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KIM [24]
2 years ago
8

A car leaving a stop sign accelerates constantly from a speed of 0 feet per second to reach a speed of 44 feet per second. The d

istance of the car from the stop sign, d d , in feet, at time t t , in seconds, can be found using the equation d=1.1t2 d = 1 . 1 t 2 . What is the average speed of the car, in feet per second, between t=2 t = 2 and t=5 t = 5 ?
Mathematics
1 answer:
dusya [7]2 years ago
8 0

Answer:

c

Step-by-step explanation:

The distance of the car from the stop sign, d , in feet, at time t , in seconds, can be found using the equation

The average rate of change of a function f(x) on [a,b] is

We need to find the average speed of the car, in feet per second, between t=2 and t=5.

At t=2,

At t=5,

The average speed of the car, in feet per second, between t=2 and t=5 is

The average speed of the car, in feet per second, between t=2 and t=5 is 7.7 feet per second.

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Marcia bought a 50 foot roll she used two 8 and 5/6 how much if left​
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Answer:

vvvvv

Step-by-step explanation:

I suspect two 8 means 8x2, but if it means 2+8, I'll add another answer.

1. 8x2= 16 + 5/6(.83) = 16.83 = 50-16.83= 33.17

ORRRRR

2. 2+8=10+.83= 10.83 = 39.17

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The formula P = 6s gives the formula for the perimeter of a regular hexagon with side length s. What is the perimeter of a regul
stellarik [79]
YOU NEED A BETTER SCHOOL
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3 years ago
Go through my questions there a surprise:)
gtnhenbr [62]

Answer:

I will do that buddy. I'll have fun

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2 years 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

10.7

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

,10.7

,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
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ValentinkaMS [17]

Answer:

thanks:)

Step-by-step explanation:

hehe

5 0
3 years ago
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