Holly works at a garage.Mr Bakir brings his car in for a service.
He also buys 2 tyres and a battery.
Holly has this list of costs
• service £110
• tyres £49.99 each
• battery £89
She uses this list to start to work out the bill for Mr Bakir.
Holly also has to charge 20% VAT on the costs to work out the total bill.
Work out the total bill. (You MUST show your working in the box below)
(4 Points)

Answers

Answer 1

Answer:

The total bill that Mr Bakir paid was £358.78, using mathematical operations.

How the total bill is determined:

The total bill can be determined using mathematical operations, including multiplication and addition as follows:

The cost of car service = £110

The cost of 2 tyres bought = £99.98 (£49.99 x 2)

The cost of a battery = £89

Sub- total = £298.98

VAT (20%) = £59.80

Total = £358.78

Thus, including 20% VAT, the total bill for Mr. Bakir was £358.78.

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Answers

Answer:

it most likely would be 260 degrees

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Answers

180-140= 40

Im not so sure of my answer so I might be wrong so enjoy!!

In a recent survey it was found that Americans drink an average of 23.2 gallons of bottled water in a year. If the standard deviation is 2.7 gallons and the variable is normally distributed, find the probability that a randomly selected American drinks more than 25 gallons of bottled water in a year. What is the probability that the selected person drinks between 22 and 30 gallons

Answers

Answer:

a) 0.25249

b) 0.66575

Step-by-step explanation:

We solve this question using z score formula

= z = (x-μ)/σ, where

x is the raw score

μ is the population mean = 23.2 gallons

σ is the population standard deviation = 2.7 gallons

a) Find the probability that a randomly selected American drinks more than 25 gallons of bottled water in a year.

For x = 25 gallons

z = 25 - 23.2/2.7

z = 0.66667

Probability value from Z-Table:

P(x<25) = 0.74751

P(x>25) = 1 - P(x<25)

1 - 0.74751

= 0.25249

The probability that a randomly selected American drinks more than 25 gallons of bottled water in a year is 0.25249

2) What is the probability that the selected person drinks between 22 and 30 gallons

For x = 22 gallons

z = 22 - 23.2/2.7

z = -0.44444

Probability value from Z-Table:

P(x = 22) = 0.32836

For x = 30 gallons

z = 30 - 23.2/2.7

z =2.51852

Probability value from Z-Table:

P(x = 30) = 0.99411

The probability that the selected person drinks between 22 and 30 gallons is

P(x = 30) - P(x = 22)

= 0.99411 - 0.32836

= 0.66575

Final answer:

The probability that a randomly selected American drinks more than 25 gallons of bottled water in a year is approximately 0.2514, while the probability that they will drink between 22 and 30 gallons is approximately 0.6643.

Explanation:

This is a statistics question about probability distribution, specifically, normal distribution. You need to find the z-scores and use the standard normal distribution table to find the probabilities.

The average or mean (μ) consumption is 23.2 gallons and standard deviation (σ) is 2.7 gallons.

First, we use the z-score formula: z = (X - μ) / σ

To find out the probability that a selected American drinks more than 25 gallons annually, we substitute X = 25, μ = 23.2 and σ = 2.7 into the z-score formula to get z = (25 - 23.2) / 2.7 ≈ 0.67. Z value of 0.67 corresponds to the probability of 0.7486 in standard normal distribution table, but this is the opposite of what we want. We need to subtract this probability from 1 to find the probability that a person drinks more than 25 gallons annually. So 1 - 0.7486 = 0.2514.

Second, to find the probability an individual drinks between 22 and 30 gallons, we calculate two z-scores: For X = 22, z = (22 - 23.2) / 2.7 ≈ -0.44 with corresponding probability 0.3300, and for X = 30, z = (30 - 23.2) / 2.7 ≈ 2.52 with corresponding probability 0.9943. We find the probability of someone drinking between these quantities by subtracting the smaller probability from the larger, 0.9943 - 0.3300 = 0.6643.

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A researcher wishes to estimate, with 90% confidence, the population proportion of likely U.S. voters who think Congress is doing a good or excellent job. Her estimate must be accurate within 2% of the true proportion(a) No preliminary estimate is available. Find the minimum sample size needed.
(b) Find the minimum sample size needed, using a prior study that found that 28% of the respondents said they think Congress is doing a good or excellent job.
(c) Compare the results from parts (a) and (b).

Answers

The sample size just be 752.

What is confidence interval?

A confidence interval (CI) for an unknown parameter in frequentist statistics is a range of estimations. The most popular confidence level is 95%, but other levels, such 90% or 99%, are occasionally used for computing confidence intervals. The fraction of related CIs over the long run that actually contain the parameter's true value is what is meant by the confidence level. The degree of confidence, sample size, and sample variability are all factors that might affect the width of the CI. A larger sample would result in a narrower confidence interval if all other factors remained constant. A wider confidence interval would also be required by a higher confidence level and would be produced by a sample with (1.645 / 0.03)2 * 0.5 * 0.5

=752

Sample size = 752

= 1 - 0.42 = 0.58

margin of error = E =3 % = 0.03

At 90% confidence level z

Hence, The sample size just be 752.

Learn more about confidence interval, by the following link.

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In a poll conducted by the Gallup organization in April 2013, 48% of a random sample of 1022 adults in the U.S. responded that they felt that economic growth is more important than protecting the environment. We can use this information to calculate a 95% confidence interval for the proportion of all U.S. adults in April 2013 who felt that economic growth is more important than protecting the environment. Make sure to include all steps.

Answers

Answer:

The 95% confidence interval is $0.449 < p < 0.48 + 0.511$

Step-by-step explanation:

From the question we are told that

The sample proportion is $\r p = 0.48$

The sample size is $n = 1022$

Given that the confidence level is 95% then the level of significance is mathematically evaluated as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha = 0.05$

Next we obtain the critical value of $(\alpha )/(2)$ from the z-table , the value is

$Z_{(\alpha )/(2) } =Z_{(0.05 )/(2) }= 1.96$

The reason we are obtaining critical value of $(\alpha )/(2)$ instead of $\alpha$ is because

$\alpha$ represents the area under the normal curve where the confidence level interval ( $1-\alpha$ ) did not cover which include both the left and right tail while $(\alpha )/(2)$ is just the area of one tail which what we required to calculate the margin of error