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Tresset [83]
3 years ago
12

How many different three-digit numbers can be written using digits from the set 5, 6, 7, 8, 9 without any repeating digits?

Mathematics
2 answers:
Law Incorporation [45]3 years ago
8 0
567
598
578
596
597
Keep doing this and you will get your answer. It is could trial and error.
Alina [70]3 years ago
4 0

Answer:

It can be written 60 different three digit numbers.

Step-by-step explanation:

For, calculate how many different three digit numbers can be written, we can use the rule of multiplication as:

<u>     5              </u>* <u>         4        </u> *  <u>           3              </u>  =      60

   1st digit         2nd digit          3rd digit

Taking into account that there is no repeating digits, we have 5 options for the first digits, this options are the number 5, 6, 7, 8 or 9. Then, we have 4 options for the second digit and then we have 3 options for the third digit.

So, there are 60 different three digit numbers that we can create with the set  5, 6, 7, 8, 9 without any repeating digits.

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The length of a rectangular garden is: 13 feet

Step-by-step explanation:

You have 2 lengths that are 2x-3 feet, and 2 widths that are x.

2x-3 + 2x-3 +x +x =42 feet

4x -6 +2x -42 feet

6x -6 = 42 feet

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x = 8 feet (so, 2x-3 = 16-3 = 13 feet)

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The size of the left upper chamber of the heart is one measure of cardiovascular health. When the upper left chamber is enlarged
Temka [501]

Answer:

a) 0.283  or 28.3%

b) 0.130 or 13%

c) 0.4 or 40%

d) 30.6 mm

Step-by-step explanation:

z-score of a single left atrial diameter value of healthy children can be calculated as:

z=\frac{X-M}{s} where

  • X is the left atrial diameter value we are looking for its z-score
  • M is the mean left atrial diameter of healthy children (26.7 mm)
  • s is the standard deviation (4.7 mm)

Then

a) proportion of healthy children who have left atrial diameters less than 24 mm

=P(z<z*) where z* is the z-score of 24 mm

z*=\frac{24-26.7}{4.7} ≈ −0.574

And P(z<−0.574)=0.283

b) proportion of healthy children who have left atrial diameters greater than 32 mm

= P(z>z*) = 1-P(z<z*) where z* is the z-score of 32 mm

z*=\frac{32-26.7}{4.7} ≈ 1.128

1-P(z<1.128)=0.8703=0.130

c) proportion of healthy children have left atrial diameters between 25 and 30 mm

=P(z(25)<z<z(30)) where z(25), z(30) are the z-scores of 25 and 30 mm

z(30)=\frac{30-26.7}{4.7} ≈ 0.702

z(25)=\frac{25-26.7}{4.7} ≈ −0.362

P(z<0.702)=0.7587

P(z<−0.362)=0.3587

Then P(z(25)<z<z(30)) =0.7587 - 0.3587 =0.4

d) to find the value for which only about 20% have a larger left atrial diameter, we assume

P(z>z*)=0.2 or 20% where z* is the z-score of the value we are looking for.

Then P(z<z*)=0.8 and z*=0.84. That is

0.84=\frac{X-26.7}{4.7}  solving this equation for X we get X=30.648

5 0
3 years ago
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