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2 years ago

Problem 3.2

Mathematics

1 answer:

Zielflug [23.3K]2 years ago

3 0

Answer:

1:1

Step-by-step explanation:

w = f, which means the amount of water equals the amount of flour.

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Ty has 230 pages of text to read in the next three days. If he averages 25 pages every 1

Leto [7]

Answer:

waittt

in 3 days there r only 72 hours

soo... if 25 pages every 114 hrs would be wayyy past 3 days

Ans. it will take him 1048. 8 hrs

so basically 43.7 days

7 0

3 years ago

Suppose that the mean time that visitors stay at a museum is 94.2 minutes with a standard deviation of 15.5 minutes. The standar

Maurinko [17]

Answer:

$94.2 -0.994*3.1 = 91.1186$

$94.2 +0.994*3.1 = 97.2814$

And the 68% confidence interval is given by (91.1186, 97.2814)

Step-by-step explanation:

For this case we know that mean time that visitors stay at a museum is given by:

$\bar X = 94.2$

The standard deviation is given by:

$s= 15.5$

And the standard error is given by:

$SE = \frac{s}{\sqrt{n}} =3.1$

And we want to interval captures 68% of the means for random samples of 25 scores and for this case the critical value can be founded like this using the normal standard distribution or excel:

$z_{\alpha/2}= \pm 0.994$

We can find the interval like this:

$\bar X \pm ME$

And replacing we got:

$94.2 -0.994*3.1 = 91.1186$

$94.2 +0.994*3.1 = 97.2814$

And the 68% confidence interval is given by (91.1186, 97.2814)

7 0

2 years ago

How to find matrics M

makkiz [27]

$\left[\begin{array}{ccc}9&7\\-4&2\end{array}\right] -4M=\left[\begin{array}{ccc}-3&3\\4&-18\end{array}\right]\qquad(*) \\\\M=\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]\to4M=\left[\begin{array}{ccc}4a&4b\\4c&4d\end{array}\right]$

$(*)\qquad\left[\begin{array}{ccc}9&7\\-4&2\end{array}\right] -\left[\begin{array}{ccc}4a&4b\\4c&4d\end{array}\right]=\left[\begin{array}{ccc}-3&3\\4&-18\end{array}\right]\\\\\left[\begin{array}{ccc}9-4a&7-4b\\-4-4c&2-4d\end{array}\right]=\left[\begin{array}{ccc}-3&3\\4&-18\end{array}\right]\\.\qquad\qquad\qquad\qquad\ \ \ \Downarrow\\(_{11})\qquad9-4a=-3\ \ \ |-9\\-4a=-12\ \ \ \ |:(-4)\\a=3\\\\(_{12})\qquad7-4b=3\ \ \ \ |-7\\-4b=-4\ \ \ \ |:(-4)\\b=1$

$(_{21})\qquad-4-4c=4\ \ \ \ |+4\\-4c=8\ \ \ \ |:(-4)\\c=-2\\\\(_{22})\qquad2-4d=-18\ \ \ \ |-2\\-4d=-20\ \ \ \ \ |:(-4)\\d=5\\\\Answer:\ A.\ M=\left[\begin{array}{ccc}3&1\\-2&5\end{array}\right]$