4% of what days is 56 days

If The 3rd term of an A.S is and 5th term is . What is an A.P?

guajiro [1.7K]

Answer:

you have to mention the 3rd term of AP and 5th term of AP. Then only we can find the AP.

8 0

3 years ago

2.1 & 2.2 Assessment: Vertex & Standard Form of Quadratic Functions

Brums [2.3K]

$~~~~~~\textit{vertical parabola vertex form} \\\\ y=a(x- h)^2+ k\qquad \begin{cases} \stackrel{vertex}{(h,k)}\\\\ \stackrel{"a"~is~negative}{op ens~\cap}\qquad \stackrel{"a"~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] ~\dotfill$

$\begin{cases} h=1\\ k=-9 \end{cases}\implies y=a(x-1)^2 - 9\qquad \textit{we also know that} \begin{cases} x=2\\ y=-7 \end{cases} \\\\\\ -7=a(2-1)^2-9\implies 2=a(1)^2\implies 2=a \\\\[-0.35em] ~\dotfill\\\\ ~\hfill {\Large \begin{array}{llll} y=2(x-1)^2 -9 \end{array}} ~\hfill$

6 0

1 year ago

According to a Los Angeles Times study of more than 1 million medical dispatches from to , the response time for medical aid var

-BARSIC- [3]

Answer:

A)Mean :10.65

Median =10.7

Mode : 10.7

B)Range = 3.5

Standard deviation :0.89916

C)The response time of 8.3 minutes should be considered an outlier in comparison to the other response times

Step-by-step explanation:

A)

Data : 11.8 ,10.3, 10.7, 10.6, 11.5, 8.3, 10.5, 10.9, 10.7, 11.2

$Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\Mean = \frac{11.8 +10.3+ 10.7+ 10.6+ 11.5+ 8.3+ 10.5+ 10.9+ 10.7+ 11.2}{10}$

Mean = 10.65

Median: The mid value of the data

Data in ascending order

8.3

10.3

10.5

10.6

10.7

10.9

11.2

11.5

11.8

n=10(even)

$Median = \frac{(\frac{n}{2})\text{th term}+(\frac{n}{2}+1)\text{th term}}{2}\\Median = \frac{(\frac{10}{2})\text{th term}+(\frac{10}{2}+1)\text{th term}}{2}\\Median = \frac{10.7+10.7}{2}\\Median = 10.7$

Mode : the most occurring frequency

10.7 is occurring twice while others are occurring once

So, Mode is 10.7

B) Range = Maximum - Minimum=11.8-8.3=3.5

Standard deviation : $\sqrt{\frac{\sum(x-\bar{x})^2}{n}}$

Standard deviation : $\sqrt{\frac{(11.8-10.65)^2+(10.3-10.65)^2+......+(10.9-10.65)^2+(10.7-10.65)^2+(11.2-10.65)^2}{10}}$

Standard deviation :0.89916

C)

8.3

,10.3

,10.5

,10.6

,10.7

,10.9

,11.2

,11.5

,11.8

For Q1 ( Median of lower quartile )

8.3

,10.3

,10.5

,10.6

,10.7

$Median = \frac{n+1}{2}\text{th term} =\frac{5+1}{2}=3 \text{rd term}=10.5$

For Q3( Median of Upper quartile )

10.7

,10.9

,11.2

,11.5

,11.8

$Median = \frac{n+1}{2}\text{th term} =\frac{5+1}{2}=3 \text{rd term}=11.2$

IQR = Q3-Q1=11.2-10.5=0.7

Range :(Q1-1.5IQR, Q3-1.5IQR)

Range : $(10.5-1.5 \times 0.7, 11.2-1.5 \times 0.7)$

Range :(9.45, 10.15)

8.3 does not lie in this interval

So, the response time of 8.3 minutes should be considered an outlier in comparison to the other response times

3 0

2 years ago

IF YOU GET IT RIGHT I WILL GIVE YOU 10 PTS AND BRAINLY CROWN SO PLEASE HELP ME!!!!!!

Alexandra [31]

I believe the answer to be C)81 ft2

3 0

3 years ago

Read 2 more answers

Write each of the following expression without using absolute value. <br> |6m-54|, if m<9

Simora [160]

9514 1404 393

Answer:

54 -6m

Step-by-step explanation:

For values of m < 9, the argument of the absolute value function is less than 0. Then the expression evaluates to ...

-(6m -54) = 54 -6m

3 0

2 years ago