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Keith_Richards [23]
3 years ago
13

Please could someone explain how to do question 2)a) (ii)

Mathematics
1 answer:
KonstantinChe [14]3 years ago
8 0

Answer:

15 kilometres

Step-by-step explanation:

Based on the wording of your question, I am assuming you already know that Paul ran 250 metres every 1 minute, and that the ratio is 250:1. To get the second part of the answer, convert metres to kilometres. 0.001*250= 0.25. Paul ran 0.25 kilometres per minute. The ratio of kilometres per minute is 0.25:1. Next, convert minutes to hours. Multiply the entire ratio by 60, as there are 60 minutes in an hour. The ratio of kilometeres per minute is 15:60, and the ratio of kilometres per hour is 15:1, meaning Paul ran 15 kilometres in one hour.

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The second one "Ben runs 3 miles a day"

Step-by-step explanation:

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Which value for x makes the sentence true?
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43/76

This is the answer

3 0
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Divide 25.84 divide by 4
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You are testing the claim that the mean GPA of night students is different from the mea GPA of day students. You sample 20 night
svetlana [45]

Answer:

As the calculated value of z does not lie in the critical region the null hypothesis is accepted that the GPA mean of the night students is the same as the GPA mean of the day students.

Step-by-step explanation:

Here n= 20

Sample mean GPA = x`= 2.84

Standard mean GPA = u= 2.55

Standard deviation = s=  0.45.

Level of Significance.= ∝ = 0.01

The hypothesis are formulated as

H0: u1=u2   i.e the GPA of night students is same as the mean GPA of day students

against the claim

Ha: u1≠u2

i.e the GPA of night students is different from the mea GPA of day students

For two tailed test  the critical value is  z ≥ z∝/2= ± 2.58

The test statistic

Z= x`-u/s/√n

z= 2.84-2.55/0.45/√20

z= 0.1441

As the calculated value of z does not lie in the critical region the null hypothesis is accepted that the GPA mean of the night students is the same as the GPA mean of the day students.

8 0
3 years ago
If a factory continuously dumps pollutants into a river at the rate of the quotient of the square root of t and 45 tons per day,
julsineya [31]
<h2>Hello!</h2>

The answer is:

The first option, the amount dumped after 5 days is 0.166 tons.

<h2>Why?</h2>

To solve the problem, we need to integrate the given expression and evaluate using the given time.

So, integrating we have:

\int\limits^5_0 {\frac{\sqrt{t} }{45} } \, dt=\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \, dt\\\\\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \ dt=\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt\\\\\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt=(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)\\\\(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)=(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)

(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)=(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)\\\\(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)=(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)\\\\(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)=(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})\\\\(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})=\frac{2}{135}*11.18-0=0.1656=0.166

Hence, we have that the amount dumped after 5 days is 0.166 tons.

Have a nice day!

5 0
3 years ago
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