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2 years ago

Select all the correct graphs. Choose the graphs that indicate equations with no solution

Mathematics

2 answers:

galben [10]2 years ago

7 0

Answer:

You didnt post any graphs, also the no solution graphs are graphs that have lines that never touch

Step-by-step explanation:

Kobotan [32]2 years ago

4 0

Answer:

i need a graph to help you can u plz post one

Step-by-step explanation:

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A carpenter is building a rectangular bookcase with diagonal braces across the back, as shown. The carpenter knows that angle AD

Valentin [98]

Refer to the diagram shown below.

Let x = m∠ADB.

Because m∠BDC is 32° greater than m∠ADB, therefore
m∠BDC = x + 32°

Each angle of a rectangle is 90°, therefore
x + (x+32) = 90
2x + 32 = 90
2x = 58
x = 29°
x+32 = 61°

Answer:
m∠BDC = 61°
m∠ADB = 29°

8 0

2 years ago

Write a two step equation involving division and addition that has a solution of x = 25

Tamiku [17]

Start with the solution
x=-25
use the reverse of what they asked
they wanted division and addition
we use multiplication and subtract

ok
x=-25
we can use subtraction and multiplication in any order
minus 10 from both sides
x-10=-35
multiply both sides by 4
4(x-10)=-140
4x-40=-140 is a possible equation

6 0

2 years ago

PQ= 2x +1 and QR= 5x - 44; find PQ

Lena [83]

2x+1=5x-44 that would be your equation

next start to simplify

subtract 1 from both sides so you are left with 2x=5x-45

then subtract 5x from both sides and you have -3x=-45

then finally divide -3 from -45 to get x=15

6 0

2 years ago

Read 2 more answers

According to a Los Angeles Times study of more than 1 million medical dispatches from 2007 to 2012, the 911 response time for me

Reil [10]

Answer:

a) $\bar X=10.65$

$Median =\frac{10.7+10.7}{2}=10.7$

$Mode= 10.7$

b) $Range = 11.8-8.3=3.5$

$s= 0.948$

c) $IQR = Q_3 -Q_1 = 11.05-10.55=0.5$

And we can find the usual limits with:

$Lower = Q_1 -1.5 IQR = 10.55 -1.5*0.5=9.8$

$Upperer = Q_3 +1.5 IQR = 10.55 +1.5*0.5=11.3$

And since 8.3 <9.8 we can consider this value too low or as an outlier for this case.

d) The mean for this case was 10.65 and the usual values are between 9.8 and 11.3, so as we can see all are above the specified value of 6 minutes, and we can conclude that the times are not satisfying the quality standards for this case.

And they should be considered apply some strategies to reduce the response time, adding more stations around points selected at the city could be useful in order to reduce the response time.

Step-by-step explanation:

We have the following data:

11.8 10.3 10.7 10.6 11.5 8.3 10.5 10.9 10.7 11.2

Part a

We can calculate the sample mean with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And if we replace we got: $\bar X=10.65$

For the median we need to sort the values on increasing order and we have:

8.3 10.3 10.5 10.6 10.7 10.7 10.9 11.2 11.5 11.8

Since n =10 we can calculate the median as the average between the 5th and 6th position of the dataset ordered and we got:

$Median =\frac{10.7+10.7}{2}=10.7$

The mode would be the most repeated value on this case:

$Mode= 10.7$

Part b

The range is defined as $Range =Max-Min$ and if we replace we got:

$Range = 11.8-8.3=3.5$

We can calculate the standard deviation with the following formula:

$s= sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And if we replace we got:

$s= 0.948$

Part c

For this case we can use the IQR method in order to determine if 8.3 is an outlier or not.

We can calculate the first quartile with these values: 8.3 10.3 10.5 10.6 10.7 10.7 and $Q_1= \frac{10.6+10.7}{2}=10.55$

And for the Q3 we can use: 10.7 10.7 10.9 11.2 11.5 11.8 and we got $Q_3 = \frac{10.9+11.2}{2}=11.05$

Then we can find the IQR like this:

$IQR = Q_3 -Q_1 = 11.05-10.55=0.5$

And we can find the usual limits with:

$Lower = Q_1 -1.5 IQR = 10.55 -1.5*0.5=9.8$

$Upperer = Q_3 +1.5 IQR = 10.55 +1.5*0.5=11.3$

And since 8.3 <9.8 we can consider this value too low or as an outlier for this case.

Part d

The mean for this case was 10.65 and the usual values are between 9.8 and 11.3, so as we can see all are above the specified value of 6 minutes, and we can conclude that the times are not satisfying the quality standards for this case.

And they should be considered apply some strategies to reduce the response time, adding more stations around points selected at the city could be useful in order to reduce the response time.

6 0

2 years ago

Katherine is landscaping her home with juniper trees and pansies. She wants to arrange 15 pansies around each of 8 trees. Each t

Doss [256]

Let
X-----------------> number of pansies
y-----------------> number of trees

we know that
x=15*8----------> x=120 pansies
y=8 trees

cost of each trees is----------> $<span>20.75
</span>cost of each pansies is------> $2.50/6------> $5/12

[<span>expression to find Katherine’s final cost]=[cost trees]+[cost pansies]
</span>[cost trees]=y*$20.75
[cost pansies]=x*($5/12)

[expression to find Katherine’s final cost]=y*($20.75)+x*($5/12)
[expression to find Katherine’s final cost]=8*($20.75)+120*($5/12)

[expression to find Katherine’s final cost]=$166+$50
[expression to find Katherine’s final cost]=$216

the answer is
[expression to find Katherine’s final cost]=y*($20.75)+x*($5/12)
[expression to find Katherine’s final cost]=8*($20.75)+120*($5/12)

Katherine’s final cost is $216

5 0

2 years ago

Read 2 more answers