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2 years ago

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On a number line, a number, b, is located the same distance from 0 as another number, a, but in the opposite direction. The numb

er b varies directly with the number a. For example b = 2a equals negative 2 and StartFraction 3 Over 4 EndFraction. when a = –2a equals negative 2 and StartFraction 3 Over 4 EndFraction.. Which equation represents this direct variation between a and b?

1 answer:

vfiekz [6]2 years ago

5 0

Answer:

b

Step-by-step explanation:

Guest

1 year ago

actually its A

You might be interested in

The amounts of vitamin C (in milligrams) for 100g (3.57 ounces) of various randomly selected fruits and vegetables are listed. I

34kurt

Answer:

H0: sigma=12

H1:sigma≠12

This hypothesis test is a two tailed test.

Step-by-step explanation:

The null value described in the statement is 12 mg. So, the null hypothesis would be sigma=12. As the statement states that whether there is sufficient evidence to conclude that standard deviation differs from 12 mg, so the alternative hypothesis would be sigma ≠12. The alternative hypothesis contains inequality sign so, the hypothesis test is a two tailed test.

4 0

2 years ago

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

2 years ago

A circular platform is to be built in a playground. The center of the structure is required to be equidistant from three support

castortr0y [4]

Answer:

The coordinates for the location of the center of the platform are (0, 1)

Step-by-step explanation:

The equation of the circle of center (h , k) and radius r is:

(x - h)² + (y - k)² = r²

Now,

- The center is equidistant from any point lies on the circumference of the circle

- There are three points equidistant from the center of the circle

- We have three unknowns in the equation of the circle h , k , r

Thus, let's substitute the coordinates of these point in the equation of the circle to find h , k , r.

The equation of the circle is (x - h)² + (y - k)² = r²

∵ Points A(2,−3), B(4,3), and C(−2,5)

- Substitute the values of x and y the coordinates of these points

Point A (2 , -3)

(2 - h)² + (-3 - k)² = r² - - - (1)

Point B (4 , 3)

(4 - h)² + (3 - k)² = r² - - - - (2)

Point C (-2 , 5)

(-2 - h)² + (5 - k)² = r² - - - - (3)

- To find h , k equate equation (1) and (2) and same for equation (2) and (3) because all of them equal r²

Thus;

(2 - h)² + (-3 - k)² = (4 - h)² + (3 - k)² - - - - - (4)

(4 - h)² + (3 - k)² = (-2 - h)² + (5 - k)² - - - - -(5)

- Simplify (5);

h² - 8h + 16 + k² - 6k + 9 = h² + 4h + 4 + k² - 10k + 25

h² and k² will cancel out to give;

-8h - 6k + 25 = 4h - 10k + 29

Rearranging, we have;

12h - 4k = -4 - - - - (6)

Similarly, for equation 4;

(2 - h)² + (-3 - k)² = (4 - h)² + (3 - k)²

h² - 4h + 4 + k² + 6k + 9 = h² - 8h + 16 + k² - 6k + 9

h², k² and 9 will cancel out to give;

4 - 4h + 6k = 16 - 8h - 6k

Rearranging;

4h + 12k = 12 - - - - (7)

Divide by 4 to give;

h + 3k = 3

Making h the subject;

h = 3 - 3k

Put 3 - 3k for h in eq 6;

12(3 - 3k) - 4k = -4

36 - 36k - 4k = -4

40k = 40

k = 40/40

k = 1

h = 3 - 3(1)

h = 0

The coordinates for the location of the center of the platform are (0, 1)

5 0

2 years ago

A television is 28.5 inches wide and 16 inches long.

Umnica [9.8K]

Answer:

33 in

Step-by-step explanation:

The Pythagorean theorem tells you ...

diagonal² = length² +width²

diagonal² = (16 in)² +(28.5 in)² = 1068.25 in²

diagonal = √(1068.25 in²) ≈ 32.684 in

The diagonal of the television is about 33 inches.

7 0

2 years ago

Among a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in

Hitman42 [59]

Answer:

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) We can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval

(3) A survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

Step-by-step explanation:

We are given that a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of Americans who decide to not go to college = 48%

n = sample of American adults = 331

p = population proportion of Americans who decide to not go to

college because they cannot afford it

<em>Here for constructing a 90% confidence interval we have used a One-sample z-test for proportions.</em>

<em />

<u>So, 90% confidence interval for the population proportion, p is ;</u>

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < $\hat p-p$ < $1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

P( $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

<u>90% confidence interval for p</u> = [ $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.48 -1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ , $0.48 +1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ ]

= [0.4348, 0.5252]

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) The interpretation of the above confidence interval is that we can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval.

3) Now, it is given that we wanted the margin of error for the 90% confidence level to be about 1.5%.

So, the margin of error = $Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$

$0.015 = 1.645 \times \sqrt{\frac{0.48(1-0.48)}{n} }$

$\sqrt{n} = \frac{1.645 \times \sqrt{0.48 \times 0.52} }{0.015}$

$\sqrt{n}$ = 54.79

n = $54.79^{2}$

n = 3001.88 ≈ 3002

Hence, a survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

5 0

2 years ago