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schepotkina [342]
3 years ago
8

Each student in your class must do 14 homework problems. There are 23 students in the class. How many problems will your teacher

have to grade?
Mathematics
1 answer:
Ganezh [65]3 years ago
4 0

Answer: 322

Step-by-step explanation:

23 students × 14 homework

=322 problems

The teacher will have to grade 322 problems

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mr_godi [17]

Roxy: 42/9 = 4.67

Jordan: 79/18 = 4.39

Rickie: 123/27 = 4.56


Jordan has the lowest number of strokes per hole


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Jordan

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3 years ago
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In the figure below, the segments overline RS and overline RT are tangent to the circle centered at Given that OS = 3.6 and RT =
ExtremeBDS [4]

Answer:

RS = 4.8

OT = 3.6

From the Pythagorean Theorem we know:

OR^2 = ST^2 + OT^2

OR^2 = 4.8^2 + 3.6^2

OR ^2 = 23.04 + 12.96

OR^2 = 36

OR = 6

Step-by-step explanation:

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Assume that lines
lora16 [44]

Answer:

b

Step-by-step explanation:

The inscribed angle 36°m is half the measure of its intercepted arc, that is

b = 2 × 36 = 72

c = 360 - (108 + 72) = 360 - 180 = 180 ( total sum of arcs in a circle )

The angle between a tangent and the radius of a circle at the point of contact is 90° , then

d = 90

Thus b = 72, c = 180, d = 90 → b

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2 years ago
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An article in The Engineer (Redesign for Suspect Wiring," June 1990) reported the results of an investigation into wiring errors
GarryVolchara [31]

Answer:

a) The 99% confidence interval on the proportion of aircraft that have such wiring errors is (0.0005, 0.0095).

b) A sample of 408 is required.

c) A sample of 20465 is required.

Step-by-step explanation:

Question a:

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}.

Of 1600 randomly selected aircraft, eight were found to have wiring errors that could display incorrect information to the flight crew.

This means that n = 1600, \pi = \frac{8}{1600} = 0.005

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}} = 0.005 - 2.575\sqrt{\frac{0.005*0.995}{1600}} = 0.0005

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.005 + 2.575\sqrt{\frac{0.005*0.995}{1600}} = 0.0095

The 99% confidence interval on the proportion of aircraft that have such wiring errors is (0.0005, 0.0095).

b. Suppose we use the information in this example to provide a preliminary estimate of p. How large a sample would be required to produce an estimate of p that we are 99% confident differs from the true value by at most 0.009?

The margin of error is of:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

A sample of n is required, and n is found for M = 0.009. So

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.009 = 2.575\sqrt{\frac{0.005*0.995}{n}}

0.009\sqrt{n} = 2.575\sqrt{0.005*0.995}

\sqrt{n} = \frac{2.575\sqrt{0.005*0.995}}{0.009}

(\sqrt{n})^2 = (\frac{2.575\sqrt{0.005*0.995}}{0.009})^2

n = 407.3

Rounding up:

A sample of 408 is required.

c. Suppose we did not have a preliminary estimate of p. How large a sample would be required if we wanted to be at least 99% confident that the sample proportion differs from the true proportion by at most 0.009 regardless of the true value of p?

Since we have no estimate, we use \pi = 0.5

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.009 = 2.575\sqrt{\frac{0.5*0.5}{n}}

0.009\sqrt{n} = 2.575*0.5

\sqrt{n} = \frac{2.575*0.5}{0.009}

(\sqrt{n})^2 = (\frac{2.575*0.5}{0.009})^2

n = 20464.9

Rounding up:

A sample of 20465 is required.

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2 years ago
John has 34 bottle caps Logan has 4 times as many bottle caps as John ?
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Answer:

136

Step-by-step explanation: 34 multiplied by 4

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