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cestrela7 [59]
3 years ago
5

-2/3 divided by -4/9

Mathematics
1 answer:
Anna35 [415]3 years ago
4 0

Answer: 1.5

Step-by-step explanation:

You might be interested in
How to find the vertex calculus 2What is the vertex, focus and directrix of x^2 = 6y
son4ous [18]

Solution:

Given:

x^2=6y

Part A:

The vertex of an up-down facing parabola of the form;

\begin{gathered} y=ax^2+bx+c \\ is \\ x_v=-\frac{b}{2a} \end{gathered}

Rewriting the equation given;

\begin{gathered} 6y=x^2 \\ y=\frac{1}{6}x^2 \\  \\ \text{Hence,} \\ a=\frac{1}{6} \\ b=0 \\ c=0 \\  \\ \text{Hence,} \\ x_v=-\frac{b}{2a} \\ x_v=-\frac{0}{2(\frac{1}{6})} \\ x_v=0 \\  \\ _{} \\ \text{Substituting the value of x into y,} \\ y=\frac{1}{6}x^2 \\ y_v=\frac{1}{6}(0^2) \\ y_v=0 \\  \\ \text{Hence, the vertex is;} \\ (x_v,y_v)=(h,k)=(0,0) \end{gathered}

Therefore, the vertex is (0,0)

Part B:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

\begin{gathered} 4p(y-k)=(x-h)^2 \\  \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}

Rewriting the equation in standard form,

\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}

Since the parabola is symmetric around the y-axis, the focus is a distance p from the center (0,0)

Hence,

\begin{gathered} Focus\text{ is;} \\ (0,0+p) \\ =(0,0+\frac{3}{2}) \\ =(0,\frac{3}{2}) \end{gathered}

Therefore, the focus is;

(0,\frac{3}{2})

Part C:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

\begin{gathered} 4p(y-k)=(x-h)^2 \\  \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}

Rewriting the equation in standard form,

\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}

Since the parabola is symmetric around the y-axis, the directrix is a line parallel to the x-axis at a distance p from the center (0,0).

Hence,

\begin{gathered} Directrix\text{ is;} \\ y=0-p \\ y=0-\frac{3}{2} \\ y=-\frac{3}{2} \end{gathered}

Therefore, the directrix is;

y=-\frac{3}{2}

3 0
1 year ago
Verify that 2,-1 and 1⁄2 are the zeroes of the cubic polynomial
Mariulka [41]

Answer:

i) Since P(2), P(-1) and P(½) gives 0, then it's true that 2,-1 and 1⁄2 are the zeroes of the cubic polynomial.

ii) - the sum of the zeros and the corresponding coefficients are the same

-the Sum of the products of roots where 2 are taken at the same time is same as the corresponding coefficient.

-the product of the zeros of the polynomial is same as the corresponding coefficient

Step-by-step explanation:

We are given the cubic polynomial;

p(x) = 2x³ - 3x² - 3x + 2

For us to verify that 2,-1 and 1⁄2 are the zeroes of the cubic polynomial, we will plug them into the equation and they must give a value of zero.

Thus;

P(2) = 2(2)³ - 3(2)² - 3(2) + 2 = 16 - 12 - 6 + 2 = 0

P(-1) = 2(-1)³ - 3(-1)² - 3(-1) + 2 = -2 - 3 + 3 + 2 = 0

P(½) = 2(½)³ - 3(½)² - 3(½) + 2 = ¼ - ¾ - 3/2 + 2 = -½ + ½ = 0

Since, P(2), P(-1) and P(½) gives 0,then it's true that 2,-1 and 1⁄2 are the zeroes of the cubic polynomial.

Now, let's verify the relationship between the zeros and the coefficients.

Let the zeros be as follows;

α = 2

β = -1

γ = ½

The coefficients are;

a = 2

b = -3

c = -3

d = 2

So, the relationships are;

α + β + γ = -b/a

αβ + βγ + γα = c/a

αβγ = -d/a

Thus,

First relationship α + β + γ = -b/a gives;

2 - 1 + ½ = -(-3/2)

1½ = 3/2

3/2 = 3/2

LHS = RHS; So, the sum of the zeros and the coefficients are the same

For the second relationship, αβ + βγ + γα = c/a it gives;

2(-1) + (-1)(½) + (½)(2) = -3/2

-2 - 1½ + 1 = -3/2

-1½ - 1½ = -3/2

-3/2 = - 3/2

LHS = RHS, so the Sum of the products of roots where 2 are taken at the same time is same as the coefficient

For the third relationship, αβγ = -d/a gives;

2 * -1 * ½ = -2/2

-1 = - 1

LHS = RHS, so the product of the zeros(roots) is same as the corresponding coefficient

7 0
3 years ago
A researcher records the amount of time (in minutes) that parent child pairs spend on social networking sites to test whether th
ki77a [65]

Answer:

(a) A matched pair design is used to perform the test.

(b) Each sample is of size 30.

(c) The null hypothesis is rejected.

(d) The effect size of the test is medium.

Step-by-step explanation:

A statistical experiment is performed to determine whether the amount of time (in minutes) that parent-child pairs spend on social networking sites to test show any generational differences.

<u>Given</u>:

MD = -42 minutes

t₍₂₉₎ = 4.021

p < 0.05

d = 0.49

The hypothesis is:

<em>H₀</em>: There is no difference between the means, i.e. <em>μ₁</em> = <em>μ₂</em>.

<em>Hₐ</em>: There is a significant difference between the means, i.e. <em>μ₁</em> ≠ <em>μ₂</em>.

(a)

A matched pair design is used to perform the test.

In a matched pair design the participant of the two groups are related in some way. For example, the same sample is tested before and after applying the treatment.

In this case the matched are pairs are of children and their parents. And the treatment is the amount of time spent on social networking sites.

(b)

The degrees of freedom of a matched pair design is:

df = n - 1

29 = n - 1

n = 29 + 1

n = 30

Thus, each sample is of size 30.

(c)

The <em>p</em>-value of the test statistic is < 0.05.

If the <em>p</em>-value is less than the significance level <em>α </em>(say, 0.05) then the null hypothesis is rejected and vise-versa.

Since <em>p</em> < 0.05 the null hypothesis is rejected concluding that there is a significant difference between the amount of time spent on social networking sites by parents and children.

(d)

The estimated value of Cohen's d is, <em>d</em> = 0.49.

A Cohen's d value between the range 0.50 to 0.80 is considered as medium.

Thus, the effect size of the test is medium.

3 0
3 years ago
In a recent contest, the mean score was 210 and the standard deviation was 25. a) Find the z-score of John who scored 190 b) Fin
denpristay [2]

Answer:

a) Z = -0.8

b) Z = 2.4

c) Mary's score was 241.25.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 210, \sigma = 25

a) Find the z-score of John who scored 190

Z = \frac{X - \mu}{\sigma}

Z = \frac{190 - 210}{25}

Z = -0.8

b) Find the z-score of Bill who scored 270

Z = \frac{X - \mu}{\sigma}

Z = \frac{270 - 210}{25}

Z = 2.4

c) If Mary had a score of 1.25, what was Mary’s score?

Z = \frac{X - \mu}{\sigma}

1.25 = \frac{X - 210}{25}

X - 210 = 25*1.25

X = 241.25

Mary's score was 241.25.

3 0
3 years ago
Read 2 more answers
3. The graph of an equation intersects the y-axis at some point. What do the coordinates of the intersection indicate?
Vikki [24]

Answer:

Area: ½ × base × height

Perimeter: sum of side lengths of the triangle

of vertices: 3

Number of edges: 3

Internal angle: 60° (for equilateral)

Sum of interior angles: 180°

Step-by-step explanation:

5 0
3 years ago
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