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3 years ago

11

For every two students in a class who eats a hot lunch at school, three other students purchase from the a la carte line and 1 o

ther student packs a lunch. If there were 25 students in class one day, how many students would you expect to order a hot meal

1 answer:

salantis [7]3 years ago

5 0

The ratio is 2:3 so... the total of them is 5

25/5=5 Therefore we can multiply 5x2 and 5x3

The new ratio is 10:15

Since the ratio is hot lunch:a la carte, The 10 is the amount of hot lunches expected and 15 is amount of students expected to eat a la carte

So 10 students eat hot lunch

15 students eat a la carte

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In a Gallup poll, 26.6% of 5000 people reported a Body Mass Index (BMI) greater than 30, which is classified as obese. The 5000

PSYCHO15rus [73]

Answer:

The 99% confidence interval for the proportion of the population who are obese is (0.25, 0.282)

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

$n = 5000, \pi = 0.266$

99% confidence level

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.266 - 2.575\sqrt{\frac{0.266*0.734}{5000}} = 0.25$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.266 + 2.575\sqrt{\frac{0.266*0.734}{5000}} = 0.282$

The 99% confidence interval for the proportion of the population who are obese is (0.25, 0.282)

6 0

3 years ago

A research center claims that at least 4545% of adults in a certain country think the government is not aggressive enough in pu

frez [133]

Answer:

$z=\frac{0.39 -0.45}{\sqrt{\frac{0.45(1-0.45)}{700}}}=-3.19$

$p_v =P(z$

So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of interst is significantly lower than 0.45 and the claim is not appropiate

Step-by-step explanation:

Data given and notation

n=700 represent the random sample taken

$\hat p=0.39$ estimated proportion of interest

$p_o=0.7$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion is at least 0.45.:

Null hypothesis: $p\geq 0.45$

Alternative hypothesis: $p < 0.45$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.39 -0.45}{\sqrt{\frac{0.45(1-0.45)}{700}}}=-3.19$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.01$ . The next step would be calculate the p value for this test.

Since is a lef tailed test the p value would be:

$p_v =P(z$

So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of interst is significantly lower than 0.45 and the claim is not appropiate

3 0

4 years ago

At Burnt Mesa Pueblo, archaeological studies have used the method of tree-ring dating in an effort to determine when prehistoric

Marina CMI [18]

Answer:

a) The range is (1199, 1267)

b) The range is (1165, 1301)

c) The range is (1131, 1335)

Step-by-step explanation:

Previous concepts

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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

$X \sim N(1233,34)$

Where $\mu=1233$ and $\sigma=34$

The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).

Part a

For this case we can use the statement from the empirical rule "68% of the data falls within the first standard deviation (µ ± σ)", and we can find the limits like this:

$\mu -\sigma= 1233-34=1199$

$\mu +\sigma=1233+34=1267$

The range is (1199, 1267)

Part b

For this case we can use the statement from the empirical rule "95% of the data within the first two standard deviations (µ ± 2σ)", and we can find the limits like this:

$\mu -2\sigma= 1233-(2*34)=1165$

$\mu +2\sigma=1233+(2*34)=1301$

The range is (1165, 1301)

Part c

For this case we can use the statement from the empirical rule "99.7% of the data within the first three standard deviations (µ ± 3σ)" and that represent almost all the data, and we can find the limits like this:

$\mu -3\sigma= 1233-(3*34)=1131$

$\mu +3\sigma=1233+(3*34)=1335$

The range is (1131, 1335)

5 0

3 years ago

Can someone please help me?

Hitman42 [59]

Answer:

its A I promise! good luck!

5 0

3 years ago

If f(x) = 6 + 5x– 3x^2”, find f'(4).

Serjik [45]

Answer:

f'(4)= -19

Step-by-step explanation:

I hope it will help you:

3 0

3 years ago