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2 years ago

There were three parts to Rita's race. She ran the first part, which was 4/9

Mathematics

1 answer:

Igoryamba2 years ago

5 0

The second part of Rita's race is about 1800 meters at a speed of 150 meters per minute.

<h3>Speed</h3>

let

distance covered in first Part = d1
distance covered in second Part = d2
distance covered in third Part = d3
speed covered in first Part = s1
speed covered in second Part = s2
speed covered in third Part = s3

Distance in second Part:

d₃ = 15 x 300

= 4500 m

d₂ = 2/5 × d₃

= 2/5 x 4500

d₂ = 1800 meters

Speed in second part:

S₂ = 1800 m / 12 min

S₂ = 150 meters per minute

Learn more about distance:

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A company surveyed 2400 men where 1248 of the men identified themselves as the primary grocery shopper in their household. a) E

polet [3.4K]

Answer:

a) With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

b) The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

c) $\alpha =1-0.98=0.02$

Step-by-step explanation:

If np' and n(1-p') are higher than 5, a confidence interval for the proportion is calculated as:

$p'-z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }\leq p\leq p'+z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }$

Where p' is the proportion of the sample, n is the size of the sample, p is the proportion of the population and $z_{\alpha/2}$ is the z-value that let a probability of $\alpha/2$ on the right tail.

Then, a 98% confidence interval for the percentage of all males who identify themselves as the primary grocery shopper can be calculated replacing p' by 0.52, n by 2400, $\alpha$ by 0.02 and $z_{\alpha/2}$ by 2.33

Where p' and $\alpha$ are calculated as:

$p' = \frac{1248}{2400}=0.52\\\alpha =1-0.98=0.02$

So, replacing the values we get:

$0.52-2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\leq p\leq 0.52+2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\\0.52-0.0238\leq p\leq 0.52+0.0238\\0.4962\leq p\leq 0.5438$

With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

Finally, the level of significance is the probability to reject the null hypothesis given that the null hypothesis is true. It is also the complement of the level of confidence. So, if we create a 98% confidence interval, the level of confidence $1-\alpha$ is equal to 98%

It means the the level of significance $\alpha$ is:

$\alpha =1-0.98=0.02$

4 0

3 years ago

How long does it take for the account to be worth about $750?

pochemuha

Answer:

if look at the graph and find the point on the graph where the y-value is $750 then you will find your answer. I can give u any more than that cuz i cant really read the graph. look where the y-value is $750 and write the x-value that matches it.

Step-by-step explanation:

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3 years ago

Two liabraries donated to a local school. One library donated 786 books and the other library donated 1245 books. The school div

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Each classroom has received 184 & 8 books are being left

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3 years ago

Can you help me please?

irga5000 [103]

Answer:

36.6