3X+4Y=-8<br> -9X-12Y=-10

How do you find the area. My brain is mush... And please do explain.

mel-nik [20]

The formula for the area is l • w
( length • width) 4•12= 48

8 0

3 years ago

Read 2 more answers

In space, the intersection of a line and a plane is

vredina [299]

I believe the answer is A) Not possible not sure

5 0

4 years ago

Question 2 (1 point)

vlada-n [284]

(3,0)

Step-by-step explanation:

Different - add

2x - y = 6

+

x + y. = 3

3x = 9

÷3

x = 3

Substitute x = 3 into one of the equation:

x + y = 3

3 + y = 3

-3

y = 0

Solution is (3,0)

Hope this helps!

4 0

3 years ago

Bodytemperatures ofhealthy koalasarenormally distributed with a mean of 35.6°C and a standard deviation of 1.3°C. a.What is the

Licemer1 [7]

Answer:

a

$P(X < 35) = 32.3 \%$

b

$P(\= X < 35) = 0.6 \%$

Here the probability of koalas mean temperature being less than 35 °C is very small hence the koalas are not healthy

c

A potential confounding variable for this study is the population of the koalas because in the first question the population was not taken into account and the probability was $P(X < 35) = 32.3 \%$ but when the population was taken into account (i.e n = 30) the probability became

$P(\= X < 35) = 0.6 \%$

Step-by-step explanation:

From the question we are told that

The mean is $\mu = 35.6^oC$

The standard deviation is $s = 1.3^oC$

The sample size is n = 30

Generally the probability that a health koala has a body temperature less than 35.0°C is mathematically represented as

$P(X < 35) = P(\frac{X - \mu }{s} < \frac{35 - 35.6}{1.3} )$

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

So

$P(X < 35) = P(Z < -0.46)$

From the z-table P(Z < -0.46) = 0.323

So

$P(X < 35) = 0.323$

Converting to percentage

$P(X < 35) = 0.323 * 100$

$P(X < 35) = 32.3 \%$

considering question b

The sample mean is $\= x = 35$

Generally the standard error of the mean is mathematically represented as

$\sigma_{\= x} = \frac{s}{\sqrt{n} }$

=> $\sigma_{\= x} = \frac{1.3}{\sqrt{30} }$

=> $\sigma_{\= x} = 0.2373$

Generally the probability of the mean body temperature of koalas being less than 35.0°C is mathematically represented as

$P(\= X < 35) = P(\frac{\= X - \mu }{\sigma_{\= x }} < \frac{35 -35.6}{0.2373 } )$

$P(\= X < 35) = P(Z< -2.53 )$

From the z-table we have that

$P(Z< -2.53 ) = 0.006$

So

$P(\= X < 35) = 0.006 /tex] Converting to percentage [tex]P(\= X < 35) = 0.006 * 100$

$P(\= X < 35) = 0.6 \%$

Here the probability of koalas mean temperature being less than 35 °C is very small hence the koalas are not healthy

A potential confounding variable for this study is the population of the koalas because in the first question the population was not taken into account and the probability was $P(X < 35) = 32.3 \%$ but when the population was taken into account (i.e n = 30) the probability became

$P(\= X < 35) = 0.6 \%$

4 0

4 years ago

Andrew picked vegetables from his garden and put them into bags. He put 10 carrots into each of 3 bags, 8 tomatoes into each of

xz_007 [3.2K]

10 carrots in 3 bags
8 tomatoes in 4 bags
6 turnips in 5 bags

10*3= 30 carrots
8*4= 32 tomatoes
6*5= 30 turnips

(30*2)+30= 92 vegetables

Answer: C

3 0

3 years ago

Read 2 more answers

3X+4Y=-8 -9X-12Y=-10