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vampirchik [111]
3 years ago
10

According to the Centers for Disease control and prevention, 20.8 of high school students currently use electronic cigarettes. A

high school counselor is concerned that the use of e-cigs at her school is higher.
(a) Determine the null and alternative hypotheses.
(b) If the sample data indicate that the null hypothesis should not be rejected, state the conclusion of the school counselor.
(c) Suppose, in fact, that the proportion of students at the counselor's high school who use electronic cigarettes is 0.034. Was a Type I or Type II error committed?
Mathematics
1 answer:
Setler79 [48]3 years ago
6 0

Answer:

H0 : p = 0.028

H1: p > 0.028

We conclude that there is not enough evidence to support the school counselor's claim they the use of electronic cigarettes in high school is higher Than 2.8%

Type 11 error

Step-by-step explanation:

H0 : p = 0.028

H1: p > 0.028

If null should not be rejected then;

We conclude that there is not enough evidence to support the school counselor's claim they the use of electronic cigarettes in high school is higher Than 2.8%

3.)

If the actual proportion is 0.034, which means 3.4%, this means it is actually higher. For failing to reject a false Null, then a type 11 error has been committed

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How many ways can a group of three boys and three girls be seated in a row of six seats if boys and girls must alternate in the​
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Answer:

24

Step-by-step explanation:

there are 6 seats and 3 boys, 3 girls.

First seat can be given to either boy or a girl.

So there are two ways.

If a boy is selected say, then he has to occupy 3rd and 5th only.

There are 2 ways

Now available positions are 2nd, 4th and 6th, in this girls can be arranged in 3! ways = 6

So total number of ways =

2x2x6 = 24

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At a cost of $4 per pound, how many ounces of rolled oats can be bought for $1.20?
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Since a pound has 16 ounces, each ounce would cost 4 divided by 16, or 25 cents. 1.20 divided by .25 = 4 4/5. 4.8 ounces.

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3 years ago
Circle any equivalent ratios from the list below.
marysya [2.9K]

Answer:

Equivalent ratios: we can that the first ratio is equivalent to the second,  then third ratio is equivalent to the forth.  

Ratio: 1: 2 Value of the Ratio:  1/2

Ratio: 5: 10 Value of the Ratio:  1/2

Ratio: 6: 16 Value of the Ratio:  3/8

Ratio: 12: 32 Value of the Ratio: 3/8

We notice that if the values are equivalent the ratios are equivalent

Step-by-step explanation:

Equivalent ratios:

To get if ratios are equivalent we look for the constant between ratios

a.Ratio: 1: 2  and Ratio: 5: 10  

We apply the method of comparing the first term of both ratios ,  and the second term of both ratios.  We see the constant is 5 ( 1/5 is equal to 2/10)

We do the same with third and forth ratio

Ratio: 6: 16 compare to Ratio: 12: 32

6/12 is equal to 16/32 the constant is 2

<u>So,  we can that the first ratio is equivalent to the second,  then, third ratio is equivalent to the forth.  </u>

Value of the Ratio:  The value is a ratio written as a fraction.

Ratio: 1: 2 Value of the Ratio:  1/2

Ratio: 5: 10 Value of the Ratio:  5/10 if we divide both sides by 5,  we can say  Value of the Ratio:  1/2

Ratio: 6: 16 Value of the Ratio:  6/16 if we divide both sides by 2,  we can say the value is 3/ 8

Ratio: 12: 32 Value of the Ratio: 12/32 if we divide both sides by 4,  we can say the value is 3/ 8

<u>If the values are equivalent the ratios are equivalent.</u>

7 0
3 years ago
Read 2 more answers
A container initially containing10 L of water in which there is 20 g of salt dissolved. A solution containing 4 g/L of salt is p
Kay [80]

Answer:

A(40)= \frac{-200}{10+40} +4 (10 +40)=-4+200 = 196

Step-by-step explanation:

For this case the solution flows at a rate of 2L/min and leaves at 1L/min. So then we can conclude the volume is given by V= 10 +t

Since the initial volume is 10 L and the volume increase at a rate of 1L/min.

For this case we can define A as the concentration for the salt in the container. And for this case we can set up the following differential equation:

\frac{dA}{dt}= 4 \frac{gr}{L} *2 \frac{L}{min} - \frac{A}{10+t}

Because at the begin we have a concentration of 8 gr/L and would be decreasing at a rate of \frac{A}{10+t}

So then we can reorder the differential equation like this:

\frac{dA}{dt} +\frac{A}{10+t} =8

We find the solution using the integration factor:

\mu = -\int \frac{1}{10+t} dt = -ln(10+t)

And then the solution would be given by:

A = e^{-ln (10+t)} (\int e^{\int \frac{1}{10+t} dt})

And if we simplify this we got:

A= \frac{1}{10+t} (c + \int (10 +t) 8 dt)

And after do the integral we got:

A= \frac{c}{10+t} +4 (10 +t)

And using the initial condition t=0 A= 20 we have this:

20 = \frac{c}{10} +40

c= -200

So then we have this function for the solution of A:

A= \frac{-200}{10+t} +4 (10 +t)

And now replacinf t= 40 we got:

A(40)= \frac{-200}{10+40} +4 (10 +40)=-4+200 = 196

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