One ounce is equal to 28.35 grams. convert 16 ounces to grams round your answe to the nearest tenth

traffig75
2022-09-27
Answered

One ounce is equal to 28.35 grams. convert 16 ounces to grams round your answe to the nearest tenth

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r2t1orrso

Answered 2022-09-28
Author has **8** answers

16 ounces is equivalent to 453.6g.

Data;

1 ounce = 28.35 grams

Conversion of ounce to grams

This is the conversion of a mass unit from ounce to grams. Given that 1 ounce is equal to 28.35 grams, let us calculate how grams would be in 16 ounces.

The easiest way about this is simply multiply 16 by 28.25g

$1\text{ounce}=28.35g\phantom{\rule{0ex}{0ex}}16\text{ounce}=x\phantom{\rule{0ex}{0ex}}x=16\cdot 28.35\phantom{\rule{0ex}{0ex}}x=453.6g$

From the calculations above, 16 ounces is equivalent to 453.6g.

Data;

1 ounce = 28.35 grams

Conversion of ounce to grams

This is the conversion of a mass unit from ounce to grams. Given that 1 ounce is equal to 28.35 grams, let us calculate how grams would be in 16 ounces.

The easiest way about this is simply multiply 16 by 28.25g

$1\text{ounce}=28.35g\phantom{\rule{0ex}{0ex}}16\text{ounce}=x\phantom{\rule{0ex}{0ex}}x=16\cdot 28.35\phantom{\rule{0ex}{0ex}}x=453.6g$

From the calculations above, 16 ounces is equivalent to 453.6g.

asked 2022-08-16

Convert -15 degrees Celsius into degrees Fahrenheit using the formula $F=\frac{9}{5}C+32$Round your answer to one decimal place.

asked 2022-09-17

How many quarts are in 814 gallons?

Note: 1 gallon = 4 quarts

A. 2 1/8 qt

B.16 1/2 qt

C.33 qt

D.66 qt

Note: 1 gallon = 4 quarts

A. 2 1/8 qt

B.16 1/2 qt

C.33 qt

D.66 qt

asked 2022-05-27

I have to prove the following:

Let $X,Y$ be countable sets. Show that $\mathcal{P}(X)\otimes \mathcal{P}(Y)=\mathcal{P}(X\times Y)$.

I'm not sure if in my case, $\mathcal{P}(X)\otimes \mathcal{P}(Y)$ is a product $\sigma $-algebra and thus defined as $\mathcal{P}(X)\otimes \mathcal{P}(Y)=\sigma (\{{B}_{1}\times {B}_{2}|{B}_{1}\in \mathcal{P}(X),{B}_{2}\in \mathcal{P}(Y)\})$.

I know that the power set of a set is a $\sigma $-algebra. So, it would make sense that $\mathcal{P}(X)\otimes \mathcal{P}(Y)$ is a product $\sigma $-algebra as defined above.

Can somebody confirm this or explain it if it's not the case?

Let $X,Y$ be countable sets. Show that $\mathcal{P}(X)\otimes \mathcal{P}(Y)=\mathcal{P}(X\times Y)$.

I'm not sure if in my case, $\mathcal{P}(X)\otimes \mathcal{P}(Y)$ is a product $\sigma $-algebra and thus defined as $\mathcal{P}(X)\otimes \mathcal{P}(Y)=\sigma (\{{B}_{1}\times {B}_{2}|{B}_{1}\in \mathcal{P}(X),{B}_{2}\in \mathcal{P}(Y)\})$.

I know that the power set of a set is a $\sigma $-algebra. So, it would make sense that $\mathcal{P}(X)\otimes \mathcal{P}(Y)$ is a product $\sigma $-algebra as defined above.

Can somebody confirm this or explain it if it's not the case?

asked 2022-06-15

Let's say we conduct a random experiment. The possible outcomes may be too complicated to describe, so we instead take some measurement (in terms of real numbers) from it which are of interest to us. Mathematically we have defined a random variable X from the sample space/measurable space $(\mathrm{\Omega},\mathcal{A})$ to $(\mathbb{R},\mathcal{B})$ where $\mathcal{B}$ is the Borel sigma field. We then choose an appropriate probability for each Borel set, via a distribution function. Then, how does that distribution function determine the probability space $(\mathrm{\Omega},\mathcal{A},P)$ in theory or proves its existence? What result is implicitly used here?

I understand the probability space on $(\mathbb{R},\mathcal{B})$ is well defined, and that is what practically concerns us, but in theory we also have a probability space $(\mathrm{\Omega},\mathcal{A},P)$ which pushes forward its measure to the Borel sets. Hence given the distribution on $X$ we are pulling back to $(\mathrm{\Omega},\mathcal{A})$ via $P(X\in B)={P}_{X}(B)$. But what theorem guarantees that such a space exists?

Edit for clarification:For example, let us take weather $\mathrm{\Omega}$ of a city as the original sample space and the recorded temperature $X$ as the random variable. Suppose the distribution of $X$ is decided upon, based on empirical data. That's fine and now we can answer questions like 'what is $P(a<X<b)$'. We may not care about the probability space on $\mathrm{\Omega}$ since all the probabilities we wish to know relate to $X$, but how do we know for sure that there exists a probability space on $\mathrm{\Omega}$ in the first place, and further, how do we know that a probability space on $\mathrm{\Omega}$ exists which would push its probability to yield the distribution of $X$? Unless we know that there exists such a probability space in theory, we will not be able to talk about expectation of $X$, so the knowledge that such a space exists is crucial.

*Further edit: Note that we may not have an explicit complete mathematical model of the weather ever, but that does not bother us. We do have the ability however, of taking measurements of different types and thereby getting distributions related to temperature, pressure, wind speed, historical records etc. So practically we can answer questions related to these measurements. This is so what happens in practice I think. What I want is some mathematical theorem which assures me that there exists a model of the weather space, even though it is hard/difficult/impossible to find, which pushes its probability on to these distributions.

I understand the probability space on $(\mathbb{R},\mathcal{B})$ is well defined, and that is what practically concerns us, but in theory we also have a probability space $(\mathrm{\Omega},\mathcal{A},P)$ which pushes forward its measure to the Borel sets. Hence given the distribution on $X$ we are pulling back to $(\mathrm{\Omega},\mathcal{A})$ via $P(X\in B)={P}_{X}(B)$. But what theorem guarantees that such a space exists?

Edit for clarification:For example, let us take weather $\mathrm{\Omega}$ of a city as the original sample space and the recorded temperature $X$ as the random variable. Suppose the distribution of $X$ is decided upon, based on empirical data. That's fine and now we can answer questions like 'what is $P(a<X<b)$'. We may not care about the probability space on $\mathrm{\Omega}$ since all the probabilities we wish to know relate to $X$, but how do we know for sure that there exists a probability space on $\mathrm{\Omega}$ in the first place, and further, how do we know that a probability space on $\mathrm{\Omega}$ exists which would push its probability to yield the distribution of $X$? Unless we know that there exists such a probability space in theory, we will not be able to talk about expectation of $X$, so the knowledge that such a space exists is crucial.

*Further edit: Note that we may not have an explicit complete mathematical model of the weather ever, but that does not bother us. We do have the ability however, of taking measurements of different types and thereby getting distributions related to temperature, pressure, wind speed, historical records etc. So practically we can answer questions related to these measurements. This is so what happens in practice I think. What I want is some mathematical theorem which assures me that there exists a model of the weather space, even though it is hard/difficult/impossible to find, which pushes its probability on to these distributions.

asked 2022-06-24

Pratt's Lemma is : $\xi ,\eta ,\zeta $ and ${\xi}_{n},{\eta}_{n},{\zeta}_{n}$ such that:

${\xi}_{n}\to \xi ,{\eta}_{n}\to \eta ,{\zeta}_{n}\to \zeta ,\text{convergence in probability}$

and ${\eta}_{n}\le {\xi}_{n}\le {\zeta}_{n}$, $E{\zeta}_{n}\to E\zeta ,E{\eta}_{n}\to E\eta $, and $E\zeta ,E\eta ,E\xi $ are finite, prove :

If ${\eta}_{n}\le 0\le {\zeta}_{n}$, then $E|{\xi}_{n}-\xi |\to 0$.

I know how to prove $E{\xi}_{n}\to E\xi $, but $E|{\xi}_{n}-\xi |\to 0$ seems not easy proved from it.

My first question is how to prove $E|{\xi}_{n}-\xi |\to 0$.

And I don't know why should emphasize the condition "${\eta}_{n}\le 0\le {\zeta}_{n}$", so my second question is: Is there any example that $E|{\xi}_{n}-\xi |\to 0$ is wrong if the condition is not fullfilled.

${\xi}_{n}\to \xi ,{\eta}_{n}\to \eta ,{\zeta}_{n}\to \zeta ,\text{convergence in probability}$

and ${\eta}_{n}\le {\xi}_{n}\le {\zeta}_{n}$, $E{\zeta}_{n}\to E\zeta ,E{\eta}_{n}\to E\eta $, and $E\zeta ,E\eta ,E\xi $ are finite, prove :

If ${\eta}_{n}\le 0\le {\zeta}_{n}$, then $E|{\xi}_{n}-\xi |\to 0$.

I know how to prove $E{\xi}_{n}\to E\xi $, but $E|{\xi}_{n}-\xi |\to 0$ seems not easy proved from it.

My first question is how to prove $E|{\xi}_{n}-\xi |\to 0$.

And I don't know why should emphasize the condition "${\eta}_{n}\le 0\le {\zeta}_{n}$", so my second question is: Is there any example that $E|{\xi}_{n}-\xi |\to 0$ is wrong if the condition is not fullfilled.

asked 2022-06-22

Referring to the Doob's Maximal inequality we have that letting ${M}_{t}$ be a continous martingale, $\lambda >0$, $p>1$ and $T>0$ we have $P(\underset{0\le t\le T}{sup}|{M}_{t}|>\lambda )\le \frac{1}{{\lambda}^{p}}\mathbb{E}[|{M}_{T}{|}^{p}]$. I did not understand if the sup is the sup of the events ($|{M}_{t}|>\lambda $ : $t\in [0,T]$) or if it is the sup of the random variable $|{M}_{t}|$ itself.

asked 2022-08-10

Which is greater 2 miles or 3 kilometers