Ask question

Ask question

All categories

2 years ago

13

Use the given margin of error, confidence level, and population standard deviation, sigma, to find the minimum sample size r

equired to estimate an unknown population mean, mu. Margin of error: 1.1 inches, confidence level: 90%, sigmaequals2.9 inches A confidence level of 90% requires a mimimum sample size of ______. (Round up to the nearest integer.)

1 answer:

lidiya [134]2 years ago

8 0

Answer:

19

Step-by-step explanation:

Margin of Error (E) = t × sd/√n

√n = t ×sd/E

n = (t × sd/E)^2

Using a 90% confidence level, t = 1.645, sd = 2.9 inches, E = 1.1 inches

n = (1.645 × 2.9/1.1)^2 = 4.34^2 = 18.807

Minimum sample size required = 19 (to the nearest integer)

You might be interested in

Serenity is going to invest $850 and leave it in an account

Nat2105 [25]

Answer:

5.86

Step-by-step explanation: its right on delta math

4 0

2 years ago

I need help with this tysm

erica [24]

9514 1404 393

Answer:

-51
-30
-60
2 or -2

Step-by-step explanation:

The function puts a minus sign on 3 times the magnitude of the input.

1. y = -|3×17| = -51

2. y = -|3×10| = -30

3. y = -|3×20| = -60

4. -6 = -|3x|

2 = |x| . . . . divide by -3

x = ±2 . . . . two possibilities

8 0

2 years ago

A cafe offers senior citizens a 15% discount off its regular price of $8.95 for the dinner buffet.

jonny [76]

a) 75 percent of the regular price is the price for senior citizens.

b) $6.712 is the price for senior citizens.

<u>Step-by-step explanation</u>:

The regular price is the 100% = $8.95
The offer price for senior citizen = 15% discount from 100%

∴ Percent of the regular price for senior citizens = 100% - 15% = 75%

The price for the senior citizens = 75% of $8.95

⇒ (75/100) $\times$ 8.95

⇒ 0.75 $\times$ 8.95

⇒ 6.712 dollars.

∴ The price for senior citizens = $6.712

7 0

3 years ago

Find the length of segment EG. Show all the work

svp [43]

Answer:

Hi I love you and I miss you

I don't know the answear sorry

5 0

2 years ago

Problem 10: A tank initially contains a solution of 10 pounds of salt in 60 gallons of water. Water with 1/2 pound of salt per g

AysviL [449]

Answer:

The quantity of salt at time t is $m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })$ , where t is measured in minutes.

Step-by-step explanation:

The law of mass conservation for control volume indicates that:

$\dot m_{in} - \dot m_{out} = \left(\frac{dm}{dt} \right)_{CV}$

Where mass flow is the product of salt concentration and water volume flow.

The model of the tank according to the statement is:

$(0.5\,\frac{pd}{gal} )\cdot \left(6\,\frac{gal}{min} \right) - c\cdot \left(6\,\frac{gal}{min} \right) = V\cdot \frac{dc}{dt}$

Where:

$c$ - The salt concentration in the tank, as well at the exit of the tank, measured in $\frac{pd}{gal}$ .

$\frac{dc}{dt}$ - Concentration rate of change in the tank, measured in $\frac{pd}{min}$ .

$V$ - Volume of the tank, measured in gallons.

The following first-order linear non-homogeneous differential equation is found:

$V \cdot \frac{dc}{dt} + 6\cdot c = 3$

$60\cdot \frac{dc}{dt} + 6\cdot c = 3$

$\frac{dc}{dt} + \frac{1}{10}\cdot c = 3$

This equation is solved as follows:

$e^{\frac{t}{10} }\cdot \left(\frac{dc}{dt} +\frac{1}{10} \cdot c \right) = 3 \cdot e^{\frac{t}{10} }$

$\frac{d}{dt}\left(e^{\frac{t}{10}}\cdot c\right) = 3\cdot e^{\frac{t}{10} }$

$e^{\frac{t}{10} }\cdot c = 3 \cdot \int {e^{\frac{t}{10} }} \, dt$

$e^{\frac{t}{10} }\cdot c = 30\cdot e^{\frac{t}{10} } + C$

$c = 30 + C\cdot e^{-\frac{t}{10} }$

The initial concentration in the tank is:

$c_{o} = \frac{10\,pd}{60\,gal}$

$c_{o} = 0.167\,\frac{pd}{gal}$

Now, the integration constant is:

$0.167 = 30 + C$

$C = -29.833$

The solution of the differential equation is:

$c(t) = 30 - 29.833\cdot e^{-\frac{t}{10} }$

Now, the quantity of salt at time t is:

$m_{salt} = V_{tank}\cdot c(t)$

$m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })$

Where t is measured in minutes.

7 0

3 years ago