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2 years ago

Which statement best compares the variability of the number of snacks grabbed for Class A and Class B?

Mathematics

1 answer:

UNO [17]2 years ago

5 0

*see attachment below for the dot plots

Answer:

A. The number of snacks grabbed for Class A has less variability than the number of snacks grabbed for Class B.

Step-by-step explanation:

The extent of variability of the data represented on a dot plot can be determined by the range value.

Range is the difference between maximum value of the data set and the minimum value of the data set.

The larger the range, the more the data has more variability, while the smaller the range, the less the variability of the data.

Range for Class A = 37 - 15 = 22

Range for Class B = 45 - 12 = 33

Therefore, class B has more variability in the number of snacks grabbed compared to Class A.

Correct statement is: "The number of snacks grabbed for Class A has less variability than the number of snacks grabbed for Class B."

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Joy makes 6.5 litres of soup, correct to the nearest 0.5 litre. She serves the soup in 280 ml portions, correct to the nearest 1

Levart [38]

Answer:

Joy definitely does have enough soup for 22 people.

Step-by-step explanation:

Upper Bound: 285ml turn it to 0.285 L 10÷2=5

280ml :

Lower Bound: 275ml turn it to 0.275 L

Upper Bound: 6.75 L 0.5÷2=0.25

6.5 Litters :

Lower Bound:6.25 L

6.75÷0.275=24.54 or 25

6.25÷0.285=21.92 or 22

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2 years ago

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Lenny sold 576 tacos in 48 hours what was lenny's average rate of tacos sales

cupoosta [38]

If your looking for hourly they sold a total of 12 tacos per hour.
If your looking for daily they sold a total of 288 tacos per day.

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2 years ago

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If a hurricane was headed your way, would you evacuate? The headline of a press release states, "Thirty-four Percent of People o

IgorC [24]

Answer:

The 98% confidence interval would be given by (0.326;0.354)

Step-by-step explanation:

1) Notation and definitions

$X=63$ number of people who live within 20 miles of the coast in high hurricane risk counties of eight southern states

$n=6138$ random sample taken

$\hat p=0.34$ estimated proportion of people who live within 20 miles of the coast in high hurricane risk counties of eight southern states

$p$ true population proportion of people who live within 20 miles of the coast in high hurricane risk counties of eight southern states

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

2) Confidence interval

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 98% of confidence, our significance level would be given by $\alpha=1-0.98=0.02$ and $\alpha/2 =0.01$ . And the critical value would be given by:

$z_{\alpha/2}=-2.33, z_{1-\alpha/2}=2.33$

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.34 - 2.33\sqrt{\frac{0.34(1-0.34)}{6138}}=0.326$

$0.34 + 2.33\sqrt{\frac{0.34(1-0.34)}{6138}}=0.354$

The 98% confidence interval would be given by (0.326;0.354)

8 0

2 years ago

The time for a visitor to read health instructions on a Web site is approximately normally distributed with a mean of 10 minutes

klio [65]

Answer:

a) The mean is 10 and the variance is 0.0625.

b) 0.6826 = 68.26% probability that the mean time of the visitors is within 15 seconds of 10 minutes.

c) 10.58 minutes.

Step-by-step explanation:

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

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Normally distributed with a mean of 10 minutes and a standard deviation of 2 minutes.

This means that $\mu = 10, \sigma = 2$

Suppose 64 visitors independently view the site.

This means that $n = 64, = \frac{2}{\sqrt{64}} = 0.25$

a. The expected value and the variance of the mean time of the visitors.

Using the Central Limit Theorem, mean of 10 and variance of (0.25)^2 = 0.0625.

b. The probability that the mean time of the visitors is within 15 seconds of 10 minutes.

15 seconds = 15/60 = 0.25 minutes, so between 9.75 and 10.25 seconds, which is the p-value of Z when X = 10.25 subtracted by the p-value of Z when X = 9.75.

X = 10.25

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

By the Central Limit Theorem

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

$Z = \frac{10.25 - 10}{0.25}$

$Z = 1$

$Z = 1$ has a p-value of 0.8413.

X = 9.75

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

$Z = \frac{9.75 - 10}{0.25}$

$Z = -1$

$Z = -1$ has a p-value of 0.1587.

0.8413 - 0.1587 = 0.6826.

0.6826 = 68.26% probability that the mean time of the visitors is within 15 seconds of 10 minutes.

c. The value exceeded by the mean time of the visitors with probability 0.01.

The 100 - 1 = 99th percentile, which is X when Z has a p-value of 0.99, so X when Z = 2.327.

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

$2.327 = \frac{X - 10}{0.25}$

$X - 10 = 2.327*0.25$

$X = 10.58$

So 10.58 minutes.

6 0

2 years ago

In 2011, a train carried 8% more passengers than in 2010. In 2012 , it carried 8% more passengers than in 2011. Find the percent

konstantin123 [22]

Answer:

there was a 16% increase since 2010 to 2012

Step-by-step explanation:

lets say the train had 100 people on it in 2010 and in 2011 it had an 8% increase which then brings you to 8 plus 100 and you get 108.

Now in 2012 there was an increase 8%, so you will add 8 to your previous answer 108 and receive 116. So now there was 16 more passengers in 2012 than in 2010. Transfer the 16 passengers into 16% and you now have your answer.

5 0

2 years ago