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Viefleur [7K]
3 years ago
12

You measure 22 textbooks' weights, and find they have a mean weight of 64 ounces. Assume the population standard deviation is 5.

1 ounces. Based on this, construct a 90% confidence interval for the true population mean textbook weight.
Give your answers as decimals, to two places

Mathematics
1 answer:
lara31 [8.8K]3 years ago
3 0

Answer:

Step-by-step explanation:

We want to determine a 90% confidence interval for the true population mean textbook weight.

Number of sample, n = 22

Mean, u = 64 ounces

Standard deviation, s = 5.1 ounces

For a confidence level of 90%, the corresponding z value is 1.645. This is determined from the normal distribution table.

We will apply the formula

Confidence interval

= mean ± z ×standard deviation/√n

It becomes

64 ± 1.645 × 5.1/√22

= 64 ± 1.645 × 1.087

= 64 ± 1.788

The lower end of the confidence interval is 64 - 1.788 = 62.21 ounces

The upper end of the confidence interval is 64 + 1.788 = 65.79 ounces

Therefore, with 90% confidence interval, the true population mean textbook weight is between 62.21 ounces and 65.79 ounces

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viva [34]
Pls. see attachment. 

We need to solve for the angles of the smaller triangle in order to solve for the angle of the larger triangle which would help us solve the missing measurement of a side.

Given:

51 degrees.

Cut the triangle into two equal sides and it forms a right triangle. All interior angles of a triangle sums up to 180 degrees.

180 – 51 – 90 = 39 degrees

39 degrees * 2 = 78 degrees.

Angle Q is 78 degrees.

In the bigger triangle, 4.3 is the hypotenuse. We need to solve for the measurement of the long leg which is the opposite of the 78 degree angle.

We will use the formula:

Sine theta = opposite / hypotenuse

Sin(78 deg) = opposite / 4.3

Sin(78 deg) * 4.3 = opposite

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Area of a triangle = ½ * base * height

A = ½ * 3units * 4.21units

A = 6.315 square units.

7 0
3 years ago
Use a Venn diagram to answer the question. A survey of 180 families showed that 67 had a​ dog; 52 had a​ cat; 22 had a dog and a
Mice21 [21]

Answer:

There are 13 families had a parakeet only

Step-by-step explanation:

* Lets explain the problem

- There are 180 families

- 67 families had a dog

- 52 families had a cat

- 22 families had a dog and a cat

- 70 had neither a cat nor a​ dog, and in addition did not have a​

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- 4 had a​ cat, a​ dog, and a parakeet (4 is a part of 22 and 22 is a part

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* We will explain the Venn-diagram

- A rectangle represent the total of the families

- Three intersected circles:

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- The common part of the three circle had 4 families

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- The common part between the circle of the dog and the circle of the

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- The common part between the circle of the cat and the circle of the

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- The non-intersected part of the circle of the dog had 67 - 22 - a =

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- The non-intersected part of the circle of the cat had 52 - 22 - b =  

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- The part out side the circles and inside the triangle has 70 families

- Look to the attached graph for more under stand

∵ The total of the families is 180

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∴ (45 + 18 + 4 + 30 + 70) + (-a + a) + (-b + b) + c = 180

∴ 167 + c = 180 ⇒ subtract 167 from both sides

∴ c = 180 - 167 = 13 families

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5 0
3 years ago
What is the sum of the counting numbers 1 - 40
AVprozaik [17]
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And finally the last number not used, 20

Therefore 800 + 20 = 820
5 0
3 years ago
Read 2 more answers
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Firlakuza [10]

Answer:

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2 years ago
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A container is the shape of an inverted right circular cone has a radius of 4.00 inches at the top and a height of 5.00 inches.
Len [333]

The rate at which the water from the container is being drained is 24 inches per second.

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Learn more about differentaiation at brainly.com/question/954654

#SPJ4

5 0
2 years ago
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