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tino4ka555 [31]
3 years ago
9

Find the greatest integer value of $b$ for which the expression $\frac{9x^3+4x^2+11x+7}{x^2+bx+8}$ has a domain of all real numb

ers.
Mathematics
1 answer:
vovangra [49]3 years ago
3 0

Consider fraction \dfrac{9x^3+4x^2+11x+7}{x^2+bx+8}. The domain of this fraction is x^2+bx+8\neq 0.

Find the discriminant of the the equation x^2+bx+8=0:

D=b^2-4\cdot 1\cdot 8=b^2-32.

There are choices:

1. If D>0, there are two solutions of the equation;

2. D=0, there is only unique solution of the equation;

3. D<0, there are no solutions.

If x^2+bx+8\neq 0 you should consider case 3, then
b^2-32.

Therefore, the greatest integer is 5.

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A test preparation claims that more than​ 50% of the students who take their test prep course improve their scores by at least 1
satela [25.4K]

Answer:

Step-by-step explanation:

Corresponding scores before and after taking the course form matched pairs.

The data for the test are the differences between the scores before and after taking the course.

μd = scores before taking the course minus scores before taking the course.

a) For the null hypothesis

H0: μd ≥ 0

For the alternative hypothesis

H1: μd < 0

b) We would assume a significance level of 0.05. The​ P-value from the test is 0.65. The p value is high. It increases the possibility of accepting the null hypothesis.

Since alpha, 0.05 < than the p value, 0.65, then we would fail to reject the null hypothesis. Therefore, it does not provide enough evidence that scores after the course are greater than the scores before the course.

c) The mean difference for the sample scores is greater than or equal to zero

3 0
3 years ago
A regular hexagon is dilated by a scale factor of 7/5 to create a new hexagon. How does the perimeter of the new hexagon compare
Airida [17]
<span>It's the same ratio, 7/5.</span>
3 0
3 years ago
Read 2 more answers
2/5 of the members of a school band are 6th graders. What percent of
faust18 [17]

Answer:

60%

Step-by-step explanation:

3/5 is 60%

7 0
3 years ago
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A random experiment was conducted where a Person A tossed five coins and recorded the number of ""heads"". Person B rolled two d
cestrela7 [59]

Answer:

(10) Person B

(11) Person B

(12) P(5\ or\ 6) = 60\%

(13) Person B

Step-by-step explanation:

Given

Person A \to 5 coins (records the outcome of Heads)

Person \to Rolls 2 dice (recorded the larger number)

Person A

First, we list out the sample space of roll of 5 coins (It is too long, so I added it as an attachment)

Next, we list out all number of heads in each roll (sorted)

Head = \{5,4,4,4,4,4,3,3,3,3,3,3,3,3,3,3,2,2,2,2,2,2,2,2,2,2,1,1,1,1,1,0\}

n(Head) = 32

Person B

First, we list out the sample space of toss of 2 coins (It is too long, so I added it as an attachment)

Next, we list out the highest in each toss (sorted)

Dice = \{2,2,3,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,5,5,6,6,6,6,6,6,6,6,6,6\}

n(Dice) = 30

Question 10: Who is likely to get number 5

From person A list of outcomes, the proportion of 5 is:

Pr(5) = \frac{n(5)}{n(Head)}

Pr(5) = \frac{1}{32}

Pr(5) = 0.03125

From person B list of outcomes, the proportion of 5 is:

Pr(5) = \frac{n(5)}{n(Dice)}

Pr(5) = \frac{8}{30}

Pr(5) = 0.267

<em>From the above calculations: </em>0.267 > 0.03125<em> Hence, person B is more likely to get 5</em>

Question 11: Person with Higher median

For person A

Median = \frac{n(Head) + 1}{2}th

Median = \frac{32 + 1}{2}th

Median = \frac{33}{2}th

Median = 16.5th

This means that the median is the mean of the 16th and the 17th item

So,

Median = \frac{3+2}{2}

Median = \frac{5}{2}

Median = 2.5

For person B

Median = \frac{n(Dice) + 1}{2}th

Median = \frac{30 + 1}{2}th

Median = \frac{31}{2}th

Median = 15.5th

This means that the median is the mean of the 15th and the 16th item. So,

Median = \frac{5+5}{2}

Median = \frac{10}{2}

Median = 5

<em>Person B has a greater median of 5</em>

Question 12: Probability that B gets 5 or 6

This is calculated as:

P(5\ or\ 6) = \frac{n(5\ or\ 6)}{n(Dice)}

From the sample space of person B, we have:

n(5\ or\ 6) =n(5) + n(6)

n(5\ or\ 6) =8+10

n(5\ or\ 6) = 18

So, we have:

P(5\ or\ 6) = \frac{n(5\ or\ 6)}{n(Dice)}

P(5\ or\ 6) = \frac{18}{30}

P(5\ or\ 6) = 0.60

P(5\ or\ 6) = 60\%

Question 13: Person with higher probability of 3 or more

Person A

n(3\ or\ more) = 16

So:

P(3\ or\ more) = \frac{n(3\ or\ more)}{n(Head)}

P(3\ or\ more) = \frac{16}{32}

P(3\ or\ more) = 0.50

P(3\ or\ more) = 50\%

Person B

n(3\ or\ more) = 28

So:

P(3\ or\ more) = \frac{n(3\ or\ more)}{n(Dice)}

P(3\ or\ more) = \frac{28}{30}

P(3\ or\ more) = 0.933

P(3\ or\ more) = 93.3\%

By comparison:

93.3\% > 50\%

Hence, person B has a higher probability of 3 or more

7 0
3 years ago
For the graph x = 12 find the slope of a line that is perpendicular to it and the slope of a line parallel to it. Explain your a
blagie [28]
Check the picture below.

that's the line of x = 12, just a straight vertical line, notice the green line, that's parallel to it, and the red line, that's perpendicular to it.

let's pick two points for each to get their slopes, hmm say for the green one (5,2) and (5,4)

\bf (\stackrel{x_1}{5}~,~\stackrel{y_1}{2})\qquad &#10;(\stackrel{x_2}{5}~,~\stackrel{y_2}{4})&#10;\\\\\\&#10;% slope  = m&#10;slope =  m\implies &#10;\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{4-2}{5-5}\implies \stackrel{und efined}{\cfrac{2}{0}}

and for the red one hmmm (3,2) and (7,2)

\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{2})\qquad &#10;(\stackrel{x_2}{7}~,~\stackrel{y_2}{2})&#10;\\\\\\&#10;% slope  = m&#10;slope =  m\implies &#10;\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{2-2}{7-3}\implies \cfrac{0}{4}\implies 0

8 0
3 years ago
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