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saul85 [17]
3 years ago
10

Which of the following is NOT true when testing a claim about a proportion

Mathematics
1 answer:
marysya [2.9K]3 years ago
7 0

Question: Which of the following is NOT true when testing a claim about a proportion?

Answer: A conclusion based on a confidence interval estimate will be the same as a conclusion based on a hypothesis test.

I don't really know the answer choices, but I think that's right. Sorry if not.

Hope this helps. c;


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Cho phương trình: x1 + x2 + x3 + x4 + x5 = 30 (*)
4vir4ik [10]

Answer:

Step-by-step explanation:

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6 0
3 years ago
I WILL MARKED BRAINLIEST IF YOU COULD ANSWER THIS!
tester [92]

Answer:

s = -2400t + 17400

Step-by-step explanation:

Let's say t is the x value on a coordinate plane, and s is the y. Then, we have the points (0, 17400) and (6, 3000). The slope of these is 14400/-6 or -2400.

Now we just have the equation y = -2400x + b, and from the point (0, 17400) we can find that b is 17400. So, we have y = -2400x + 17400. Convert these back into t and s and you get your answer, s = -2400t + 17400.

6 0
3 years ago
5m^{2}\sqrt{25m^{8}+15m^{6}+5m^{2}}
xxTIMURxx [149]
You can square the whole problem to cancel out the square root. Might make things easier.

5 0
3 years ago
A tank with a capacity of 500 gal originally contains 200 gal of water with 100 lb of salt in solution. Water containing 1 lb of
saw5 [17]

Let A(t) denote the amount of salt in the tank at time t.

Salt flows in at a rate of

(1 lb/gal) * (3 gal/min) = 3 lb/min

and flows out at a rate of

(A(t)/(200 + t) lb/gal) * (2 gal/min) = 2 A(t)/(500 + t)

(in case you're unsure about the denominator: the tank starts off with 200 gal of solution, and each minute solution flows in at a rate of 3 gal/min and thus the tank gains (3 gal/min) * (1 min) = 3 gal. At the same time, solution flows out at a rate of 2 gal/min and thus the tank loses 2 gal, giving a net change in volume of (3 - 2)*t = t gal)

Then the net rate of salt flow is given by the ODE,

\dfrac{\mathrm dA(t)}{\mathrm dt}-\dfrac{2A(t)}{200+t}=3

Multiply both sides by (200+t)^{-2}:

(200+t)^{-1}\dfrac{\mathrm dA(t)}{\mathrm dt}-2(200+t)^{-3}A()=3(200+t)^{-2}

\implies\dfrac{\mathrm d}{\mathrm dt}\bigg((200+t)^{-2}A(t)\bigg)=3(200+t)^{-2}

Integrating both sides and solving for A(t) gives

(200+t)^{-2}A(t)=-\dfrac3{200+t}+C

A(t)=-2(200+t)+C(200+t)^2

The tank starts off with 100 lb of salt in solution, so A(0)=100 and we find

100=-2(200)+C(200)^2\implies C=\dfrac1{80}

and so

A(t)=-2(200+t)+\dfrac{(200+t)^2}{80}=\dfrac{(200+t)(40+t)}{80}

The tank will begin to overflow once the volume of solution reaches 500 gal; this happens when

500=200+t\implies t=300

or 300 minutes or 5 hours after solution starts flowing. At this point, the tank will contain

A(300)=2125

or 2125 lb of salt.

Theoretically, the amount of salt in the tank will increase forever, since A(t)\to\infty as t\to\infty.

6 0
4 years ago
The dog has gone 5/8 of the way across the yard. How much farther does the it have to go to reach the gate?
Alekssandra [29.7K]

3/8 s


It already went 5/8. 8/8 is the whole yard. 8/8 - 5/8 is 3/8

4 0
3 years ago
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