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Sarah wants to arrive at her friend's wedding at 3:00. The distance from Sarah's house to the wedding is 97 miles. Based on usua

atroni [7]

Answer:

First, you need to remember the relation:

speed = distance/time

We know that she needs to arrive at her friend's wedding at 3:00.

We also know that the total distance between her house and the wedding is 97 miles.

Now we need to find the total time that she needs to arrive at the wedding, so we can estimate at which hour she needs to start the travel.

The first 20 miles are driven att 30 miles per hour.

Then:

distance = 20mi

speed = 30mi/h

30mi/h = 20mi/T

T = 20mi/(30mi/h) = 0.67 hours.

The next 12 miles are driven at 45 mi/h

then:

45mi/h = 12mi/T

T = (12mi/(45mi/h)) = 0.27 hours

And the remainder distance is traveled at 65 mi/h

First, at this point she did travel 20mi + 12mi = 32mi

And we know that the total distance was 97mi, then the remaining distance is:

97mi - 32mi = 65mi

Then now she travels 65 miles at the speed of 65mi/h

then:

65mi/h = 65mi/(T)

T = (65mi)/(65mi/h) = 1 hour

Then the total time needed for the travel is:

Time = 0.67 hours + 0.27 hours + 1 hour = 1.94 hours.

knowing that 1 hour = 60 min

0.94 hours = 0.94*(60 min) = 56.4 min

time = 1 hour and 56.4 min

Meaning that she needs to leave her house around two hours before 3:00

So she would need to leave her house at around 1:03

8 0

3 years ago

For the data set below, compute the standard deviation. Round to the nearest hundredth. 2, 3, 5, 8 2.03 2.20 2.6 4.5

exis [7]

$The \ standard \ deviation = 
 \sqrt{ \frac{1}{n} \sum ( x_{i} - \mu )^2 } 
\\ where: \\
n \ is\ the \ number \ of \ elements \\
 x_{i} \ element \ number \ i \\
\mu \ is \ the \ mean \ of \ the \ elements 
$

The elements are :
 2,3,5,8,2.03,2.20,2.6,4.5
n = 8

μ = ( 2 + 3 + 5 + 8 + 2.03 + 2.20 + 2.6 + 4.5 )/8 = 3.66625

sum = ( 2 - 3.66625)² + ( 3 - 3.66625)² + ( 5 - 3.66625)² + ( 8 - 3.66625)² +
( 2.03 - 3.66625)² + ( 2.20 - 3.66625)² + ( 2.6 - 3.66625)² + ( 4.5 - 3.66625)²
 = 30.44

∴ The standard deviation = √ (30.44 / 8)
 ≈ 1.95 ( to the nearest hundredth )

7 0

4 years ago

what is the probability of randomly selecting a score between 1 and 2 standard deviations below the mean?

velikii [3]

Answer:

0.13591.

Step-by-step explanation:

We re asked to find the probability of randomly selecting a score between 1 and 2 standard deviations below the mean.

We know that z-score tells us that a data point is how many standard deviation above or below mean.

To solve our given problem, we need to find area between z-score of -2 and -1 that is $P(-2$ .