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77julia77 [94]
3 years ago
5

The "off-peak" weekly average cost of hiring a caravan in Wales is 520. In July the average cost increases to 800. calculate the

percentage increase in cost. Give your answer to 2 significant figures.
Mathematics
2 answers:
Artemon [7]3 years ago
8 0
For this case you can solve the problem by applying a simple rule of three in which we have
 520 ---> 100%
 800 ---> x

 Clearing x we have
 x = (800/520) * 100 = 154%
 the percentage increase in cost was
 154-100 = 54%
 answer
 the percentage increase in cost was 54%
pochemuha3 years ago
5 0

Answer:

54%

Step-by-step explanation:

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Read 2 more answers
According to a Yale program on climate change communication survey, 71% of Americans think global warming is happening.† (a) For
SpyIntel [72]

Answer:

a) 0.2741 = 27.41% probability that at least 13 believe global warming is occurring

b) 0.7611 = 76.11% probability that at least 110 believe global warming is occurring

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

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:

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.

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:

p = 0.71

(a) For a sample of 16 Americans, what is the probability that at least 13 believe global warming is occurring?

Here n = 16, we want P(X \geq 13). So

P(X \geq 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 13) = C_{16,13}.(0.71)^{13}.(0.29)^{3} = 0.1591

P(X = 14) = C_{16,14}.(0.71)^{14}.(0.29)^{2} = 0.0835

P(X = 15) = C_{16,15}.(0.71)^{15}.(0.29)^{1} = 0.0273

P(X = 16) = C_{16,16}.(0.71)^{16}.(0.29)^{0} = 0.0042

P(X \geq 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16) = 0.1591 + 0.0835 + 0.0273 + 0.0042 = 0.2741

0.2741 = 27.41% probability that at least 13 believe global warming is occurring

(b) For a sample of 160 Americans, what is the probability that at least 110 believe global warming is occurring?

Now n = 160. So

\mu = E(X) = np = 160*0.71 = 113.6

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{160*0.71*0.29} = 5.74

Using continuity correction, this is P(X \geq 110 - 0.5) = P(X \geq 109.5), which is 1 subtracted by the pvalue of Z when X = 109.5. So

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

Z = \frac{109.5 - 113.6}{5.74}

Z = -0.71

Z = -0.71 has a pvalue of 0.2389

1 - 0.2389 = 0.7611

0.7611 = 76.11% probability that at least 110 believe global warming is occurring

3 0
3 years ago
A store manager did a study to determine the amount of money the first 50 customers spent in her store. The data are approximate
MatroZZZ [7]
P(x > 35) = 1 - P(x < 35) = 1 - P[z < (35 - 29.60)/10.50] = 1 - P(z < 0.5143) = 1 - 0.6965 = 0.3035

Therefore, answer is 0.30 to the hundredths place.
6 0
4 years ago
Twelve and three tenths more than five and thirteen thousandths of a number d is equal to fifteen and three hundred two thousand
Masja [62]
12 and 3/10 more than 5 and 13/1000 of d equals 15 and 302/1000
12 and 3/10+(5 and 13/1000 times d)=15 and 302/1000
convert to improper fractions
12 and 3/10=123/10
5 and 13/1000=5013/1000
15 and 302/1000=15302/1000

123/10+(5013/1000 times d)=15302/1000
subtract 123/10 from both sides
123/10=12300/1000
(15302-123000)/1000=2698/1000
5013/1000 times d=2698/1000
multiply both sides by 1000/5013 to clear fraction
d=2698/5013

7 0
3 years ago
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