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svetoff [14.1K]

3 years ago

6

Ok so I I don’t feel like doing math work while I’m sick can someone else figure out the answer because if I don’t turn this in

I get an F but I’m to sick to want to figure out the maze. Please help me

1 answer:

Serggg [28]3 years ago

5 0

Answer:

down, right, bottom left(1), right, right, right, finish

Step-by-step explanation:

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Find the midpoint of R(5,3) and S(-4,6)

irinina [24]

Answer:

(1/2, 9/2)

Step-by-step explanation:

Using the midpoint formula

$\left(\frac{-4+5}{2},\:\frac{6+3}{2}\right)$

= $\left(\frac{1}{2},\:\frac{9}{2}\right)$

7 0

3 years ago

Read 2 more answers

HELP PLEASE 23 POINTS I NEED U RN I NEED U GORL

artcher [175]

Answer:

No

Step-by-step explanation:

He is not correct because paper is a one-denominational surface, and volume only comes from 3 denominational objects.

6 0

3 years ago

Do you guys know the answer?

Andrews [41]

EFGH is twice the size of ABCD

6 0

3 years ago

If z1= 3+3i and z2=7(cos(5pi/9) + i sin (5pi/9)), then z1/z2= blank

mixas84 [53]

$z1=\stackrel{a}{3}+\stackrel{b}{3}i~~ \begin{cases} r = \sqrt{a^2+b^2}\\ r = \sqrt{18}\\[-0.5em] \hrulefill\\ \theta =\tan^{-1}\left( \frac{b}{a} \right)\\ \theta =\frac{\pi }{4} \end{cases}~\hfill z1=\sqrt{18}\left[\cos\left( \frac{\pi }{4} \right) i\sin\left( \frac{\pi }{4} \right) \right] \\\\[-0.35em] ~\dotfill$

$\cfrac{z1}{z2}\implies \cfrac{\sqrt{18}\left[\cos\left( \frac{\pi }{4} \right) i\sin\left( \frac{\pi }{4} \right) \right]} {7\left[\cos\left( \frac{5\pi }{9} \right) i\sin\left( \frac{5\pi }{9} \right) \right]} \\\\[-0.35em] ~\dotfill\\\\ \qquad \textit{division of two complex numbers} \\\\ \cfrac{r_1[\cos(\alpha)+i\sin(\alpha)]}{r_2[\cos(\beta)+i\sin(\beta)]}\implies \cfrac{r_1}{r_2}[\cos(\alpha - \beta)+i\sin(\alpha - \beta)] \\\\[-0.35em] ~\dotfill$

$\cfrac{z1}{z2}\implies \cfrac{\sqrt{18}}{7}\left[\cos\left( \frac{\pi }{4}-\frac{5\pi }{9} \right)+i\sin\left( \frac{\pi }{4}-\frac{5\pi }{9} \right) \right] \\\\\\ \cfrac{\sqrt{18}}{7}\left[\cos\left( \frac{-11\pi }{36} \right) +i\sin\left( \frac{-11\pi }{36} \right) \right]\implies \cfrac{\sqrt{18}}{7}\left[\cos\left( \frac{83\pi }{36} \right) +i\sin\left( \frac{83\pi }{36} \right) \right] \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \cfrac{z1}{z2}\approx 0.348~~ + ~~0.496i~\hfill$

6 0

3 years ago

Suppose 250 randomly selected people are surveyed to determine if they own a tablet. Of the 250 surveyed, 98 reported owning a t

SashulF [63]

Answer:

The true proportion for the population would be between 0.386 and 0,398 (0.392±0.06) in 95% confidence level.

Step-by-step explanation:

Confidence interval for true population proportion can be computed as p±ME where

p is the sample proportion ( $\frac{98}{250}=0.392$ )
ME is the margin of Error from the mean

and margin of error (ME) can be found using the formula

ME= $\frac{z*\sqrt{p*(1-p)}}{\sqrt{N} }$ where

z is the corresponding statistic in 95% confidence level (1.96)
p is the sample proportion ( $\frac{98}{250}=0.392$ )

N is the sample size (250)

Using the numbers

ME= $\frac{1.96*\sqrt{0.392*0.608}}{\sqrt{250} }$ ≈ 0.06

Then the true proportion for the population would be p±ME = 0.392±0.06 in 95% confidence level.

6 0

3 years ago