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3 years ago

If i were to yeet a rock the size of Greg Heffley at mach 8, what would be the velocity?

Mathematics

1 answer:

monitta3 years ago

5 0

9514 1440 393

Answer:

2744 m/s

Step-by-step explanation:

8 times the sea-level speed of sound is ...

8(343 m/s) = 2744 m/s

_____

At higher altitude, it will be slower.

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Answer as soon as possible

alexandr1967 [171]

Answer:

C −2a^3+9a^2+45a+6ab^2+18b^2

Step-by-step explanation:

(a+3) ( −2a^2+15a+6b^2)

Distribute the a to the large term in parentheses and the 3 to the large term in parentheses

a ( −2a^2+15a+6b^2)+3 ( −2a^2+15a+6b^2)

−2a^3+15a^2+6ab^2 −6a^2+45a+18b^2

Combine like terms

−2a^3+15a^2−6a^2+6ab^2 +45a+18b^2

-2 a^3 + 9 a^2 + 6 a b^2 + 45 a + 18 b^2

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2 years ago

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In triangle LMN, P lies on LM and Q lies on LN. If Pq is 1/2Mn, PQ is called what

xz_007 [3.2K]

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6 0

2 years ago

7/13 + (-4/13) the answer has to be in simplest form.

Digiron [165]

Answer:

short form:0.23 Long Form:0.23076923076

Step-by-step explanation:

8 0

3 years ago

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omeli [17]

Step-by-step explanation:

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2 years ago

Recent census data indicated that 14.2% of adults between the ages of 25 and 34 live with their parents. A random sample of 125

kicyunya [14]

Answer:

The probability is $P(14 < X < 20 ) = 0.5354$

Step-by-step explanation:

From the question we are told that

The proportion that live with their parents is $\r p = 0.142$

The sample size is n = 125

Given that there are two possible outcomes and that this outcomes are independent of each other then we can say the Recent census data follows a Binomial distribution

i.e

$X \ \~ \ B( \mu , \sigma )$

Now the mean is evaluated as

$\mu = n * \r p$

$\mu = 125 * 0.142$

$\mu = 17.75$

Generally the proportion that are not staying with parents is

$\r q = 1 - \r p$

= > $\r q = 0.858$

The standard deviation is mathematically evaluated as

$\sigma = \sqrt{n * \r p * \r q }$

$\sigma = \sqrt{ 125 * 0.142 * 0.858 }$

$\sigma = 3.90$

Given the n is large then we can use normal approximation to evaluate the probability as follows

$P(14 < X < 20 ) = P( \frac{ 14 - 17.75}{3.90}$

Now applying continuity correction

$P(14 < X < 20 ) = P( \frac{ 13.5 - 17.75}{3.90} < \frac{ X - \mu }{\sigma } < \frac{ 19.5 - 17.75}{3.90} )$

Generally

$\frac{ X - \mu }{\sigma } = Z ( The \ standardized \ value \ of X )$

$P(14 < X < 20 ) = P( \frac{ 13.5 - 17.75}{3.90} < Z< \frac{ 19.5 - 17.75}{3.90} )$

$P(14 < X < 20 ) = P( -1.0897 < Z< 0.449 } )$

$P(14 < X < 20 ) = P( Z< 0.449 ) - P(Z < -1.0897)$

So for the z - table

$P( Z< 0.449 ) = 0.67328$

$P(Z < -1.0897) = 0.13792$

$P(14 < X < 20 ) = 0.67328 - 0.13792$

$P(14 < X < 20 ) = 0.5354$

6 0

3 years ago