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2 years ago

Solve x2 - 8x - 9 = 0. Rewrite the equation so that it is of the form x2 + bx = c.

Mathematics

2 answers:

barxatty [35]2 years ago

4 0

Answer:

x=9,-1 and $x^2+(-8)x=9$

Explanation:

we have been given with the quadratic equation $x^2-8x-9$

we compare the given quadratic equation with general quadratic equation

general quadratic is $ax^2+bx+c=0$

from given quadratic equation a=1,b= -8,c= -9

substituting these values in the formula for discriminant $D=b^{2}-4ac$

$D=(-8)^2-4(1)(-9)=100$

Now, to find the value of x

Formula is $x=\frac{-b\pm\sqrt{D}}{2a}$

Now, substituting the values we will get

$x=\frac{-(-8)\pm\sqrt{100}} {2}= \frac{8\pm10}{2}=9,-1$

And rewritting the given equation by shifting 9 to right hand side of the given equation and taking minus inside the bracket so as to convert it in the form of

$x^2+bx=c$

sweet [91]2 years ago

4 0

It can be noted from the given that a = 1, b = -8 and c = -9. To transpose c to the other side of the equation, all that needs to be done is to add 9 to both sides of the equation,
x² - 8x -9 + 9 = 0 + 9
The answer would be,
x² - 8x = 9

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£1 (pound) is worth 1.64 euros. Jane changes £400 into euros. how many euros does she recieve?

Dimas [21]

1.64 euros times by 400 pounds is 656 euros

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2 years ago

Of the 400 eighth-graders at pascal middle school, 117 take algebra, 109 take advanced computer, and 114 take industrial technol

Hatshy [7]

If 164 of the 400 eighth-graders take none of these courses, then 400-164=236 students take some courses.

Let x students take all three courses, then:

1. 70-x take only both algebra and advanced computer,

2. 34-x take only both algebra and industrial technology,

3. 29-x take only both advanced computer and industrial technology.

Now let's count how many student take only one course:

1. 117-(x+34-x+70-x)=13+x take algebra,

2. 109-(x+29-x+70-x)=10+x take advanced computer,

3. 114-(x+29-x+34-x)=51+x take industrial technology.

Now count all students that take some courses:

51+x+10+x+13+x+29-x+34-x+70-x+x=236,

x=236-51-10-13-29-34-70,

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Answer: all three courses take 29 students.

6 0

2 years ago

Read 2 more answers

A series contains 18 numbers and has a sum of 4,185. The last number of the series is 275. What is the first number?

dusya [7]

The sum of arithmetic series is given by:
Sn=n/2(a1+an)
where:
n=number of terms
a1=first term
an=nth term
but
n=18, an=275, Sn=4185
plugging the values in the formula we get:
4185=18/2(a1+275)
simplifying this we get:
4185=9(a1+275)
dividing through by 9 we get:
465=a1+275
thus
a1=465-275
a1=190

Answer: first number is 190

3 0

2 years ago

A certain article indicates that in a sample of 1,000 dog owners, 610 said that they take more pictures of their dog than of the

lbvjy [14]

Answer:

(a) The 90% confidence interval is: (0.60, 0.63) Correct interpretation is (3).

(b) The 95% confidence interval is: (0.42, 0.46) Correct interpretation is (1).

(c) First, the confidence level in part(b) is more than the confidence level in part(a) is, so the critical value of part(b) is more than the critical value of part(a), Second the margin of error in part(b) is more than than in part(a).

Step-by-step explanation:

(a)

Let X = number of dog owners who take more pictures of their dog than of their significant others or friends.

Given:

X = 610

n = 1000

Confidence level = 90%

The (1 - α)% confidence interval for population proportion is:

$CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

The sample proportion is:

$\hat p=\frac{X}{n}=\frac{610}{1000}=0.61$

The critical value of z for a 90% confidence level is:

$z_{\alpha/2}=z_{0.10/2}=z_[0.05}=1.645$

Construct a 90% confidence interval for the population proportion of dog owners who take more pictures of their dog than their significant others or friends as follows:

$CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.61\pm 1.645\sqrt{\frac{0.61(1-0.61}{1000}}\\=0.61\pm 0.015\\=(0.595, 0.625)\\\approx(0.60, 0.63)$

Thus, the 90% confidence interval for the population proportion of dog owners who take more pictures of their dog than their significant others or friends is (0.60, 0.63).

Interpretation:

There is a 90% chance that the true proportion of dog owners who take more pictures of their dog than of their significant others or friends falls within the interval (0.60, 0.63).

Correct option is (3).

(b)

Let X = number of dog owners who are more likely to complain to their dog than to a friend.

Given:

X = 440

n = 1000

Confidence level = 95%

The (1 - α)% confidence interval for population proportion is:

$CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

The sample proportion is:

$\hat p=\frac{X}{n}=\frac{440}{1000}=0.44$

The critical value of z for a 95% confidence level is:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

Construct a 95% confidence interval for the population proportion of dog owners who are more likely to complain to their dog than to a friend as follows:

$CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.44\pm 1.96\sqrt{\frac{0.44(1-0.44}{1000}}\\=0.44\pm 0.016\\=(0.424, 0.456)\\\approx(0.42, 0.46)$

Thus, the 95% confidence interval for the population proportion of dog owners who are more likely to complain to their dog than to a friend is (0.42, 0.46).

Interpretation:

There is a 95% chance that the true proportion of dog owners who are more likely to complain to their dog than to a friend falls within the interval (0.42, 0.46).

Correct option is (1).

(c)

The confidence interval in part (b) is wider than the confidence interval in part (a).

The width of the interval is affected by:

The confidence level
Sample size
Standard deviation.

The confidence level in part (b) is more than that in part (a).

Because of this the critical value of z in part (b) is more than that in part (a).

Also the margin of error in part (b) is 0.016 which is more than the margin of error in part (a), 0.015.

First, the confidence level in part(b) is more than the confidence level in part(a) is, so the critical value of part(b) is more than the critical value of part(a), Second the margin of error in part(b) is more than than in part(a).

4 0

1 year ago

A superintendent of schools wonders if scores on a test of science achievement differ for female and male 8th grade students. To

lina2011 [118]

Answer:

b. 2-independent sample t-test

Correct for this case is the best test since the result from each group are independent and we can compare the sample means between th two groups

Step-by-step explanation:

For this case the intention is try to test if scores on a test of science achievement differ for female and male 8th grade students. let's analyze the possible options for this case:

a. 2-dependent sample t-test

False is not possible since the nature of the gender is not possible to conclude tha the results from boys and girls are dependent

b. 2-independent sample t-test

Correct for this case is the best test since the result from each group are independent and we can compare the sample means between the two groups

c. Correlation

False we can't compare the means of interest with a correlation coefficient since that's not the purpose of the study

d. 2-way ANOVA

False we have just one variable between two groups so is not possible to apply a 2 way ANOVA

5 0

2 years ago