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elena55 [62]
3 years ago
13

Mr Abraham is able to divide the learners in his mathematics class into equal groups of 2, 5 or 6 without leaving any learners o

ut. What is the smallest possible number of learners in the class​
Mathematics
1 answer:
IRISSAK [1]3 years ago
3 0

Answer:

30

Step-by-step explanation:

Find the lowest number divisible by both 6, 5, and 2.

To start, let's list off the numbers divisible by six, and see if we can check any of them off for 5. Since both 2 and 6 are even numbers, we know that if a number is divisible by 6, it's divisible by 2.

6 12 18 24 30 36 42 48 54 60

All of these numbers are divisible by two. Let's find the lowest one that is divisible by 5. We know this by either the umber ending in 5 or 0.

30 is the lowest number that is divisible by 2, 5, or 6.

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A baby was born on November 1. The table gives the length of the baby on three
ch4aika [34]

Answer:

The first one to your left

Step-by-step explanation:

6 0
2 years ago
Find the slope of the line using the table. (keep as a fraction)
attashe74 [19]
The slope is -2/3 or -2 over 3
8 0
3 years ago
ing needs 3 cups of milk to make 30 cookies. How many ounces of milk would he need to make 120 cookies?
Llana [10]

Answer:

12 cup of milk.

Step-by-step explanation:

if 3 cups of milk can make 30 cookies it means 1 cup of milk can make 10 cookies then for 100 cookies we need 10 cup of milk and 30 more means 3 more so answer is 12

8 0
3 years ago
A large electronic office product contains 2000 electronic components. Assume that the probability that each component operates
KIM [24]

Answer:

The probability is 0.971032

Step-by-step explanation:

The variable that says the number of components that fail during the useful life of the product follows a binomial distribution.

The Binomial distribution apply when we have n identical and independent events with a probability p of success and a probability 1-p of not success. Then, the probability that x of the n events are success is given by:

P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}

In this case, we have 2000 electronics components with a probability 0.005 of fail during the useful life of the product and a probability 0.995 that each component operates without failure during the useful life of the product. Then, the probability that x components of the 2000 fail is:

P(x)=\frac{2000!}{x!(2000-x)!}*0.005^{x}*(0.995)^{2000-x}     (eq. 1)

So, the probability that 5 or more of the original 2000 components fail during the useful life of the product is:

P(x ≥ 5) = P(5) + P(6) + ... + P(1999) + P(2000)

We can also calculated that as:

P(x ≥ 5) = 1 - P(x ≤ 4)

Where P(x ≤ 4) = P(0) + P(1) + P(2) + P(3) + P(4)

Then, if we calculate every probability using eq. 1, we get:

P(x ≤ 4) = 0.000044 + 0.000445 + 0.002235 + 0.007479 + 0.018765

P(x ≤ 4) = 0.028968

Finally, P(x ≥ 5) is:

P(x ≥ 5) = 1 - 0.028968

P(x ≥ 5) = 0.971032

3 0
3 years ago
What is 2×2hhhhggggggggggggggg
TiliK225 [7]

Answer:

2*2h4g15

Step-by-step explanation:

count how many "g"s there are and same with "h"

5 0
3 years ago
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