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3 years ago

16.437 rounded to the neatest tenth is 16.4 true or false

2 answers:

7 0

Answer=True

When rounding 16.437 to the nearest tenth, we need to look to the hundreds place.

16.437

The number 3 is in the hundreds place

The rules of rounding are:

If the number is 5 or greater, then you round up.

If the number is 4 or less, Then you round down.

Since 3 is less than 4, we would round down. So...

16.437 becomes 16.4

Pavlova-9 [17]3 years ago

7 0

It is true because after the decimal the digit is in the tenths catagory then hundreths and thousandths etc. That is why it is tenths.
:)

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To measure the distance across a pond, Sandy and Garrett mark a spot at on

DerKrebs [107]

Answer:

distance across the pond: 13 yd

distance walked by Garret: 12 yd

Step-by-step explanation:

Data:

distance across the pond: x yd

distance walked by Sandy: 5 yd

distance walked by Garret: x-1 yd

A right triangle is formed where the distance across the pond is the hypotenuse. From Pythagorean theorem:

x² = 5² + (x-1)²

x² = 25 + x² - 2x + 1

0 = 26 - 2x

x = 13

4 0

3 years ago

Rectangle J'K'L'M' shown on the grid is the image of rectangle JKLM after transformation. The same transformation will be applie

-BARSIC- [3]

Translation is $(h, k)$ → $(h+7, k+5)$
T is $(6, -6)$
So T' is $(6+7, -6+5)=(13, -1)$

4 0

3 years ago

Read 2 more answers

What is 6 tenths 20 hundredths 4 ones and 50 hundreds

Elina [12.6K]

Answer:

The answer is 25064.

Step-by-step explanation:

There is a mistake in the question so the correct is below:

What is 6 tenths 20 thousandths 4 ones and 50 hundreds.

Now, to solve it:

So, we arrange it in sequence:

20 thousandths 50 hundreds 6 tenths and 4 ones.

Now,

20 thousandths = 20 000

50 hundreds = 50 00

6 tenths = 6 0

4 ones = 4

Thus, 20 thousandths 50 hundreds 6 tenths and 4 ones is :

20,000 + 5000 + 60 + 4.

= 25064.

Therefore, the answer is 25064.

3 0

3 years ago

Whats the roots of the following three problems:

klemol [59]

QUESTION 1

The given function is

$f(x)=\frac{(x-3)(x-1)}{(x-1)(x+2)}$

We simplify to get;

$f(x)=\frac{x-3}{x+2}$

This function is equal to zero when $x-3=0$

$\Rightarrow x=3$

QUESTION 2

The given function is

$f(x)=\frac{5x^2-10x+5}{2x^2-5x+3}$

$f(x)=\frac{5(x^2-2x+1)}{2x^2-5x+3}$

We factor to get;

$f(x)=\frac{5(x-1)^2}{(2x-3)(x-1)}$

$f(x)=\frac{5(x-1)}{(2x-3)}$

This function equals zero when

$5(x-1)=0$

$x=1$

QUESTION 3

The given function is

$f(x)=\frac{x^3-8}{x^2-6x+8}$

We factor to get,

$f(x)=\frac{(x-2)(x^2+2x+4)}{(x-2)(x+4)}$

$f(x)=\frac{x^2+2x+4}{x+4}$

The function equals zero when $x^2+2x+4=0$

$D=b^2-4ac$

$D=2^2-4(1)(4)$

$D=-12$

Hence the equation has no real roots.

The complex roots are

$x=-\sqrt{3}i-1$ or $x=-1+\sqrt{3}i$

4 0

3 years ago

In a jar, there is a green, a blue, and a red marble. You draw (with replacement) a marble from the jar until you get the green

denpristay [2]

Answer:

The value is $P( X = m ) = (\frac{2}{3} )^{m-1 } * \frac{1}{3}$

Step-by-step explanation:

From question we are told that

The number of types of marbles present jar is n = 3

Generally the probability of drawing a green marble is

$P(g) = \frac{1}{3}$

Generally the probability of drawing a marble that is not green is

$P(g') = 1 - \frac{1}{3} = \frac{2}{3}$

From the question we are told that there will be continuous drawing of marbles from the jar (in such a way that after each marble is drawn it is being replaced) until a green marble drawn

Let m be the number of times marbles has been drawn when a green marble was gotten

it then means that for m - 1 times the marbles where drawn a green marble was not obtain.

Generally the probability drawing m times is mathematically is mathematically represented as

$P( X = m ) = (P(g'))^{m-1 } * P(g)$

=> $P( X = m ) = (\frac{2}{3} )^{m-1 } * \frac{1}{3}$

7 0

3 years ago