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MatroZZZ [7]
3 years ago
11

A train starts from rest and accelerates uniformly until it has traveled 5.6 km and acquired a velocity of 42 m/s. What is the a

pproximate acceleration of the train during this time?
Mathematics
1 answer:
shepuryov [24]3 years ago
4 0

Answer:

0.16 m/s²

Step-by-step explanation:

Given:

Δx = 5600 m

v₀ = 0 m/s

v = 42 m/s

Find: a

v² = v₀² + 2aΔx

(42 m/s)² = (0 m/s)² + 2a (5600 m)

a = 0.16 m/s²

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stira [4]

Answer:

The 't' test statistic = 1.46 < 1.69

The test of hypothesis is H 0 :μ = 800 is accepted

A sample of 30 bulbs are found came from average µ= 800

Step-by-step explanation:

Step 1:-

Given population of mean μ = 800

given size of small sample n =30

sample standard deviation 'S' = 45

Mean value of the sample χ = 788

Null hypothesis H_{0} =µ  =800

alternative hypothesis  H_{1} = µ ≠ 800

<u>Step 2</u>:-

The 't' test statistic t =  \frac{x-μ}{\frac{sig}{\sqrt{n} } }

                             t = \frac{788-800}{\frac{45}{\sqrt{30} } }

                       t = \frac{12}{8.215} = 1.4607

<u>Step 3</u>:-

The degrees of freedom γ = n-1 = 30-1 =29

From "t" value from table at 0.05 level of significance ( t = 1.69)

The calculated value t = 1.4607 < 1.69

Therefore The null hypothesis H_{0} ' is accepted.

<u>conclusion</u>:-

A sample of 30 bulbs are came found from average µ= 800

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Step-by-step explanation:

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