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Eddi Din [679]
3 years ago
10

It takes Lucas 3 hours to cycle from his home to his friends village but if travels 5 mph slower, it will take him 1.5 hours lon

ger. How far is the friends village from Lucas’s house
Mathematics
1 answer:
Alika [10]3 years ago
6 0

The distance from friend's village to Lucas's house is 45 miles

Step-by-step explanation:

Lets assume the distance from Lucas home to friend's village is x miles

The time taken to cycle from Lucas home to friend's village is 3 hrs....t₁

The time taken if travelling at a 5 mph reduced speed is 1.5 hrs longer......t₂

t₂=4.5 hrs

t₁= 3 hrs

Speed for cycle in first instance, where t₁=3 hrs and distance is x miles

s₁=d/t = x/3 mph

Speed for cycle if travelling at a slower speed,

s=\frac{x}{3}-5

Time taken in the second instance when travelling at a slower speed

=\frac{x}{\frac{x}{3}-5 }=4.5\\ \\\frac{3x}{x-15} =4.5\\\\\\3x=4.5(x-15)\\\\3x=4.5x-67.5\\\\\\67.5=4.5x-3x\\\\\\67.5=1.5x\\\\\\67.5/1.5=x\\\\\\45=x

The distance from friend's village to Lucas's house is 45 miles

Learn More

Formula for speed:brainly.com/question/13664094

Keywords: cycle,home, village, travels,slower,longer

#LearnwithBrainly

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Complete question :

Sweets are sold loose, or pre-packed in 120g bags.The 120g bags are £1.49 each.The loose sweets are £0.89 for 100g.

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Answer:

100 g bag

Step-by-step explanation:

For the 120g bag:

Price per gram :

Price / Numbe of gram

1.49 / 120 = 0.0124 gram per bag

For the 100 g bag :

0.89 / 100

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Since $0.0089 is < $0.0124

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Step-by-step explanation:

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3 years ago
For a sample of nequals37​, find the probability of a sample mean being less than 12 comma 751 or greater than 12 comma 754 when
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Answer:

50% probability of a sample mean being less than 12,751 or greater than 12,754

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation, which is also called standard error s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 12751, \sigma = 2.1, n = 37, s = \frac{2.1}{\sqrt{37}} = 0.3456

Find the probability of a sample mean being less than 12,751 or greater than 12,754

Less than 12,751

pvalue of Z when X = 12751.

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{12751 - 12751}{0.3456}

Z = 0

Z = 0 has a pvalue of 0.5.

50% probability of the sample mean being less than 12,751.

Greater than 12,754

1 subtracted by the pvalue of Z when X = 12,754.

Z = \frac{X - \mu}{s}

Z = \frac{12754 - 12751}{0.3456}

Z = 8.68

Z = 8.68 has a pvalue of 1

1 - 1 = 0

0% probability of the sample mean being greater than 12754

Less than 12,751 or greater than 12,754

50 + 0 =50

50% probability of a sample mean being less than 12,751 or greater than 12,754

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Move all terms not containing x to the right:

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