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3 years ago

Dan's science magazine has a mass of 256.674 grams. What is the mass of his magazine rounded to the nearest tenth?

2 answers:

4 0

The tenth place is the .6. Look over at the number next to it, the 7. It's greater than 5 so make the .6 one digit greater. Now your answer is 256.7

Serjik [45]3 years ago

3 0

The answer is 256.7

You round the .67 off to .7

The .6 in 256.674 is in the tenth place. You are rounding off that number. If the number next to it is greater then 5, you round up. If it's less then 5 you keep it the same.

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One serving of granola provides 4% of the protein you need daily. You must get the remaining 48 grams from other sources. How ma

Inessa [10]

X=grams of protein you need daily.
Remember (4%=4/100=0.04).

x=4% of x + 48 g

therefore, we can suggest this equation:
x=0.04x+ 48
x-0.04x=48
0.96x=48
x=48/0.96
x=50

Answer: 50 grams of proteins you need daily.

Other method:
the remaning is equal to =100%-4%=96%

Therefore: 96% of x=48
Remember (96%=96/100=0.96)

we can suggest this equation too.
0.96x=48
x=48/0.96
x=50

Answer:50 grams of proteins you need daily.

3 0

3 years ago

The mass of a grain of sand is about 103 gram.

KatRina [158]

1 grain = 103g

2kg = 2000g

grains = 2000g/103g = 19.417

Answer:

About 19.417 grains

3 0

4 years ago

If 6x -1 = 29, what is the value of x² + x?

irina1246 [14]

Answer: <em>The answer would be x (x+1)</em>

7 0

4 years ago

Read 2 more answers

Consider a discrete random variable x with pmf px (1) = c 3 ; px (2) = c 6 ; px (5) = c 3 and 0 otherwise, where c is a positive

PtichkaEL [24]

Looks like the PMF is supposed to be

$\mathbb P(X=x)=\begin{cases}\dfrac c3&\text{for }x\in\{1,5\}\\\\\dfrac c6&\text{for }x=2\\\\0&\text{otherwise}\end{cases}$

which is kinda weird, but it's not entirely clear what you meant...

Anyway, assuming the PMF above, for this to be a valid PMF, we need the probabilities of all events to sum to 1:

$\displaystyle\sum_{x\in\{1,2,5\}}\mathbb P(X=x)=\dfrac c3+\dfrac c6+\dfrac c3=\dfrac{5c}6=1\implies c=\dfrac65$

Next,

$\mathbb P(X>2)=\mathbb P(X=5)=\dfrac c3=\dfrac25$

$\mathbb E(X)=\displaystyle\sum_{x\in\{1,2,5\}}x\,\mathbb P(X=x)=\dfrac c3+\dfrac{2c}6+\dfrac{5c}3=\dfrac{7c}3=\dfrac{14}5$

$\mathbb V(X)=\mathbb E\bigg((X-\mathbb E(X))^2\bigg)=\mathbb E(X^2)-\mathbb E(X)^2$
$\mathbb E(X^2)=\displaystyle\sum_{x\in\{1,2,5\}}x^2\,\mathbb P(X=x)=\dfrac c3+\dfrac{4c}6+\dfrac{25c}3=\dfrac{28c}3=\dfrac{56}5$
$\implies\mathbb V(X)=\dfrac{56}5-\left(\dfrac{14}5\right)^2=\dfrac{84}{25}$

If $Y=X^2+1$ , then $X^2=Y-1\implies X=\sqrt{Y-1}$ , where we take the positive root because we know $X$ can only take on positive values, namely 1, 2, and 5. Correspondingly, we know that $Y$ can take on the values $1^2+1=2$ , $2^2+1=5$ , and $5^2+1=26$ . At these values of $Y$ , we would have the same probability as we did for the respective value of $X$ . That is,

$\mathbb P(Y=y)=\begin{cases}\dfrac c3&\text{for }y=2\\\\\dfrac c6&\text{for }y=5\\\\\dfrac c3&\text{for }y=26\\\\0&\text{otherwise}\end{cases}$

Part (5) is incomplete, so I'll stop here.

8 0

3 years ago

Me pueden ayudar? Yo no entiendo nada :(

V125BC [204]

yo tampoco :( pero pone lo que sea lo siento trate de poner 890 pero creo que ya las hkgsxv gfubi Yu hiuvv bien va vuub que n20 vuvkboonni 80? hjohgdw

4 0

3 years ago