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3 years ago

If ABCD is dilated by a factor of 3, the

Mathematics

1 answer:

Vlad [161]3 years ago

7 0

Answer:

(6,-6)

Step-by-step explanation:

First let's identify the current coordinates of D

It appears that D is located at (2 , -2)

Now let's find the coordinate of D if it were dilated by a scale factor of 3.

To find the coordinates of a point after a dilation you simply multiply the x and y values of the pre image coordinates by the scale factor

In this case the scale factor is 3 and the coordinates are (2,-2)

That being said let's apply the dilation rule

Current coordinates: (2,-2)

Scale factor:3

Multiply x and y values by scale factor

(2 * 3 , -2 * 3) --------> (6 , -6)

The coordinates of D' would be (6,-6)

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Assume that z scores are normally distributed with a mean of 0 and a standard deviation of 1. if p(z >

madreJ [45]

Given that μ=0, and σ=1, P(z>c)=0.1093
thus
P(z<c)=1-0.1093=0.8907
hence the value of c will be:
z-score=(c-μ)/σ
thus
1.23=(c-0)/1
c=1.23

8 0

4 years ago

Many elementary school students in a school district currently have ear infections. A random sample of children in two different

Marta_Voda [28]

Answer:

Step-by-step explanation:

The summary of the given data includes;

sample size for the first school $n_1$ = 42

sample size for the second school $n_2$ = 34

so 16 out of 42 i.e $x_1$ = 16 and 18 out of 34 i.e $x_2$ = 18 have ear infection.

the proportion of students with ear infection Is as follows:

$\hat p_1 = \dfrac{16}{42}$ = 0.38095

$\hat p_2 = \dfrac{18}{34}$ = 0.5294

Since this is a two tailed test , the null and the alternative hypothesis can be computed as :

$H_0 :p_1 -p_2 = 0 \\ \\ H_1 : p_1 - p_2 \neq 0$

level of significance ∝ = 0.05,

Using the table of standard normal distribution, the value of z that corresponds to the two-tailed probability 0.05 is 1.96. Thus, we will reject the null hypothesis if the value of the test statistics is less than -1.96 or more than 1.96.

The test statistics for the difference in proportion can be achieved by using a pooled sample proportion.

$\bar p = \dfrac{x_1 +x_2}{n_1 +n_2}$

$\bar p = \dfrac{16 +18}{42 +34}$

$\bar p = \dfrac{34}{76}$

$\bar p = 0.447368$

$\bar p + \bar q = 1 \\ \\ \bar q = 1 -\bar p \\ \\\bar q = 1 - 0.447368 \\ \\\bar q = 0.552632$

The pooled standard error can be computed by using the formula:

$S.E = \sqrt{ \dfrac{ \bar p \bar q}{ n_1} + \dfrac{\bar p \bar p}{n_2} }$

$S.E = \sqrt{ \dfrac{ 0.447368 * 0.552632}{ 42} + \dfrac{ 0.447368 * 0.447368}{34} }$

$S.E = \sqrt{ \dfrac{ 0.2472298726}{ 42} + \dfrac{ 0.2001381274}{34} }$

$S.E = \sqrt{ 0.01177284105}$

$S.E = 0.1085$

The test statistics is ;

$z = \dfrac{\hat p_1 - \hat p_2}{S.E}$

$z = \dfrac{0.38095- 0.5294}{0.1085}$

$z = \dfrac{-0.14845}{0.1085}$

z = - 1.368

Decision Rule: Since the test statistics is greater than the rejection region - 1.96 , we fail to reject the null hypothesis.

Conclusion: There is insufficient evidence to support the claim that a difference exists between the proportions of students who have ear infections at the two schools

5 0

3 years ago

The probability that a data communications system will have high selectivity is 0.72, and the probability that it will have high

34kurt

Let us assume that even F represents fidelity and event S represents selectivity.

We have been given the probabilities:

P(S) = 0.72, P(F) = 0.59 and $P(S\cap F)=0.33$

We need to find the conditional probability that a system with high fidelity will also have high selectivity. We know the conditional probability formula:

$P(\frac{S}{F})=\frac{P(S\cup F)}{P(F)}$