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Mumz [18]
4 years ago
8

What is the volume of containers that are 55 mm wide, 94 mm long, and 184 mm tall?

Mathematics
1 answer:
Juliette [100K]4 years ago
8 0
Assume this container is rectangular prism, so the volume of the rectangular prism equal length*width*height, or:
V=l*w*h
V=55*94*184
V=951280 cubic mm . As a result, the volume of the container is 951280 cubic mm. Hope it help!
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Step-by-step explanation:

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Step-by-step explanation:

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The SAT Reasoning Test (formerly called the Scholastic Aptitude Test) is perhaps the most widely used standardized test for coll
mihalych1998 [28]

Given Information:  

Mean SAT score = μ = 1500

Standard deviation of SAT score = σ = 3 00

Required Information:  

Minimum score in the top 10% of this test that qualifies for the scholarship = ?

Answer:

\bar{x} = 1884\\

Step-by-step explanation:

What is Normal Distribution?

We are given a Normal Distribution, which is a continuous probability distribution and is symmetrical around the mean. The shape of this distribution is like a bell curve and most of the data is clustered around the mean. The area under this bell shaped curve represents the probability.  

We want to find out the minimum score that qualifies for the scholarship by scoring in the top 10% of this test.

P(X > \bar{x} )= P(Z > \bar{x}) = 0.10\\P(X < \bar{x} )= P(Z < \bar{x}) = 1 - 0.10\\P(X < \bar{x} )= P(Z < \bar{x}) = 0.90\\

The z-score corresponding to the probability of 0.90 is 1.28 (from the z-table)

\bar{x} = \mu + z(\sigma) \\\bar{x} = 1500 + 1.28(300)\\\bar{x} = 1500 + 384\\\bar{x} = 1884\\

Therefore, you need to score 1884 in order to  qualify for the scholarship.

How to use z-table?

Step 1:

In the z-table, find the probability value of 0.90 and note down the value of the that row which is 1.2

Step 2:

Then look up at the top of z-table and note down the value of the that column which is 0.08

Step 3:

Finally, note down the intersection of step 1 and step 2 which is 1.28

4 0
3 years ago
Assume that different groups of couples use a particular method of gender selection and each couple gives birth to one baby. Thi
Dovator [93]

Complete question:

Assume that different groups of couples use a particular method of gender selection and each couple gives birth to one baby this mention is designed to increase the likelihood that each baby will be a girl, but assume that the method has no effect so the probability of a girl is 0.5. Assume that the group consists of 36 couples.

A) Find the mean and standard deviation for the number of girls in groups of 36 births.

B) Use the range rule of thumb to find the values separating results that are significantly low and significantly high.

C) Is the result of 33 girls significantly high? A result of 33 girls would suggest the method is effective or is not effective?

Answer:

a) mean = 18

Standard deviation =3

b) low range = 12

High range = 24

c) The result of 33 girls is significantly high. Yes, the method is effective.

Step-by-step explanation:

Given:

p = 0.5

n = 36

a) The mean is the product of n and p

Mean u = np

u = 36 * 0.5 = 18

The standard deviation is the square root of the product of n and p&q.

S.d ó = \sqrt{npq}

= \sqrt{np(1-p)}

= \sqrt{36(0.5)(1-0.5)} = \sqrt{9} = 3

b) To find the range rule of thumb:

• For low range

Low range = u - 2ó

= 18 - (2 * 3)

= 12

• High range

= u + 2ó

= 18 + (2*3)

= 24

c) The result is significantly high, because 33 is greater than 24 girls.

A result of 33 girls would prove the method as effective.

8 0
3 years ago
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