All categories

3 years ago

Sixty-five hundred in standard form

Mathematics

2 answers:

Xelga [282]3 years ago

8 0

In standard form, it would look like this: 6,500

Nata [24]3 years ago

7 0

In standard form it will be 6,500

You might be interested in

Below is the graph of y=b^x. Find b.

omeli [17]

Y = b^x

when x = 1
y = b^1
y = b

Therefore, the value of b is the same as the value of y when x =1
From the graph,
When x = 1, y = 0.5
Therefore, b = 0.5

To confirm this

From the graph,

When x = -1, y = 2
Since y = b^x
2 = b^-1
2 = 1/b
2b = 1
b = 0.5

When x = -2, y = 4
Since y = b^x
4 = b^-2
4 = 1/(b^2)
b^2 = 1/4
b = √(1/4)
b = 1/2
b = 0.5

Therefore, it is conformed that b = 0.5

5 0

3 years ago

Read 2 more answers

Any gírl here . . . . let's chàt

Rufina [12.5K]

Answer:

uhm no stop this is an app for people and work so please go

3 0

3 years ago

What is the value of the function y=3x−1 when x=−1 ? −8 −4 2 6

Paraphin [41]

Substitute x for -1 in y=3x-1
y = (3) (-1) -1
Solve
y = (3) (-1) -1
y = -3 - 1
y = -4

B.) -4 is your answer

3 0

4 years ago

Read 2 more answers

In a cinema theatre, why do we normally not hear echoes even though the sounds are loud?

ikadub [295]

Answer:

An echo is formed as a result of reflection of the original sound which returns to the listener. You might have experienced Echo in an EMPTY ROOM but no echo is produced after the room is occupied. To avoid echo from being produced the echo has to be absorbed. Echo in a cinema theatre is absorbed by the WALLS of the theatre.

They are not plain. The walls are corrugated or shaped to form tiny holes or empty spaces to ABSORB the ECHO. Modern Theatre's have ABSORBENT materials sticked to the walls of the theatre.

4 0

3 years ago

Identify the lower class limits, upper class limits, class width, class midpoints, and class boundaries for the given freque

bezimeni [28]

Answer:

See explanation and results below.

Step-by-step explanation:

We have the following data:

Age (yr) when award was won Frequency 25-34 31 35-44 36 45-54 12 55-64 4 65-74 6 75-84 1 85-94 2

If we order the data we got the following result. We assume that the midpoint is given by:

$Midpoint= \frac{L.Limit +U.Limit}{2}$

The $Class.Width = U.Limit-L.Limit$

And the limits are given by the lower and upper limit.

L.Limit U.Limit Class W Midpoint Frequency Limits

25 34 34-25=9 29.5 31 [25-34]

35 44 44-35=9 39.5 36 [35-44]

45 54 54-45=9 49.5 12 [45-54]

55 64 64-55=9 59.5 4 [55-64]

65 74 74-65=9 69.5 6 [65-74]

75 84 84-75=9 79.5 1 [75-84]

85 94 94-85=9 89.5 2 [85-94]

8 0

3 years ago