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3 years ago

Which is the best estimate of the number of beads on a cord that is 16 inches long?

Mathematics

2 answers:

Kitty [74]3 years ago

8 0

Answer:

the most sensible answer is probably 48

Step-by-step explanation:

16 inches is a feet and 4 inches long that's long for beads

s2008m [1.1K]3 years ago

3 0

Answer: The correct answer will be D:48

Step-by-step explanation:

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For the data set 2.5, 6.5, 9, 19, 20, 2.5. What is the mean?

Ahat [919]

Answer:

9.9166667

Step-by-step explanation:

All of the numbers equal 59.5 then divide by 6 equals the answer above.

3 0

2 years ago

A statistician chooses 27 randomly selected dates, and when examining the occupancy records of a particular motel for those date

Anni [7]

Answer:

Confidence interval variance [21.297 ; 64.493]

Confidence interval standard deviation;

4.615, 8.031

Step-by-step explanation:

Given :

Variance, s² = 34.34

Standard deviation, s = 5.86

Sample size, n = 27

Degree of freedom, df = 27 - 1 = 26

Using the relation for the confidence interval :

[s²(n - 1) / X²α/2, n-1] ; [s²(n - 1) / X²1-α/2, n-1]

From the chi distribition table :

X²α/2, n-1 = 41.923 ; X²1-α/2, n-1 = 13.844

Hence,

[34.34*26 / 41.923] ; [34.34*26 / 13.844]

[21.297 ; 64.493]

The 95% confidence interval for the population variance is :

21.297 < σ² < 64.493

Standard deviation is the square root of variance, hence,

The 95% confidence interval for the population standard deviation is :

4.615 < σ < 8.031

3 0

2 years ago

A tank contains 1600 L of pure water. Solution that contains 0.04 kg of sugar per liter enters the tank at the rate 2 L/min, and

goldfiish [28.3K]

Let S(t) denote the amount of sugar in the tank at time t. Sugar flows in at a rate of

(0.04 kg/L) * (2 L/min) = 0.08 kg/min = 8/100 kg/min

and flows out at a rate of

(S(t)/1600 kg/L) * (2 L/min) = S(t)/800 kg/min

Then the net flow rate is governed by the differential equation

$\dfrac{\mathrm dS(t)}{\mathrm dt}=\dfrac8{100}-\dfrac{S(t)}{800}$

Solve for S(t):

$\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{S(t)}{800}=\dfrac8{100}$

$e^{t/800}\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{e^{t/800}}{800}S(t)=\dfrac8{100}e^{t/800}$

The left side is the derivative of a product:

$\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/800}S(t)\right]=\dfrac8{100}e^{t/800}$

Integrate both sides:

$e^{t/800}S(t)=\displaystyle\frac8{100}\int e^{t/800}\,\mathrm dt$

$e^{t/800}S(t)=64e^{t/800}+C$

$S(t)=64+Ce^{-t/800}$

There's no sugar in the water at the start, so (a) S(0) = 0, which gives

$0=64+C\impleis C=-64$

and so (b) the amount of sugar in the tank at time t is

$S(t)=64\left(1-e^{-t/800}\right)$

As $t\to\infty$ , the exponential term vanishes and (c) the tank will eventually contain 64 kg of sugar.

7 0

3 years ago

60 minutes is 20% of ? minute s

Hatshy [7]

60 minutes is 20% of 300 minutes

3 0

2 years ago

Read 2 more answers

Applying a quadratic regression model y = ax2 + bx + C,

Semmy [17]

I think the answer is c= 0.248

8 0

2 years ago