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2 years ago

Sanjit drives the 45 km distance between York and Leeds he then drives a further 42 km from Leeds to Blackpool

2 answers:

8 0

45 km at 54 km/hr

45 / 54 = 5/6 

It took him 5/6 of an hour, or 50 minute 

61.4117 km/hr

I am Lyosha [343]2 years ago

3 0

Answer: His average speed for his total journey from York to Blackpool is approximately 61.41 km/h.

Step-by-step explanation:

$\text{ The average speed}=\frac{\text{Total distance covered}}{\text{Total time taken}}$

Since, The distance covered by him from York to Leeds = 45 km,

The speed when he covered this distance = 54 km/ h

Thus, the time taken by him in travelling from York to Leeds = 45/54 hours (Because, Time = Distance/speed)

Now, The distance covered by him from Leeds to Blackpool = 42 km,

The time taken by him in travelling from Leeds to Blackpool = 35 minutes = 35/60 hours

Hence, the total time taken by him in this journey

$=\frac{45}{54}+\frac{35}{60}=\frac{450+315}{540}=\frac{765}{540}=\frac{17}{12}$ hours

And, the total distance he covered= 45 + 42 = 87 km

Thus, His average speed

$=\frac{87}{17/12}= \frac{12\times 87}{17}=\frac{1044}{17}=61.4117647\approx 61.41\text{ km per hour}$

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Answer:

120*2.75*60/5280

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Step-by-step explanation:

So James is walking at 120 steps a minute, there are 60 minutes in one hour, and 5,280 feet in a mile. So, to find how quickly James is walking, we multiply how many steps he is taking a minute by how much distance each step is (120 * 2.75) that gives us how much distance in one minute, so we then multiply by 60 because there are 60 minutes in one hour. Then divide the entire equation by how many feet are in one mile (5,280) giving us the answer 3.75 miles per hour. And the equation 120*2.75*60/5280

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1 year ago

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A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 449 gram setting. It is

stealth61 [152]

Answer:

We accetp H₀

Step-by-step explanation:

Information:

Normal distribution

Population mean = μ₀ = 449

Population standard deviation σ unknown

Sample size n = 23 n < 30 we use t-student test

so n = 23 degree of fredom df = n - 1 df = 23- 1 df = 22

Sample mean μ = 448

Sample standard deviation s = 20

Significance level α = 0,05

1.-Hypothesis Test

Null hypothesis H₀ μ₀ = 449

Alternative hypothesis Hₐ μ₀ ≠ 449

Problem statement ask for determine decision rule for rejecting the null hypothesis. For rejecting the null hypothesis we have to get an statistic parameter wich implies that μ is bigger or smaller than μ₀

2.-Significance level α = 0,05 ; as we have a two tail test

α/2 = 0,025

Then from t - student table for df = 22 and 0,025 (two tail-test)

t(c) = ± 2.074

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t(s) = ( μ - μ₀ ) / s /√n

plugging in values

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t(s) = - 0.2398

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2 years ago

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