Round 387.869911589to 3 decimal places

Hatshy [7]

<span>Since this is an SAT Math Level 2 problem derivatives should not be required to find the solution. To find "How many more hours of daylight does the day with max sunlight have than May 1," all you need to understand is that sin(x) has a maximum value of 1.

The day with max sunlight will occur when sin(2*pi*t/365) = 1, giving the max sunlight to be 35/3 + 7/3 = 14 hours

Evaluating your equation for sunlight when t = 41, May 1 will have about 13.18 hours of sunlight.

The difference is about 0.82 hours of sunlight.

Even though it is unnecessary for this problem, finding the actual max sunlight day can be done by solving for t when d = 14, of by the use of calculus. Common min/max problems on the SAT Math Level 2 involve sin and cos, which both have min values of -1 and max values of 1, and also polynomial functions with only even powered variables or variable expressions, which have a min/max when the variable or variable expression equals 0.

For example, f(x) = (x-2)^4 + 4 will have a min value of 4 when x = 2. Hope this helps</span>

3 0

3 years ago

Indicate which property is illustrated in step 1

Vedmedyk [2.9K]

Answer:

Step-by-step explanation:

communicative because you moved the three to the other side

6 0

2 years ago

Which expression is equivalent to third- 32^3/5

lara [203]

Answer:

Step-by-step explanation:

(-32)^3/5 =((-32)^1/5)^3 = (-2)^3 = -8. QED

6 0

2 years ago

For the following telescoping series, find a formula for the nth term of the sequence of partial sums

gtnhenbr [62]

I'm guessing the sum is supposed to be

$\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}$

Split the summand into partial fractions:

$\dfrac1{(5k-1)(5k+4)}=\dfrac a{5k-1}+\dfrac b{5k+4}$

$1=a(5k+4)+b(5k-1)$

If $k=-\frac45$ , then

$1=b(-4-1)\implies b=-\frac15$

If $k=\frac15$ , then

$1=a(1+4)\implies a=\frac15$

This means

$\dfrac{10}{(5k-1)(5k+4)}=\dfrac2{5k-1}-\dfrac2{5k+4}$

Consider the $n$ th partial sum of the series:

$S_n=2\left(\dfrac14-\dfrac19\right)+2\left(\dfrac19-\dfrac1{14}\right)+2\left(\dfrac1{14}-\dfrac1{19}\right)+\cdots+2\left(\dfrac1{5n-1}-\dfrac1{5n+4}\right)$

The sum telescopes so that

$S_n=\dfrac2{14}-\dfrac2{5n+4}$

and as $n\to\infty$ , the second term vanishes and leaves us with

$\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}=\lim_{n\to\infty}S_n=\frac17$

7 0

3 years ago

Wires manufactured for use in a computer system are specified to have resistances between 0.11 and 0.13 ohms. The actual measure

Marina CMI [18]

Answer:

a) $P(0.11$

And we can find this probability with this difference and with the normal standard table or excel:

$P(-1.11$

b) $P(0.11 < \bar X < 0.13)$

And we can use the z score defined by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And using the limits we got:

$z = \frac{0.11-0.12}{\frac{0.009}{\sqrt{4}}}= -2.22$

$z = \frac{0.13-0.12}{\frac{0.009}{\sqrt{4}}}= 2.22$

And we want to find this probability:

$P(-2.22< Z$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Part a

Let X the random variable that represent the resitances of a population, and for this case we know the distribution for X is given by:

$X \sim N(0.12,0.009)$

Where $\mu=0.12$ and $\sigma=0.009$

We are interested on this probability

$P(0.11$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(0.11$

And we can find this probability with this difference and with the normal standard table or excel:

$P(-1.11$

Part b

We select a sample size of n =4. And since the distribution for X is normal then we know that the distribution for the sample mean $\bar X$ is given by: