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olchik [2.2K]
4 years ago
11

Is 32643 divisble by both 3 and 9

Mathematics
1 answer:
telo118 [61]4 years ago
6 0
 Hey yo!!!!!!!!!!!!!!!!

yes because if you add up the digits, you get 18, which is divisible by 3 and 9. that is the divisibility rule of 3 and 9. 

  hope this helped:)

  HAGD
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Put the triangle inside of each other lining up the 90° angle corners.
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3 years ago
If m< 3=74 find each measure​
frez [133]

Answer:

for you

Step-by-step explanation:

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3 years ago
The Johnson twins were born nine years after their older sister. This year, the product of the three siblings" Ages is exactly 2
larisa [96]

Answer:

ages of the twins are 11 years

Step-by-step explanation:

let the age of the twins each be X

Age of the older sister = X+9

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3 years ago
A batch of 25 injection-molded parts contains 5 that have suffered excessive shrinkage. Round your answers to four decimal place
Firdavs [7]

Answer:

(a) The probability that the second part is the one with excessive shrinkage is 0.2000.

(b) The probability that the third part is the one with excessive shrinkage is 0.2000.

Step-by-step explanation:

Let the variable <em>X </em>ₙ denote the <em>n</em>th part that has  suffered excessive shrinkage.

(a)

It is provided that two parts are selected at random.

The parts are selected without replacement.

Now, for the two parts selected it is possible that either both have excessive shrinkage or only the second part has excessive shrinkage.

The probability that the second part is the one with excessive shrinkage is:

P (X₂) = P (X₂ ∩ X₁) + P (X₂ ∩ X₁')

         =[\frac{4}{24}\times \frac{5}{25}]+[\frac{5}{24}\times \frac{20}{25}]               (without replacement)

         =\frac{1}{30}+\frac{1}{6}

         =0.2000

Thus, the probability that the second part is the one with excessive shrinkage is 0.2000.

(b)

It is provided that three parts are selected at random.

The parts are selected without replacement.

Now, for the three parts selected it is possible that either all three have excessive shrinkage or any two of the three has excessive shrinkage or only the third part has excessive shrinkage.

The probability that the third part is the one with excessive shrinkage is:

P (X₃) = P (X₃ ∩ X₂ ∩ X₁) + P (X₃ ∩ X₂ ∩ X₁')

                 + P (X₃ ∩ X₂' ∩ X₁) + P (X₃ ∩ X₂' ∩ X₁')

         =[\frac{3}{23}\times \frac{4}{24}\times \frac{5}{25}]+[\frac{4}{23}\times \frac{5}{24}\times \frac{20}{25}]+[\frac{4}{23}\times \frac{20}{25}\times \frac{5}{24}]+[\frac{5}{23}\times \frac{19}{24}\times \frac{20}{25}]  

         =0.2000

Thus, the probability that the third part is the one with excessive shrinkage is 0.2000.

6 0
3 years ago
Please help, no links, please explain below ​
Bogdan [553]

Answer:

A- mean

median/2 = 2,000

mean/4= 6,000

Step-by-step explanation:

A- mean would should the drastic difference in the salaries, which would help to justify why they want a raise.

B-  median = 4,000.      4,000/2 = 2,000

C-  mean =    24,000/4 = 6,000

let me know if you need more help :)

7 0
3 years ago
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