Express Rs 1.50 as a percentage of 200 cents

A second grade student is 4 feet tall. Her teacher is 5 2/3 feet tall. How many times as tall as the student is the teacher

Alchen [17]

Answer:

The teacher is 1.416667 times taller than the student

Step-by-step explanation:

(5 2/3)/4

=1.416667

4 0

1 year ago

Use the drop-down menus to complete the proof. Given that w ∥ x and y is a transversal, we know that ∠1 ≅∠5 by the . Therefore,

ololo11 [35]

Answer:

From the graph attached, we know that $\angle 1 \cong \angle 5$ by the corresponding angle theorem, this theorem is about all angles that derive form the intersection of one transversal line with a pair of parallels. Specifically, corresponding angles are those which are placed at the same side of the transversal, one interior to parallels, one exterior to parallels, like $\angle 1$ and $\angle 5$ .

We also know that, by definition of linear pair postulate, $\angle 3$ and $\angle 1$ are linear pair. Linear pair postulate is a math concept that defines two angles that are adjacent and for a straight angle, which is equal to 180°.

They are supplementary by the definition of supplementary angles. This definition states that angles which sum 180° are supplementary, and we found that $\angle 3$ and $\angle 1$ together are 180°, because they are on a straight angle. That is, $m \angle 3 + m \angle 1 = 180\°$

If we substitute $\angle 5$ for $\angle 1$ , we have $m \angle 3 + m \angle 5 = 180\°$ , which means that $\angle 3$ and $\angle 5$ are also supplementary by definition.

4 0

1 year ago

Read 2 more answers

A government official is in charge of allocating social programs throughout the city of Vancouver. He will decide where these so

otez555 [7]

Answer: (0.132132, 0.274368)

Step-by-step explanation:

Given : A simple random sample of 123 people living in Gastown and finds that 25 have an annual income that is below the poverty line.

i.e. n= 123

$\hat{p}=\dfrac{25}{123}\approx0.203252$

Critical value for 95% confidence interval : $z_{\alpha/2}=1.96$

Confidence interval for population :

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

i.e. $0.203252\pm (1.96)\sqrt{\dfrac{0.203252(1-0.203252)}{123}}$

$0.203252\pm (1.96)\sqrt{\dfrac{0.203252(1-0.203252)}{123}}\\\\\0.203252\pm0.071118\\\\=(0.20325-0.071118,\ 0.203252+0.071118)\\\\=(0.132132,\ 0.274368)$

Hence, the 95% confidence interval for the true proportion of Gastown residents living below the poverty line : (0.132132, 0.274368)

5 0

2 years ago

Oishi and Schimmack (2010) report that people who move from home to home frequently as children tend to have lower than average

user100 [1]

Answer:

The well-being for frequent movers is significantly different from well-being in the general population. ( Alternate Hypothesis accepted )

cohen's d = -0.91 , ( Large Effect )

Step-by-step explanation:

Given:-

- A sample of size n = 12

- The population mean u_p = 40

- The sample was taken as:

38, 37, 41, 35, 42, 40, 33, 33, 36, 38, 32, 39

Find:-

On the basis of this sample, is well-being for frequent movers significantly different from well-being in the general population? Use a two-tailed test with α = 0.05.

Solution:-

- State the hypothesis for sample mean u_s is same as population mean u_p.

Null Hypothesis: u_s = 40

Alternate Hypothesis: u_s ≠ 40

- The rejection criteria for the Null hypothesis can be modeled by T-value ( n < 30 ) with significance level α = 0.05.

DOF = n - 1 = 12 - 1 = 11

Significance level α = 0.05

t_α/2 = t_0.025 = +/- 2.201

- For the statistic value we have to compute sample mean u_s given by:

u_s = Σ xi / n

u_s = (38 + 37 + 41 + 35 + 42 + 40 + 33 + 33 + 36 + 38 + 32 + 39) / 12

u_s = 37

- For the statistic value we need population standard deviation S_p given by:

S_p = S_s / √n

Where, S_s : Sample standard deviation.

S_s^2 = Σ (xi - u_s)^2 / (n-1)

=[ 2*(38-37)^2 + (37-37)^2 + (41-37)^2 + (35-37)^2 + (42-37)^2 + (40-37)^2 + 2*(33-37)^2 + (36-37)^2 + (32-37)^2 + (39-37)^2 ] / ( 11 )

S_s^2 = [ 2 + 0 + 16 + 4 + 25 + 9 + 32 + 1 + 25 + 4 ] / 11

S_s^2 = 10.73

S_s = 3.28

The population standard deviation ( S_p ) is:

S_p = 3.28 / √12

S_p = 0.95

- The T-statistics value is computed as follows:

t = ( u_s - u_p ) / S_p

t = ( 37 - 40 ) / 0.95 = -3.16

- Compare the T-statistics (t) with rejection criteria (t_α/2).

-3.16 < -2.201

t < t_α/2 ...... Reject Null Hypothesis.

- The well-being for frequent movers is significantly different from well-being in the general population. ( Alternate Hypothesis accepted )

- The cohen's d is calculated as follows:

cohen's d = ( u_s - u_p ) / S_s

cohen's d = ( 37 - 40 ) / 3.28 = -0.91 , ( Large Effect )

5 0

1 year ago

Let P2 be the vector space of all polynomials of degree 2 or less, and let H be the subspace spanned by 10x2+4xâ1, 3xâ4x2+3, and

lord [1]

I suppose

$H=\mathrm{span}\{10x^2+4x-1,3x-4x^2+3,5x^2+x-1\}$

The vectors that span $H$ form a basis for $P_2$ if they are (1) linearly independent and (2) any vector in $P_2$ can be expressed as a linear combination of those vectors (i.e. they span $P_2$ ).

Independence:

Compute the Wronskian determinant:

$\begin{vmatrix}10x^2+4x-1&3x-4x^2+3&5x^2+x-1\\20x+4&3-8x&10x+1\\20&-8&10\end{vmatrix}=-6\neq0$

The determinant is non-zero, so the vectors are linearly independent. For this reason, we also know the dimension of $H$ is 3.

Span:

Write an arbitrary vector in $P_2$ as $ax^2+bx+c$ . Then the given vectors span $P_2$ if there is always a choice of scalars $k_1,k_2,k_3$ such that

$k_1(10x^2+4x-1)+k_2(3x-4x^2+3)+k_3(5x^2+x-1)=ax^2+bx+c$

which is equivalent to the system

$\begin{bmatrix}10&-4&5\\4&3&1\\-1&3&-1\end{bmatrix}\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}a\\b\\c\end{bmatrix}$

The coefficient matrix is non-singular, so it has an inverse. Multiplying both sides by that inverse gives

$\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}-\dfrac{6a-11b+19c}3\\\dfrac{3a-5b+2c}3\\\dfrac{15a-26b+46c}3\end{bmatrix}$

so the vectors do span $P_2$ .

The vectors comprising $H$ form a basis for it because they are linearly independent.

4 0

2 years ago