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3 years ago

What is the range of 24.35cm 22.51cm 20.98cm 30.33cm and 30.98cm ?

1 answer:

6 0

The range is the variance between the smallest and largest value in the data set. The range is 10; calculated by taking 20.98 and subtracting it from 30.98.

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In numb3rs, they use math to solve crimes. One of the ways they use Math is to reveal patterns. What is the pattern in this sequ

Mama L [17]

Step-by-step explanation:

I think the first number is 2

Then 2 x 2 = 4 so there is a difference of 4 number

Then in between 6 and 14 there is a difference of 8 so 4 x 2 = 8

Then 16 so 8 x 2 = 16

So after 126 the number is 254

5 0

3 years ago

I need help ASAP <br> Integrated math II

Molodets [167]

Answer:

$\boxed{ \bold{ \sf{x = 16}}}$

$\boxed{ \bold{ \sf{m < AFD = 82}}}$

Step-by-step explanation:

$\sf{5x + 18 = 3x + 50}$

( Being vertically opposite angles)

Vertically opposite angles are always equal.

Move variable to L.H.S and change it's sign

Similarly, Move constant to R.H.S and change it's sign

⇒ $\sf{5x - 3x = 50 - 18}$

Collect like terms

⇒ $\sf{2x = 50 - 18}$

Subtract 18 from 50

⇒ $\sf{2x = 32}$

Divide both sides of the equation by 2

⇒ $\sf{ \frac{2x}{2} = \frac{32}{2} }$

Calculate

⇒ $\sf{x = 16}$

The value of x is 16

Now, let's find value of m<AFD :

$\sf{m < afd + 5x + 18 = 180}$ ( sum of angle in straight line )

plug the value of x

⇒ $\sf{m < AFD + 5 \times 16 + 18 = 180}$

Multiply the numbers

⇒ $\sf{m < AFD + 80 + 18 = 180}$

Add the numbers

⇒ $\sf{m < AFD + 98 = 180}$

Move constant to R.H.S and change it's sign

⇒ $\sf{m < AFD = 180 - 98}$

Subtract 98 from 180

⇒ $\sf{m < AFD = 82}$

Value of m<AFD = 82

Hope I helped!

Best regards!!

3 0

3 years ago

John looked at his bank account and noticed that he had $2,029.90 in his account. All that he could remember is that he created

Over [174]

We will have that for the previous year, he had in the account:

$m_{y6}=2029.90-\frac{2029.90\cdot6}{100}\Rightarrow m_{y6}=1908.106$

For the year previous to that one, he had:

$m_{y5}=1908.106-\frac{1908.106\cdot6}{100}\Rightarrow m_{y5}=1793.61964$

For the previous year to that, he had:

$m_{y4}=1793.61964-\frac{1793.61964\cdot6}{100}\Rightarrow m_{y4}=1686.002462$

For the year previous to that one, he had:

$m_{y3}=1686.002462-\frac{1686.002462\cdot6}{100}\Rightarrow m_{y3}=1584.842314$

For the second year that the money was in the account, there was:

$m_{y2}=1584.842314-\frac{1584.842314\cdot6}{100}\Rightarrow m_{y2}=1489.751775$

And the orinial ammount of money that was in the account was:

$m_{y1}=1489.751775-\frac{1489.751775\cdot6}{100}\Rightarrow m_{y1}=1400.366669$

From this, we know that Jhon had originally $1400.37 in the bank account.

6 0

1 year ago

What is your estimate for the number of airline flights in a year?

Firdavs [7]

The answer is between 45,000,000-65,000,000

8 0

3 years ago

HELP ASAP. Two angles have to add up to 180, but have a difference of 24. What angles are they?

kykrilka [37]

Two angles that add up to 180 is supplementary

8 0

3 years ago