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2 years ago

Describe a sequence of three transformations of ABC that will produce the image A’ B’ C’.

Mathematics

2 answers:

rusak2 [61]2 years ago

8 0

Answer:

steps below

Step-by-step explanation:

A(-1,-1) B(-2,-3) C(-3,-1)

1. Reflection over y=-1 A(-1,-1) B(-2,1) C(-3,-1)

2. Shift to right 4 units A(3,-1) B(2,1) C(1,-1)

3. Dilation with factor 2 and center C'(1.-1) A'(5,-1) B'(3,3) C'(1,-1)

Leokris [45]2 years ago

6 0

Answer:

Reflect over x-axis, Translate 5 units right, Dilate by a factor of 2

Step-by-step explanation:

I think this is correct The dilation probably isn't but those are the three transformations. But i am pretty sure i am right.

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John's cell phone gives him 1250 minutes for $25. He used 1300 minutes. How much will his cell phone company bill him for? A) Si

Flauer [41]

Answer:

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

A pair of parallel lines is cut by a transversal, as shown below.

MA_775_DIABLO [31]

Answer:

$x=45\degree$

Step-by-step explanation:

In the diagram, the measure of the $(x+30\degree)$ angle is equal to the $75\degree$ angle.

These two angles are alternate angles.

We therefore set up the following equation and solve for x.

$x+30\degree =75\degree$

$\Rightarrow x =75\degree-30\degree$

$\Rightarrow x =45\degree$

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2 years ago

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The director of admissions at Kinzua University in Nova Scotia estimated the distribution of student admissions for the fall sem

Naily [24]

Answer: 1406

Step-by-step explanation:

Given Table :

Admissions Probability

1,040 0.3

1,320 0.2

1,660 0.5

Now, the expected number of admissions for the fall semester is given by :-

$E(x)=p_1x_1+p_2x_2+p_3x_3\\\\\Rightarrow\ E(x)=0.3\times1040+0.2\times1320+0.5\times1660\\\\\Rightarrow\ E(x)=1406$

Hence, the expected number of admissions for the fall semester = 1406

3 0

2 years ago

A genetic experiment involving peas yielded one sample of offspring consisting of 420 green peas and 174 yellow peas. Use a 0.01

slavikrds [6]

Answer:

a) $z=\frac{0.293 -0.23}{\sqrt{\frac{0.23(1-0.23)}{594}}}=3.649$

b) For this case we need to find a critical value that accumulates $\alpha/2$ of the area on each tail, we know that $\alpha=0.01$ , so then $\alpha/2 =0.005$ , using the normal standard table or excel we see that:

$z_{crit}= \pm 2.58$

Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 1% of significance.

Step-by-step explanation:

Data given and notation

n=420+174=594 represent the random sample taken

X=174 represent the number of yellow peas

$\hat p=\frac{174}{594}=0.293$ estimated proportion of yellow peas

$p_o=0.23$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion of yellow peas is 0.23:

Null hypothesis: $p=0.23$

Alternative hypothesis: $p \neq 0.23$

When we conduct a proportion test we need to use the z statisitc, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.293 -0.23}{\sqrt{\frac{0.23(1-0.23)}{594}}}=3.649$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

$p_v =2*P(z>3.649)=0.00026$

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

b) Critical value

For this case we need to find a critical value that accumulates $\alpha/2$ of the area on each tail, we know that $\alpha=0.01$ , so then $\alpha/2 =0.005$ , using the normal standard table or excel we see that:

$z_{crit}= \pm 2.58$

Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 1% of significance.

5 0

2 years ago

Find the value of x. 6 14 8 0 21 x = [ ? ] 3x-12

love history [14]

Answer: 7

Step-by-step explanation:

By the angle bisector theorem,

$\frac{21}{14}=\frac{3x-12}{6}\\21(6)=14(3x-12)\\126=14(3x-12)\\9=3x-12\\21=3x\\x=\boxed{7}$

4 0

1 year ago