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4 years ago

Is 645 round to 600 when rounded to the nearest ten

Mathematics

1 answer:

Juliette [100K]4 years ago

6 0

No
645 becomes 650 when rounded to the nearest ten
645 becomes 600 when rounded to the nearest hundred

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A fruit market equally divided 80 red apples into 8 containers, and equally divided 54 green apples into 6 containers. Natalie b

Goryan [66]

Answer:

10 + 9 = n

19 apples

Step-by-step explanation:

80 ÷ 8 = y and 56 ÷ 6 = a

n = y + a

80 ÷ 8 × 8 = y × 8

80 = 8y

80 ÷ 8 = 8y ÷ 8

y = 10

54 ÷ 6 = a

54 ÷ 6 × 6 = a × 6

6a = 54

6a ÷ 6 = 54 ÷ 6

a = 9

10 + 9 = n

19 = n

7 0

3 years ago

1. 3x + 7 = 13 (x = -2; x = 2; x = 5)

White raven [17]

3x + 7 = 1 3

3x = 13 - 7

3x = 6

x = 6/3

x = 2

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3 years ago

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I know its summer and everyone done with school but someone help

Gemiola [76]

Answer:

1.55

Step-by-step explanation:

3750+900 = 4650

4650/3000 = 1.55

7 0

4 years ago

Read 2 more answers

A combination lock requires a user to select three numbers to unlock it, where all three numbers are between 1 to 30, inclusive,

Sonbull [250]

Answer:

$\frac{1}{270}=0.00370$

Step-by-step explanation:

As the numbers are only between 1 and 30

Total number of possible combinations = 30×30×30 = 27000

If there are 100 attempts then

$\text{P (someone guesses correct)}=\frac{\text{Number of attempts}}{\text{Total number of possible combinations}}\\\Rightarrow \text{P (someone guesses correct)}=\frac{100}{27000}\\\Rightarrow \text{P (someone guesses correct)}=\frac{1}{270}=0.00370$

∴ Probability someone can guess the correct combination given 100 attempts is $\mathbf{\frac{1}{270}=0.00370}$

5 0

3 years ago

student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te

alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = $\frac{1}{4}$ .

Thus, the probability that the student does not receive version A = $1-\frac{1}{4}$ = $\frac{3}{4}$ .

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

$\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}$