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Brums [2.3K]
3 years ago
8

The cost of a circular table is directly proportional to the square of the radius. A circular table with a radius of 50cm costs

£60. What is the cost of a circular table with a radius of 75cm? Show all your working
Mathematics
1 answer:
Lana71 [14]3 years ago
7 0

Answer:

£135 is the correct answer.

Step-by-step explanation:

Let C be the cost of table.

And let R be the radius of table.

Cost of table is directly proportional to square of radius.

As per question statement:

C\propto R^{2} or

C=a\times R^2  ....... (1)

where a is the constant to remove the \propto sign.

It is given that

C_1 = £60 and R_1 = 50\ cm

C_2 = ? when R_2= 75\ cm

Putting the values of C_1 and R_1 in equation (1):

60=a \times 50^2  ....... (2)

Putting the values of C_2 and R_2 in equation (1):

C_2=a \times 75^2  ....... (3)

Dividing equation (2) by (3):

\dfrac{60}{C_2}= \dfrac{a \times 50^2}{a \times 75^2}\\\Rightarrow \dfrac{60}{C_2}= \dfrac{50^2}{75^2}\\\Rightarrow \dfrac{60}{C_2}= \dfrac{2^2}{3^2}\\\Rightarrow \dfrac{60}{C_2}= \dfrac{4}{9}\\\Rightarrow C_2 = 15 \times 9 \\\Rightarrow C_2 = 135

So, <em>£135 is the correct answer</em>.

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A door delivery florist wishes to estimate the proportion of people in his city that will purchase his flowers. Suppose the true
kvv77 [185]

Answer:

99.74% probability that the sample proportion will be less than 0.1

Step-by-step explanation:

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Binomial probability distribution

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Problems of normally distributed samples can be 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:

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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.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

n = 276, p = 0.06

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\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{276*0.06*0.94} = 3.9454

What is the probability that the sample proportion will be less than 0.1

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

Z = \frac{27.6 - 16.56}{3.9454}

Z = 2.8

Z = 2.8 has a pvalue of 0.9974

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