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Leya [2.2K]
3 years ago
14

Subtract 6 1/3 - 4 1/9​

Mathematics
1 answer:
Debora [2.8K]3 years ago
6 0

Answer:

2 2/9

Step-by-step explanation:

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Factorize (144z2y2 – 225a2b2).
bearhunter [10]

Answer:

sorry dont know

Step-by-step explanation:

6 0
3 years ago
The navy reports that the distribution of waist sizes among male sailors is approximately normal, with a mean of 32.6 inches and
grandymaker [24]

Answer:

a) 87.49%

b) 2.72%

Step-by-step explanation:

Mean of the waist sizes = u = 32.6 inches

Standard Deviation of the waist sizes = \sigma = 1.3 inches

It is given that the Distribution is approximately Normal, so we can use the z-distribution to answer the given questions.

Part A) A male sailor whose waist is 34.1 inches is at what percentile

In order to find the percentile score of 34.1 we need to convert it into equivalent z-scores, and then find what percent of the value lie below that point.

So, here x = 34.1 inches

The formula for z score is:

z=\frac{x-u}{\sigma}

Using the values in the above formula, we get:

z=\frac{34.1-32.6}{1.3}=1.15

Thus, 34.1 is equivalent to z score of 1.15

So,

P( X ≤ 34.1 ) =  P( z ≤ 1.15 )

From the z-table we can find the probability of a z score being less than 1.15 to be: 0.8749

Thus, the 87.49 % of the values in a Normal Distribution are below the z score of 1.15. For our given scenario, we an write: 87.49% of the values lie below 34.1 inches.

Hence, the percentile rank of 34.1 inches is 87.49%

Part B)

The regular measure of waist sizes is from 30 to 36 inches. Any measure outside this range will need a customized order.  We need to find that what percent of the male sailors will need a customized pant. This question can also be answered by using the z-distribution.

In a normal distribution, the overall percentage of the event is 100%. So if we find what percentage of values lie between 30 and 36, we can subtract that from 100% to obtain the percentage of values that are outside this range and hence will need customized pants.

First step is again to convert the values to z-scores.

30 converted to z scores will be:

z=\frac{30-32.6}{1.3}=-2

36 converted to z score will be:

z=\frac{36-32.6}{1.3}=2.62

So,

P ( 30 ≤ X ≤ 36 ) = P ( -2 ≤ z ≤ 2.62 )

From the z table, we can find P ( -2 ≤ z ≤ 2.62 )

P ( -2 ≤ z ≤ 2.62 ) = P(z ≤ 2.62) - P(z ≤ -2)

P ( -2 ≤ z ≤ 2.62 ) = 0.9956 - 0.0228

P ( -2 ≤ z ≤ 2.62 ) = 0.9728

Thus, 97.28% of the values lie between the waist sizes of 30 and 36 inches. The percentage of the values outside this range will be:

100 - 97.28 = 2.72%

Thus, 2.72% of the male sailors will need custom uniform pants.

The given scenario is represented in the image below. The black portion under the curve represents the percentage of male sailors that will require custom uniform pants.

3 0
3 years ago
If p is true and q is false, the p-&gt;q is ____<br> true.<br> never<br> always<br> sometimes
ratelena [41]

Answer:

never

Step-by-step explanation:

p->q means if p is true, then q must also be true

but since q is false, p->q is never true

5 0
3 years ago
Read 2 more answers
What is the probability of rolling a sum less than 8 with two dice if it is known that the first die rolled is a 2
Pachacha [2.7K]
The probability of rolling a sum less than 8 with two dices is 5/6
6 0
2 years ago
Grace read 9 books in 3 months what was her rate of reading Books per month​
Lorico [155]

Answer:

3 books per month

Step-by-step explanation:

9 books ÷ 3 months = 3 books per month :)

7 0
3 years ago
Read 2 more answers
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