Ask question

Ask question

All categories

vladimir2022 [97]

2 years ago

11

A teacher would like to estimate the true mean amount of time her students spend completing a particular homework assignment. Th

e day the homework is due, the teacher selects a random sample of 30 of her 75 students and records the amount of time that each of them spent completing the assignment. Are the conditions for constructing a t confidence interval met?
a. No, the random condition is not met.
b. No, the 10% condition is not met.
c. No, the Normal/large sample condition is not met.
d. Yes, the conditions for inference are met.

1 answer:

finlep [7]2 years ago

6 0

B no the 10% condition is not met

You might be interested in

A bag of marbles can be divided evenly among 2,3 or 4 friends. A) How many marbles might be in the bag? B) What is the least num

agasfer [191]

Answer:

a)12 b)12 c)36

Step-by-step explanation:

a)find least common multiple. the number of marbles is that number * n (where n is an integer)

b) the least common multiple. it is 4*3=...

c)12*3 =?

then just divide by 2,3,4 to get answer

6 0

3 years ago

An unusually wet spring has caused the size of a mosquito population in some city to increase by 8% each day. If an estimated

tatyana61 [14]

Answer:

The number of mosquitoes that will inhabit the city on May 25 is 4,471,200.

Step-by-step explanation:

Note: This question is not complete as some important figures and points are omitted. The complete question is therefore provided before answering the question as follows:

An unusually wet spring has caused the size of a mosquito population in some city to increase by 8% each day. If an estimated 180,000 mosquitoes are in the city on May 12, find how many mosquitoes will inhabit the city on May 25. Use y= 180,000(1.08)x where x is number of days since May 2.

The explanation to the answer is now given as follows:

Given;

y = 180,000(1.08)x = 180,000 * 1.08 * x .................... (1)

Where;

y = Number of mosquitoes that will inhabit the city on May 25 = ?

x = Number of days since May 2 = 25 - 2 = 23

Substituting for x = 23 into equation (1), we have:

y = 180,000 * 1.08 * 23

y = 4,471,200

Therefore, the number of mosquitoes that will inhabit the city on May 25 is 4,471,200.

3 0

2 years ago

Help please if your good at maths ?

Norma-Jean [14]

Answer:

Year 7 = 75 students

Year 9 = 25 students

Step-by-step explanation:

Year 7 has 3/8 of the total since the circle is divided into 8 sections and has 3 of the 8 sections

3/8 * 200 students = 75

Year 9 has 1/8 of the total since the circle is divided into 8 sections and has 1 of the 8 sections

1/8 * 200 =25

3 0

3 years ago

Evaluate the expression

Neporo4naja [7]

Simplify the expression.

Exact form:

$\frac{476}{25}$

Decimal form:

19.04

Mixed Number Form:

$19\frac{1}{25}$

8 0

3 years ago

Initially a tank contains 10 liters of pure water. Brine of unknown (but constant) concentration of salt is flowing in at 1 lite

zhenek [66]

Answer:

Therefore the concentration of salt in the incoming brine is 1.73 g/L.

Step-by-step explanation:

Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.

Let the concentration of salt be a gram/L

Let the amount salt in the tank at any time t be Q(t).

$\frac{dQ}{dt} =\textrm {incoming rate - outgoing rate}$

Incoming rate = (a g/L)×(1 L/min)

=a g/min

The concentration of salt in the tank at any time t is = $\frac{Q(t)}{10}$ g/L

Outgoing rate = $(\frac{Q(t)}{10} g/L)(1 L/ min)$ $\frac{Q(t)}{10} g/min$

$\frac{dQ}{dt} = a- \frac{Q(t)}{10}$

$\Rightarrow \frac{dQ}{10a-Q(t)} =\frac{1}{10} dt$

Integrating both sides

$\Rightarrow \int \frac{dQ}{10a-Q(t)} =\int\frac{1}{10} dt$

$\Rightarrow -log|10a-Q(t)|=\frac{1}{10} t +c$ [ where c arbitrary constant]

Initial condition when t= 20 , Q(t)= 15 gram

$\Rightarrow -log|10a-15|=\frac{1}{10}\times 20 +c$

$\Rightarrow -log|10a-15|-2=c$

Therefore ,

$-log|10a-Q(t)|=\frac{1}{10} t -log|10a-15|-2$ .......(1)

In the starting time t=0 and Q(t)=0

Putting t=0 and Q(t)=0 in equation (1) we get

$- log|10a|= -log|10a-15| -2$

$\Rightarrow- log|10a|+log|10a-15|= -2$

$\Rightarrow log|\frac{10a-15}{10a}|= -2$

$\Rightarrow |\frac{10a-15}{10a}|=e ^{-2}$

$\Rightarrow 1-\frac{15}{10a} =e^{-2}$

$\Rightarrow \frac{15}{10a} =1-e^{-2}$

$\Rightarrow \frac{3}{2a} =1-e^{-2}$

$\Rightarrow2a= \frac{3}{1-e^{-2}}$

$\Rightarrow a = 1.73$

Therefore the concentration of salt in the incoming brine is 1.73 g/L

8 0

3 years ago