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3 years ago

It must exceed seven

Mathematics

1 answer:

nignag [31]3 years ago

6 0

The digits that are between numbers 1 and 10 and larger than 7 are: 8, 9, 10

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76 div 4 a)15 b) 8 c) 19 d) 0 Explain

Sunny_sXe [5.5K]

Answer:

C. 19.

Step-by-step explanation:

Let's find the answer.

Step 1: divide the first number of 76, which is 7, by 4 and find the integer part.

7/4=1.75, so the integer part is 1.

Step 2: subtract from 7 the (integer part times 4)

7-(1*4)=3

Step 3: put toether the result (3) and the second number of 76, which is 6, so the result is 36.

Step 4: find how many times we need to add 4s to reach 36, so:

4+4+4+4+4+4+4+4+4=36 which means that:

36/4=9

Step 5: put together the two resulted numbers calculated in step 1, which is 1, and step 4, which is 9:

1 and 9 giving 19, so the answer is:

76/4=19.

In conclusion the answer is C. 19.

6 0

3 years ago

Every person has blood type O, A, B, or AB. A random group of people are blood-typed, and the results are shown in the table. A

S_A_V [24]

Answer:

3/25 for type B, 1/25 for type AB, and 4/25 for both

Step-by-step explanation:

O goes to 22

A goes to 20

B goes to 6

AB goes to 2

22+20+6+2 = 50, so (number of people with a certain blood type)/50 = the answers for each question.

4 0

3 years ago

Which expression represents the prime factorization of 108?

ankoles [38]

Answer:

2×2×3×3×3

Step-by-step explanation:

●108=

•2×54

•2×27

•3×9

• 3×3

Writing the prime numbers;

108=2×2×3×3×3

4 0

2 years ago

Ano ang ibig sabihin nang exsena?

zvonat [6]

Wait huh?????????????????

8 0

3 years ago

What equation best models this data?(use y to represent the population of rabbits and t to represent the year, assuming that 201

liraira [26]

If we see the data closely, a pattern emerges. The pattern is that the ratio of the population of every consecutive year to the present year is 1.6

Let us check it using a couple of examples.

The rabbit population in the year 2010 is 50. The population increases to 80 the next year (2011). Now, $\frac{80}{50}=1.6$

Likewise, the rabbit population in the year 2011 is 80. The population increases to 128 the next year (2012). Again, $\frac{128}{80}$ $=1.6$

We can verify the same ratio with all the data provided.

Thus, we know that the population in any given year is 1.6 times the population of the previous year. This is a classic case of a compounding problem. We know that the formula for compounding is as:

$F=P\times r^n$

Where $F$ is the future value of the rabbit population in any given year

$P$ is the rabbit population in the year "0" (that is the starting year 2010) and that is 50 in this question. (please note that there is just one starting year).

$r$ is the ratio multiple with which the rabbit population increases each consecutive year.

$n$ is the nth year from the start.

Let us take an example for the better understanding of the working of this formula.

Let us take the year 2014. This is the 4th year

So, the rabbit population in 2014 should be:

$F_{2014} =50\times(1.6)^4\approx328$

This is exactly what we get from the table too.

Thus, $F=P\times r^n$ aptly represents the formula that dictates the rabbit population in the present question.

4 0

3 years ago