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Tema [17]
2 years ago
5

Three drivers competed in the same fifteen drag races. The mean and standard deviation for the race times of each of the drivers

are given.
- Driver A had a mean race time of 4.01 seconds and a standard deviation of 0.05 seconds.

- Driver B had a mean race time of 3.96 seconds and a standard deviation of 0.12 seconds.

- Driver C had a mean race time of 3.99 seconds and a standard deviation of 0.19 seconds.


Which driver do you predict will win the next drag race? Support your prediction using the mean and standard deviation.
Mathematics
1 answer:
Monica [59]2 years ago
8 0

Answer:

Driver B

Step-by-step explanation:

From the given options, Driver B will win the next race because the previous result shows that Driver B has the fastest race time due to the fact that it recorded the lowest mean race time.

Similarly, the standard deviation of 0.12 seconds illustrates that the driver keep-up with the bearest minimum time frame all through the entire race time.

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sveta [45]
10.7
23.7-10=10
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2 years ago
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The success of an airline depends heavily on its ability to provide a pleasant customer experience. One dimension of customer se
tankabanditka [31]

Answer:

A)H0: μ1 − μ2 = 0

Ha: μ1 − μ2 ≠ 0

(b)  Means

Company A___50.6___ min.

Company B___52.75___ min.

c)The t-value is -0.30107.

The p-value is 0 .764815.

C) Do not reject H0. There is no statistical evidence that one airline does better than the other in terms of their population mean delay time.

Step-by-step explanation:

A)H0: μ1 − μ2 = 0

i.e there is no difference between the means of delayed flight for two different airlines

Ha: μ1 − μ2 ≠ 0

i.e there is a difference between the means of delayed flight for two different airlines

(b)

Company A___50.6___ min.

Company B___52.75___ min.

Mean of Company A = x`1= ∑x/n =34+ 59+ 43+ 30+ 3+ 32+ 42+ 85+ 30+ 48+ 110+ 50+ 10+ 26+ 70+ 52+ 83+ 78+ 27+ 70+ 27+ 90+ 38+ 52+ 76/25

= 1265/25= 50.6

Mean of Company B = x`2= ∑x/n =

=46+ 63+ 43+ 33+ 65+ 104+ 45+ 27+ 39+ 84+ 75+ 44+ 34+ 51+ 63+ 42+ 34+ 34+ 65+ 64/20

= 1055/20= 52.75

<u><em>Difference Scores Calculations</em></u>

Company A

Sample size for Company A= n1= 25

Degrees of freedom for company A= df1 = n1 - 1 = 25 - 1 = 24

Mean for Company A= x`1=  50.6

Total Squared Difference (x-x`1) for Company A= SS1: 16938

s21 = SS1/(n1 - 1) = 16938/(25-1) = 705.75

Company B

Sample size for Company B= n1= 20

Degrees of freedom for company B= df2 = n2 - 1 = 20 - 1 = 19

Mean for Company B= x`2=  52.75

Total Squared Difference (x-x`2) for Company B= SS2=7427.75

s22 = SS2/(n2 - 1) = 7427.75/(20-1) = 390.93

<u><em>T-value Calculation</em></u>

<u><em>Pooled Variance= Sp²</em></u>

Sp² = ((df1/(df1 + df2)) * s21) + ((df2/(df2 + df2)) * s22)

Sp²= ((24/43) * 705.75) + ((19/43) * 390.93) = 566.65

s2x`1 = s2p/n1 = 566.65/25 = 22.67

s2x`2 = s2p/n2 = 566.65/20 = 28.33

t = (x`1 - x`2)/√(s2x`1 + s2x`2) = -2.15/√51 = -0.3

The t-value is -0.30107.

The total degrees of freedom is = n1+n2- 2= 25+20-2=43

The critical region for two tailed test at significance level ∝ =0.05 is

t(0.025) (43) = t > ±2.017

Since the calculated value of t=  -0.30107.  does not fall in the critical region t > ±2.017, null hypothesis is not rejected that is there is no difference between the means of delayed flight for two different airlines.

The p-value is 0 .764815. The result is not significant at p < 0.05.

7 0
1 year ago
A $55 pair of shoes was discounted 15 percent, making the sale price $46.75. The discounted price was then discounted again by 1
BigorU [14]
46.75 - 0.10(46.75) = 46.75 - 4.675 = 42.075 rounds to 42.08
ur simply taking ur discounted price of $ 46.75 and subtracting 10% of 46.75....leaving u with a final price of 42.08
5 0
2 years ago
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In the month of June, the temperature in Johannesburg, South Africa, varies over the day in a periodic way that can be modeled a
UNO [17]

Answer:

T = - 7.5 Cos π/12( t - 4 ) + 10.5

Step-by-step explanation:

Given that the

Maximum temp. = 18 degree Celsius

Minimum temp. = 3 degree Celsius

The half way between 10 am and 10 pm is 4 am

The sine and cosine functions can be used to model fluctuations  in temperature data through out the year. An equation that can be used to model these data is of the form:

T = A cos B(t - C) + D, where A,B,C,D, are constants, T is the  temperature in °C and t is the hour (1–24)

A = amplitude = (Tmax - Tmin)/2

A = (3 - 18)/2 = - 15/2 = -7.5 ( note : after midnight)

B = 2π/24 = π/12

C = units translated to the right

C = 4

D = ymin + amplitude = units translated up

D = 7.5 + 3 = 10.5

The formula of the trigonometric function that models the temperature T in Johannesburg t hours after midnight we be

T = - 7.5 Cos π/12( t - 4 ) + 10.5

3 0
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a.Karim is ordering video games. Each game costs $39.99, and there is a $4.85 shipping charge per order. How much will it cost h
Cerrena [4.2K]
Each game will cost him 39.99+4.85= $44.84
7 0
2 years ago
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