions are:

$Q_1 = \frac{n + 1}{4}$

$Q_1 = \frac{15 + 1}{4}$

$Q_1 = \frac{16}{4}$

$Q_1 = 4th$

$Q_3 = Q_1 \times 3$

$Q_3 = 4th \times 3$

$Q_3 = 12th$

So, we have:

$Q_1 = 4th$

$Q_3 = 12th$

From the pet group:

The data elements at the 4th and 12th positions are 68 and 79

So, we have:

$Q_1 = 68$

$Q_3= 79$

The lower and upper limits of the outlier are:

$L = Q_1 - 1.5 \times (Q_3 - Q_1)$

$U = Q_3 + 1.5 \times (Q_3 - Q_1)$

So, we have:

$L = 68 - 1.5 \times (79 - 68)$

$L = 51.5$

$U = 79+ 1.5 \times (79 - 68)$

$U = 95.5$

This means that data below 51.5 or above 95.5 are outliers.

Hence, 99 is an outlier because 99 is greater than 95.5

(b) The parallel box plot

The three groups are:

Pets: 58 64 65 68 69 69 69 70 70 72 76 79 85 86 99

Erlento: 88 80 80 81 92 87 88 81 82 80 87 92 87 80 82

Alone: 62 70 73 75 77 80 84 84 84 87 87 87 90 91 99

See attachment for the parallel box plots

Read more about box plots and outliers at:

brainly.com/question/14940764

5 0

3 years ago