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andrew-mc [135]
3 years ago
6

How do you use graphing lines y=mx + b

Mathematics
1 answer:
melisa1 [442]3 years ago
7 0
M is the slope (rise over run). b is the Y-intercept (on the y axis). plot the y point then use your slope to find the line 
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Three students are arguing over who can run the fastest. Student A says that she can run 2/3 mile in six minutes. Student B says
Liula [17]

Answer:

student A- can run 2/3 mile in six minutes

1/9 mile/min

student B- can run 1/2 mile in 4 minutes

Speed=1/8 mile/min

student C- can run 4/5 mile in 8 minutes

speed=1/10 mile/min

Step-by-step explanation:

5 0
3 years ago
What is 4,321,109,432 rounded to the nearest ten million
SOVA2 [1]
To round to the nearest ten million, you look at the millions digit. In this case, the millions digit is 1, which is less than 5, so you round the number down to 4,320,000,000.
5 0
3 years ago
Read 2 more answers
Identify whether each variable is categorical or continuous.
balandron [24]

Answer:

u33u3373737

Step-by-step explanation:

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3 0
3 years ago
If you travel 70000 M at a rate of 25 m/s and , how long does it take to complete the trip?
ivolga24 [154]

Answer:

Time=2800 seconds.

Time in minutes= 46.67 minutes

Time in hours= 0.778 hours

Step-by-step explanation:

Distance traveled= 70000 m

Rate or speed of travel= 25m/s

To calculate the time for the travel, we multiply the rate of travel by the distance covered or traveled.

Rate= distance/time

Time= distance/rate

Time= 70000/25

Time=2800 seconds.

Time in minutes= 2800/60

Time in minutes= 46.67 minutes

Time in hours = 2800/(60*60)

Time in hours= 2800/360

Time in hours= 0.778 hours

8 0
3 years ago
An example of an early application of statistics was in the year 1817. A study of chest circumference among a group of Scottish
otez555 [7]

Answer:

Step-by-step explanation:

Hello!

The variable of interest is X: chest circumference of a Scottish man.

X≈N(μ;δ²)

μ= 40 inches

δ= 2 inches

The empirical rule states that

68% of the distribution lies within one standard deviation of the mean: μ±δ= 0.68

95% of the distribution lies within 2 standard deviations of the mean: μ±2δ= 0.95

99% of the distribution lies within 3 standard deviations of the mean: μ±3δ= 0.99

a)

The 58% that falls closest to the mean can also be referred to as the middle 58% of the distribution, assuming that both values are equally distant from the mean.

P(a≤X≤b)= 0.58

If 1-α= 0.58, then the remaining proportion α= 0.42 is divided in two equal tails α/2= 0.21.

The accumulated proportion until "a" is 0.21 and the accumulated proportion until "b" is 0.21 + 0.58= 0.79 (See attachment)

P(X≤a)= 0.21

P(X≤b)= 0.79

Using the standard normal distribution, you can find the corresponding values for the accumulated probabilities, then using the information of the original distribution:

P(Z≤zᵃ)= 0.21

zᵃ= -0.806

P(Z≤zᵇ)= 0.79

zᵇ= 0.806

Using the standard normal distribution Z= (X-μ)/δ you "transform" the values of Z to values of chest circumference (X):

zᵃ= (a-μ)/δ

zᵃ*δ= a-μ

a= (zᵃ*δ)+μ

a= (-0.806*2)+40= 38.388

and

zᵇ= (b-μ)/δ

zᵇ*δ= b-μ

b= (zᵇ*δ)+μ

b= (0.806*2)+40= 41.612

58% of the chest measurements will be within 38.388 and 41.612 inches.

b)

The measurements of the 2.5% men with the smallest chest measurements, can also be interpreted as the "bottom" 2.5% of the distribution, the value that separates the bottom 2.5% of the distribution from the 97.5%, symbolically:

P(X≤b)= 0.025 (See attachment)

Now you have to look under the standard normal distribution the value of z that accumulates 0.025 of the distribution:

P(Z≤zᵇ)= 0.025

zᵇ= -1.960

Now you reverse the standardization to find the value of chest circumference:

zᵇ= (b-μ)/δ

zᵇ*δ= b-μ

b= (zᵇ*δ)+μ

b= (-1.960*2)+40= 36.08

The chest measurement of the 2.5% smallest chest measurements is 36.08 inches.

c)

Using the empirical rule:

95% of the distribution lies within 2 standard deviations of the mean: μ±2δ= 0.95

(μ-2δ) ≤ Xc ≤ (μ+2δ)=0.95 ⇒ (40-4) ≤ Xc ≤ (40+4)= 0.95 ⇒ 36 ≤ Xc ≤ 44= 0.95

d)

The measurements of the 16% of the men with the largest chests in the population or the "top" 16% of the distribution:

P(X≥d)= 0.16

P(X≤d)= 1 - 0.16

P(X≤d)= 0.84

First, you look for the value that accumulates 0.84 of probability under the standard normal distribution:

P(Z≤zd)= 0.84

zd= 0.994

Now you reverse the standardization to find the value of chest circumference:

zd= (d-μ)/δ

zd*δ= d-μ

d= (zd*δ)+μ

d= (0.994*2)+40= 41.988

The measurements of the 16% of the men with larges chess are at least 41.988 inches.

I hope this helps!

8 0
3 years ago
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