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OLga [1]
3 years ago
14

Please give an explanation? 30 points

Mathematics
1 answer:
Leto [7]3 years ago
3 0
The composition of two rotations about the same point (Z) is the sum of the rotations.  This can be proved using the matrix versions of the rotations.

This means
R_z(t1)\circ{R_z(t2)} = R_z(t1+t2)}
where
R=rotation operator
z=centre of rotation
t1,t2 are angles of rotation
In this particular case, the composition is commutative, i.e. order does not matter because addition of the angles is also commutative.

Thus substituting values, t1=170, t2=100
R_z(170)\circ{R_z(t2)} = R_z(170+100)}=R_z(270)

So the result of the composition is a rotation of 270 degrees about Z from the original position, which is also equal to a rotation of -90 degrees.


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Increase 400litres by 20%
Sati [7]

Answer:

480 Litres

Step-by-step explanation:

Percentage increase = 20% × 400

New value =

400 + Percentage increase =

400 + (20% × 400) =

400 + 20% × 400 =

(1 + 20%) × 400 =

(100% + 20%) × 400 =

120% × 400 =

120 ÷ 100 × 400 =

120 × 400 ÷ 100 =

48,000 ÷ 100 =

480

3 0
3 years ago
In the graph line ( ) passes through the complex number 1+2i and the line( ) passes through -2+2i
Nikitich [7]
Line E, line G.

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7 0
3 years ago
Read 2 more answers
A survey collected data on annual credit card charges in seven different categories of expenditures: transportation, groceries,
Andru [333]

Answer:

A. Null and alternative hypothesis:

H_0: \mu_d=0\\\\H_a:\mu_d\neq 0

B. Yes. At a significance level of 0.05, there is enough evidence to support the claim that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

P-value = 0.00002

C. As the difference is calculated as (population 1 − population 2), being population 1: groceries and population 2: dinning out, and knowing there is evidence that the true mean difference is positive, we can say that the groceries annual credit card charge is higher than dinning out annual credit card charge.

The point estimate is the sample mean difference d=$840.

The 95% confidence interval for the mean difference between the population means is (490, 1190).

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

Then, the null and alternative hypothesis are:

H_0: \mu_d=0\\\\H_a:\mu_d\neq 0

The significance level is 0.05.

The sample has a size n=42.

The sample mean is M=840.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=1123.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{1123}{\sqrt{42}}=173.2827

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{840-0}{173.2827}=\dfrac{840}{173.2827}=4.848

The degrees of freedom for this sample size are:

df=n-1=42-1=41

This test is a two-tailed test, with 41 degrees of freedom and t=4.848, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=2\cdot P(t>4.848)=0.00002

As the P-value (0.00002) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

We have to calculate a 95% confidence interval for the mean difference.

The t-value for a 95% confidence interval and 41 degrees of freedom is t=2.02.

The margin of error (MOE) can be calculated as:

MOE=t\cdot s_M=2.02 \cdot 173.283=350

Then, the lower and upper bounds of the confidence interval are:

LL=M-t \cdot s_M = 840-350=490\\\\UL=M+t \cdot s_M = 840+350=1190

The 95% confidence interval for the mean difference is (490, 1190).

7 0
3 years ago
Jorge said that y-values would stay the same when you reflect a preimage across the line y=5 since the y-values stay the same wh
OLga [1]
No because y=5 is a horizontal line on a graph, in this scenario the x-values would stay the same
3 0
4 years ago
The area for the trapezoid is in2.
Nady [450]

Answer:

1204

Step-by-step explanation:

8 0
3 years ago
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