Can someone help me with 3 and 4

What is 231 divided by 42

luda_lava [24]

5.5 is the answer for the problem

6 0

3 years ago

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Devisibility rules for 3, 6, and 9

fenix001 [56]

Answer:

★ Divisibility rule for 3 -

If the sum of all the digits divisible by 3 then the number is divisible by 3.

Here, we take an example for better understanding -

Example 1 - 3525

Add up the digits - 3 + 5 + 2 + 5

→ 15

As, 15 is divisible by 3. So, 3525 is a factor of 3.

★ Divisibility rule for 6 -

A number is to be divisible by 6 , if the number is also divisible by 2 and 3.

Example 2 - 132

Add up the digits = 1 + 3 + 2 = 6

As, 6 is divisible by 2 and it's also divisible by 3. Then, 132 is a factor of 6.

★ Divisibility rule for 9 -

If the sum of all the digits is divisible by 9 then the number is divisible by 9. Just like we'd in 3.

Example 3 - 1052

Add up the digits - 10 + 5 + 2 = 17

As, 17 is not divisible by 9. So, 1052 is also not divisible by 9.

3 0

3 years ago

What is the answer to 125÷3

Sunny_sXe [5.5K]

Answer:

41.6666667?? I didnt know if you needed is simplified .-.

5 0

3 years ago

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Genetic Defects Data indicate that a particular genetic defect occurs in of every children. The records of a medical clinic show

Dovator [93]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The probability that there exist 60 or more defected children is $P(x \ge 60)=0.0901$

Looking at the value for this probability we see that it is not so small to the point that the observation of this kind would be a rare occurrence

Step-by-step explanation:

From the question we are told that

in every 1000 children a particular genetic defect occurs to 1

The number of sample selected is $n= 50,000$

The probability of observing the defect is mathematically evaluated as

$p = \frac{1}{1000}$

$= 0.001$

The probability of not observing the defect is mathematically evaluated as

$q = 1-p$

$= 1-0.001$

$= 0.999$

The mean of this probability is mathematically represented as

$\mu = np$

Substituting values

$\mu = 50000*0.001$

$= 50$

The standard deviation of this probability is mathematically represented as

$\sigma = \sqrt{npq}$

Substituting values

$= \sqrt{50000 * 0.001 * 0.999}$

$= \sqrt{49.95}$

$= 7.07$

the probability of detecting $x \ge60$ defects can be represented in as normal distribution like

$P(x \ge 60)$

in standardizing the normal distribution the normal area used to approximate $P(x \ge 60)$ is the right of 59.5 instead of 60 because x= 60 is part of the observation

The z -score is obtained mathematically as

$z = \frac{x-\mu }{\sigma }$

$= \frac{59.5 - 50 }{7.07}$

$=1.34$

The area to the left of z = 1.35 on the standardized normal distribution curve is 0.9099 obtained from the z-table shown z value to the left of the standardized normal curve

Note: We are looking for the area to the right i.e 60 or more

The total area under the curve is 1

So

$P(x \ge 60) \approx P(z > 1.34)$