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kondor19780726 [428]
3 years ago
12

Twelve of the 20 students in Mr. Skinner’s class brought lunch from home. Fourteen of the 21 students in Ms. Cho’s class brought

lunch from home. Siloni is using two 15-section spinners to simulate randomly selecting students from each class and predicting whether they brought lunch from home or will buy lunch in the cafeteria.
If each spinner is divided into 15 congruent sectors, how does the spinner representing Mr. Skinner’s class differ from the spinner representing Ms. Cho’s class?
One more sector of the Skinner-class spinner will represent bringing lunch from home.
One fewer sector of the Skinner-class spinner will represent bringing lunch from home.
Two more sectors of the Skinner-class spinner will represent bringing lunch from home.
Two fewer sectors of the Skinner-class spinner will represent bringing lunch from home.
Mathematics
2 answers:
babymother [125]3 years ago
9 0

Solution:

Number of students in Mr.Skinner's class who brought lunch from home if there are 20 students in the class=12

Fraction of students who brought lunch from home in Mr. Skinner's class=\frac{12}{20}=\frac{3}{5}

Number of students in Ms. Cho's class who brought lunch from home if there are 21 students in the class=14

Fraction of students who brought lunch from home in Ms. Cho's class=\frac{14}{21}=\frac{2}{3}

As Siloni is using  two 15-section spinners to simulate randomly selecting students from each class and predicting whether they brought lunch from home or will buy lunch in the cafeteria.

Number of Congruent sectors in each Spinner=15

So, if we represent students from Mr. Skinner's class who brought lunch from home in Spinner having 15 congruent Sectors =\frac{9}{15}

So, if we represent students from Mrs. Cho's class who brought lunch from home in Spinner having 15 congruent Sectors =\frac{10}{15}

Mr Skinner class +1 = Mr's Cho's Class

So Ms Cho's class =One more sector of the Skinner-class spinner will represent bringing lunch from home.

Option A which is One more sector of the Skinner-class spinner will represent bringing lunch from home represents Ms Cho's Class.

lutik1710 [3]3 years ago
7 0

Answer: A

Step-by-step explanation:

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Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 144 millimeters,
kkurt [141]

Answer:

0.0524 = 5.24% probability that the sample mean would differ from the population mean by more than 2 millimeters.

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.

Mean diameter of 144 millimeters, and a variance of 49.

This means that \mu = 144, \sigma = \sqrt{49} = 7

Sample of 46:

This means that n = 46, s = \frac{7}{\sqrt{46}}

Wat is the probability that the sample mean would differ from the population mean by more than 2 millimeters?

Above 144 + 2 = 146 or below 144 - 2 = 142. Since the normal distribution is symmetric, these probabilities are equal, which means that we find one of them and multiply by two.

Probability the sample mean is below 142:

p-value of Z when X = 142, so:

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

By the Central Limit Theorem

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

Z = \frac{142 - 144}{\frac{7}{\sqrt{46}}}

Z = -1.94

Z = -1.94 has a p-value of 0.0262

2*0.0262 = 0.0524

0.0524 = 5.24% probability that the sample mean would differ from the population mean by more than 2 millimeters.

8 0
3 years ago
A random sample of 60 suspension helmets used by motorcycle riders and automobile race-car drivers was subjected to an impact te
pentagon [3]

Answer:

The 95% confidence interval on the true proportion of helmets of this type that would show damage from this test is (0.169, 0.397).

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 = 60, \pi = \frac{17}{60} = 0.283

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.283 - 1.96\sqrt{\frac{0.283*0.717}{60}} = 0.169

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.283 + 1.96\sqrt{\frac{0.283*0.717}{60}} = 0.397

The 95% confidence interval on the true proportion of helmets of this type that would show damage from this test is (0.169, 0.397).

6 0
3 years ago
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Marina CMI [18]

The length of the band in terms of π is given by the expression:

L = 40mm + π*10mm

<h3>How to get the length of the band?</h3>

Remember that for a circle of diameter D, the circumference is:

C = π*D.

Now, if you look at the image, you can see that the length of the band will be equal to 4 times the diameter of the pencils (one time for each side). Plus 4 times one-fourth of the circumference of each pencil (for the four corners).

because the diameter is 10mm, the length of the band will be:

L = 4*10mm + 4*(π*10mm/4)

L = 40mm + π*10mm

This is the length of the band in terms of π.

If you want to learn more about circles:

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3 years ago
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Alex777 [14]

Answer:

m=5

Step-by-step explanation:

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3 years ago
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