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Molodets [167]
3 years ago
11

The language arts teacher wants to know whether the students in the entire school prefer a debate or an speech. The teacher draw

s a random sample from the following groups:
All teachers in the school
All boys in each grade
All students in the speech club
All students in each grade
Which group best represents the population she should take a random sample from to get the best results for her survey? (5 points)

Group of answer choices

All teachers in the school

All boys in each grade

All students in the speech club

All students in each grade
Mathematics
2 answers:
love history [14]3 years ago
6 0

Answer:

it D answer D and it will be correct I just did this test guys well if it isnt D then its all students in each grade I just did this test lol

Step-by-step explanation:

svp [43]3 years ago
4 0

Answer:

It would be to ask all the students in the grade.

Step-by-step explanation:

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The zeros of a quadratic function are 6 and -4. Which of these cholces could be the function
Katen [24]

Answer:

Anything that looks a little like this: y = a(x - 6)(x + 4)

I'm not sure what the choices are because you didn't include them in the question.

7 0
3 years ago
An investment of $8,000 earns interest at an annual rate of 7% compounded continuously. Complete parts (A) and (B) below. Click
Sergeeva-Olga [200]

Answer:

A.    \mathtt{\dfrac{dA}{dt}|_{t=2}=644.15}

B.    \mathtt{\dfrac{dA}{dt}|_{t = 5.79}= 839.86 }

Step-by-step explanation:

Given that:

An investment of  Amount = $8000

earns  at an annual rate of interest = 7% = 0.07 compounded continuously

The objective is to :

A)  Find the instantaneous rate of change of the amount in the account after 2 year(s).

we all know that:

A = Pe^{rt}

where;

A = (8000) \ e ^{0.7t}

The instantaneous rate of change = \dfrac{dA}{dt}

\dfrac{dA}{dt} = \dfrac{d}{dt}(8000 \ e ^{0.07t} )

= 8000 \dfrac{d}{dt}e^{0.07 \ t}

\dfrac{dA}{dt}= 8000 (0.07)e^{0.07 \ t}

\dfrac{dA}{dt}= 560 e^{0.07 \ t}

At t = 2 years; the instantaneous rate of change is:

\dfrac{dA}{dt}|_{t=2}= 560 e^{0.07 \times 2}

\mathtt{\dfrac{dA}{dt}|_{t=2}=644.15}

(B) Find the instantaneous rate of change of the amount in the account at the time the amount is equal to $12,000.

Here the amount = 12000

12000 = (8000)e^{0.07 \ t}

\dfrac{12000 }{8000}= e^{0.07 \ t}

1.5= e^{0.07 \ t}

㏑(1.5) = 0.07 t

0.405465 = 0.07 t

t = 0.405465 /0.07

t = 5.79

\dfrac{dA}{dt}= 560 e^{0.07 \ t}

At t = 5.79

\dfrac{dA}{dt}|_{t = 5.79}= 560 e^{0.07 \times 5.79}

\mathtt{\dfrac{dA}{dt}|_{t = 5.79}= 839.86 }

7 0
3 years ago
One model of Earth's population growth is P(t)= 64/(1+11e^0.8t)
Pepsi [2]

Answer:

A. In 1991, there were 5.74 billion people  

D. The carrying capacity of Earth is 64 billion people

Have a nice day! :)

5 0
3 years ago
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3 years ago
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