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Fittoniya [83]
2 years ago
8

to prepare for a downhill skiing competition ,roman completed three training sessions. the table shows his average time and dist

ance to each session. did romans rate in miles per hour increase from session to session write an agurement that can be used to defend your solution
Mathematics
1 answer:
Arte-miy333 [17]2 years ago
3 0

Answer:

  no

Step-by-step explanation:

Apparently, the three runs are ...

  (hours, miles) = (2/125, 9/10), (3/200, 11/12), (3/250, 17/30)

The speed in miles per hour is found by dividing miles by hours for each run. Of course dividing by a fraction is the same as multiplying by its inverse.

  session 1 speed = (9/10)(125/2) = 56.25

  session 2 speed = (11/12)(200/3) ≈ 61.11

  session 3 speed = (17/30)(250/3) ≈ 47.22

Roman's speed increased from the first session to the second, but <em>decreased for the third session</em>.

Roman's speed did not increase from session to session.

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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
2 years ago
Assess the extent to which bad road use has a direct impact on the physical,emotional,social,and economic aspects to the family
Dahasolnce [82]
It is true that bad road condition does have a direct impact on the physical, emotional and economic aspects to the family, the community and the country. Bad roads can cause accidents which directly impacts a person physically. Mentally it is very stressful for a person to drive on bad roads. He or she has to be extra careful regarding potholes and other damages that have happened to the road. Economically bad roads can be the reason for price rise of products as time required to move a product via road is more. Also the cost of transport increases drastically.



3 0
2 years ago
Read 2 more answers
Find the area of segment CFD given the following information: radius = 8in, area of ΔCBD = 25.9in2, and m∠CBD = 54° Round your a
Llana [10]

Answer:

A. 4.26 in^2

Step-by-step explanation:

Step 1: Find the area of the sector DBC. Here we have to use the formula.

Area of a sector = Central Angle/360 *πr^{2}

The area of the sector DBC = (54/360)*3.14*8^{2}

= 30.16

Step 2: Area of segment CFD = Area of the sector DBC - Area of the ΔCBD

= 30.17 - 25.9

Area of segment CFD = 4.27in^2

8 0
2 years ago
In parallelogram ABCD , diagonals AC⎯⎯⎯⎯⎯ and BD⎯⎯⎯⎯⎯ intersect at point E, AE=x2−16 , and CE=6x .
MAXImum [283]

Answer:

<h3>AC=96 units.</h3>

Step-by-step explanation:

We are given a parallelogram ABCD with diagonals AC and BD intersect at point E.

AE=x^2-16. , and CE=6x .

<em>Note: The diagonals of a parallelogram intersects at mid-point.</em>

Therefore, AE = EC.

Plugging expressions for AE and EC, we get

x^2-16=6x.

Subtracting 6x from both sides, we get

x^2-16-6x=6x-6x

x^2-6x-16=0

Factoriong quadratic by product sum rule.

We need to find the factors of -16 that add upto -6.

-16 has factors -8 and +2 that add upto -6.

Therefore, factor of x^2-6x-16=0 quadratic is (x-8)(x+2)=0

Setting each factor equal to 0 and solve for x.

x-8=0  => x=8

x+2=0  => x=-2.

We can't take x=-2 as it's a negative number.

Therefore, plugging x=8 in EC =6x, we get

EC = 6(8) = 48.

<h3>AC = AE + EC = 48+48 =96 units.</h3>
5 0
2 years ago
A cube-shaped water tank having 6 ft side lengths is being filled with water. The bottom is solid metal but the sides of the tan
Mekhanik [1.2K]

Answer:

0.033 ft

Step-by-step explanation:

Let g = 32.2 ft/s2 and h be the maximum height that water can be filled before the sides shatter.

Pressure is distributed from 0 to maximum at the bottom like the following equation:

P = \rho gh

So the force generated by water pressure on the side of the tank is

F = PA = \rho ghs/2

where s = 6ft is the side length of the tank. This force cannot be larger than 200lb

\rho ghs/2 \leq 200

62.4*32.2*h*6/2 \leq 200

h \leq 0.033 ft

4 0
2 years ago
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