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3 years ago

8

The following sample was collected during registration at a large middle school. At the 0.05 level of significance, can it be co

ncluded that level of math is dependent on grade level?
Honors Math Regular Math General Math
6th Grade 34 45 15
7th Grade 37 49 13
8th Grade 29 45 17
Hypotheses:
sub(H,0): Level of math is independent of/dependent on grade level.
sub(H,1): Level of math is independent of/dependent on grade level.
Enter the expected matrix - round to 4 decimal places.
Honors Math Regular Math General Math
6th Grade
7th Grade
8th Grade
After running an independence test, can it be concluded that level of math is dependent on grade level?
Yes/No

1 answer:

larisa [96]3 years ago

7 0

Answer:

use Ma.thway it helps

Step-by-step explanation:

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Alfred is taking a trip. He drove 150 mi and then stopped to refuel. He will drive 60 mph to finish his trip. Let x represent th

hodyreva [135]

Y = 60x +150

we put our ^0 mph as our Multiplier os x and since Alfred already traveled the 150 miles... that is just added on.

6 0

3 years ago

Reduce 25\ 1000 to its lowest terms

denis-greek [22]

Answer:

The answer would be 1/40 I believe. Hope this helps :)

Step-by-step explanation:

7 0

3 years ago

The average score of this quiz is 24 points with a standard deviation of 1.5 points. What is the probability that someone scored

navik [9.2K]

Answer:

8.08 %

Step-by-step explanation:

Let us denote : $X=$ score X follows a normal distribution with mean $=24$ and standard deviation $=1.5$

Step 2

The probability that someone scored 21.9 or less is :

$\mathrm{P}(\mathrm{X} \leq 21.9)$

$=\mathrm{P}\left(\mathrm{Z} \leq \frac{21.9-24}{1.5}\right)$

$=\mathrm{P}(\mathrm{Z} \leq-1.4)$

$=0.0808$

$=8.08%$

Answer : $8.08 %$

7 0

3 years ago

Work out the height of a triangle with base b= 7.7mm and area a=164.78mm2

solong [7]

Answer:

21.4 mm

Step-by-step explanation:

area ÷ base = height

164.78 mm^2 ÷ 7.7 mm

21.4 mm

Check:

base • height = area

7.7 mm • 21.4 mm = 164.78 mm^2

5 0

2 years ago

A given field mouse population satisfies the differential equation dp dt = 0.5p − 410 where p is the number of mice and t is the

ohaa [14]

Answer:

a) $t = 2 *ln(\frac{82}{5}) =5.595$

b) $t = 2 *ln(-\frac{820}{p_0 -820})$

c) $p_0 = 820-\frac{820}{e^6}$

Step-by-step explanation:

For this case we have the following differential equation:

$\frac{dp}{dt}=\frac{1}{2} (p-820)$

And if we rewrite the expression we got:

$\frac{dp}{p-820}= \frac{1}{2} dt$

If we integrate both sides we have:

$ln|P-820|= \frac{1}{2}t +c$

Using exponential on both sides we got:

$P= 820 + P_o e^{1/2t}$

Part a

For this case we know that p(0) = 770 so we have this:

$770 = 820 + P_o e^0$

$P_o = -50$

So then our model would be given by:

$P(t) = -50e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=-50 e^{1/2 t} +820$

$\frac{820}{50} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(\frac{82}{5}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(\frac{82}{5}) =5.595$

Part b

For this case we know that p(0) = p0 so we have this:

$p_0 = 820 + P_o e^0$

$P_o = p_0 -820$

So then our model would be given by:

$P(t) = (p_o -820)e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=(p_o -820)e^{1/2 t} +820$

$-\frac{820}{p_0 -820} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(-\frac{820}{p_0 -820}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(-\frac{820}{p_0 -820})$

Part c

For this case we want to find the initial population if we know that the population become extinct in 1 year = 12 months. Using the equation founded on part b we got:

$12 = 2 *ln(\frac{820}{820-p_0})$

$6 = ln (\frac{820}{820-p_0})$

Using exponentials we got:

$e^6 = \frac{820}{820-p_0}$

$(820-p_0) e^6 = 820$

$820-p_0 = \frac{820}{e^6}$

$p_0 = 820-\frac{820}{e^6}$

8 0

3 years ago