All categories

3 years ago

Which metric unit is 1000 times bigger than a gram?

Mathematics

2 answers:

11111nata11111 [884]3 years ago

6 0

Answer:

Step-by-step explanation:

a kilogram

makkiz [27]3 years ago

5 0

<h3>▬▬▬▬▬</h3><h2>Kilogram</h2>

<em>kilo</em> in math means <em>thousand.</em>

A gram is 1 metric unit of a kilogram alone, so,

$1 * 1000 = 1kilogram$

Your answer is kilogram.

Hope this helps! Have a great day. ♣

You might be interested in

Use the given graph to determine the limit, if it exists. A coordinate graph is shown with a horizontal line crossing the y axis

Lesechka [4]

Answer:

The limit of the function does not exists.

Step-by-step explanation:

From the graph it is noticed that the value of the function is 6 from all values of x which are less than 2. At x=2, the line y=6 has open circle. It means x=2 is not included.

For x<2

$f(x)=6$

The value of the function is -3 from all values of x which are greater than 2. At x=2, the line y=-3 has open circle. It means x=2 is not included.

For x>2

$f(x)=-3$

The value of y is 1 at x=2, because of he close circles on (2,1).

For x=2

$f(x)=1$

Therefore the graph represents a piecewise function, which is defined as

$f(x)=\begin{cases}6& \text{ if } x2 \end{cases}$

The limit of a function exist at a point a if the left hand limit and right hand limit are equal.

$lim_{x\rightarrow a^-}f(x)=lim_{x\rightarrow a^+}f(x)$

The function is broken at x=2, therefore we have to find the left and right hand limit at x=2.

$lim_{x\rightarrow 2^-}f(x)=6$

$lim_{x\rightarrow 2^+}f(x)=-3$

$6\neq-3$

Since the left hand limit and right hand limit are not equal therefore the limit of the function does not exists.

6 0

3 years ago

Read 2 more answers

How many units long is a segment whose endpoints are (2,3) and (7,15)?

Ahat [919]

Answer:

13 units

Step-by-step explanation:

Distance formula: $d = \sqrt{(x2-x1)^2 + (y2-y1)^2}$

Simply plug in your variables

$d = \sqrt{(7-2)^2 + (15-3)^2}$

$d = \sqrt{5^2 + 12^2}$

$d = \sqrt{25 + 144}$

$d = \sqrt{169}$

d = 13

5 0

3 years ago

Past records indicate that the probability of online retail orders that turn out to be fraudulent is 0.08. Suppose that, on a gi

Sunny_sXe [5.5K]

Answer:

The probability that there are 2 or more fraudulent online retail orders in the sample is 0.483.

Step-by-step explanation:

We can model this with a binomial random variable, with sample size n=20 and probability of success p=0.08.

The probability of k online retail orders that turn out to be fraudulent in the sample is:

$P(x=k)=\dbinom{n}{k}p^k(1-p)^{n-k}=\dbinom{20}{k}\cdot0.08^k\cdot0.92^{20-k}$

We have to calculate the probability that 2 or more online retail orders that turn out to be fraudulent. This can be calculated as:

$P(x\geq2)=1-[P(x=0)+P(x=1)]\\\\\\P(x=0)=\dbinom{20}{0}\cdot0.08^{0}\cdot0.92^{20}=1\cdot1\cdot0.189=0.189\\\\\\P(x=1)=\dbinom{20}{1}\cdot0.08^{1}\cdot0.92^{19}=20\cdot0.08\cdot0.205=0.328\\\\\\\\P(x\geq2)=1-[0.189+0.328]\\\\P(x\geq2)=1-0.517=0.483$