All categories

3 years ago

Round 10,031.464 to the nearest hundred.

Mathematics

1 answer:

insens350 [35]3 years ago

7 0

Answer:

10,031.46

Rounded to the nearest 0.01 or

the Hundredths Place.

Step-by-step explanation:

You might be interested in

Length=x width=x+16 AREA of rectangle=80X^2 FIND x

valina [46]

Will i get brainliest if i answer?
Here is same thing but different numbers

12 in2 = · 3 inSince 4 · 3 = 12, we get (4 in) · (3 in) = 12 in2. So must equal 4 in. = 4 in

4 0

3 years ago

To predict what diameter a child would draw a nickel given family income, it would be most appropriate to: ___________

e-lub [12.9K]

Answer:

a. Use the regression equation

Step-by-step explanation:

Given

See attachment for missing part of question

Required

What can be used to predict diameter, given family income

We can see (by observation) that the data on the table can only be represented on a scattered plot, and scattered plot uses regression equation for data prediction.

i.e. $y = mx + b$ --- equation of line of best fit

In this case;

$x \to income$

$y \to diameter$

When the x value (income) is known, we can simply plug in the value in the above equation to get the y value (diameter)

5 0

3 years ago

I know I need to use the change of base

Zinaida [17]

The change of base formula looks like this:
log x
(log to the base b of ) x = ----------- (you could also use ln instead of log)
log b

log 40 1.602
Then (log to the base 7 of) 40 = ------------ = ------------- = 1.896 (answer)
log 7 0.8451

check (on calculator): 7^1.896 = 40.023 (close enough to 40 to verify
that this method "worked")

7 0

3 years ago

Please help me with this :/ I will give Brainiest! I just really need help. Please

ollegr [7]

Did you ever get the answer to this?

8 0

3 years ago

Draw two segments. How many common points is it possible for the segments to have? Explore all possibilities.

Arte-miy333 [17]

A segment is bounded by two endpoints.

The two segments can have up to two common points

Assume the line segments are AB and CD where the length of AB is greater than the length of CD.

The possibilities are:

A point of segment CD lies on segment AB
Both points of segment CD lie on segment AB.

See attachment for both possibilities.

Hence, it is possible for the two segments to have two common points.