What set of transformations is performed on ABCD to form A′B′C′D′?

Which description does NOT guarantee that a quadrilateral is a square?

shutvik [7]

A. Is your answer because it’s doesn’t go that in-depth in the answer.
Hope I helped

3 0

4 years ago

Substitution method 2x-y+24 y=-3x+16

Llana [10]

- Using substitution means you are going to solve one equation for one variable and substitute with its value in the other equation in order to get also an equation with one variable.

4 0

3 years ago

g The proportion of U.S. births that result in a birth defect is approximately 1/33 according to the Centers for Disease Control

Rudiy27

Answer:

0.1426 = 14.26% probability that at least one of the births results in a defect.

Step-by-step explanation:

For each birth, there are only two possible outcomes. Either it results in a defect, or it does not. The probability that a birth results in a defect is independent of any other birth. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The proportion of U.S. births that result in a birth defect is approximately 1/33 according to the Centers for Disease Control and Prevention (CDC).

This means that $p = \frac{1}{33}$

A local hospital randomly selects five births.

This means that $n = 5$

What is the probability that at least one of the births results in a defect?

This is:

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{5,0}.(\frac{1}{33})^{0}.(\frac{32}{33})^{5} = 0.8574$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.8574 = 0.1426$

0.1426 = 14.26% probability that at least one of the births results in a defect.

4 0

3 years ago

Can someone please help me with this?

Anettt [7]

Answersorry

Step-by-step explanation:sorry idk

7 0

4 years ago

A survey was conducted two years ago asking college students their top motivations for using a credit card. To determine whether

salantis [7]

Answer:

See explanation

Step-by-step explanation:

Solution:-

- A survey was conducted among the College students for their motivations of using credit cards two years ago. A randomly selected group of sample size n = 425 college students were selected.

- The results of the survey test taken 2 years ago and recent study are as follows:

Old Survey ( % ) New survey ( Frequency )

Reward 27 112

Low rate 23 96

Cash back 21 109

Discount 9 48

Others 20 60

- We are to test the claim for any changes in the expected distribution.

We will state the hypothesis accordingly:

Null hypothesis: The expected distribution obtained 2 years ago for the motivation behind the use of credit cards are as follows: Rewards = 27% , Low rate = 23%, Cash back = 21%, Discount = 9%, Others = 20%

Alternate Hypothesis: Any changes observed in the expected distribution of proportion of reasons for the use of credit cards by college students.

( We are to test this claim - Ha )

We apply the chi-square test for independence.

- A chi-square test for independence compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each other.

- We will compute the chi-square test statistics ( X^2 ) according to the following formula:

$X^2 = Sum [ \frac{(O_i - E_i)^2}{Ei} ]$

Where,

O_i : The observed value for ith data point

E_i : The expected value for ith data point.

- We have 5 data points.

So, Oi :Rewards = 27% , Low rate = 23%, Cash back = 21%, Discount = 9%, Others = 20% from a group of n = 425.

Ei : Rewards = 112 , Low rate = 96, Cash back = 109, Discount = 48, Others = 60.

Therefore,

$X^2 = [ \frac{(112 - 425*0.27)^2}{425*0.27} + \frac{(96 - 425*0.23)^2}{425*0.23} + \frac{(109 - 425*0.21)^2}{425*0.21} + \frac{(48 - 425*0.09)^2}{425*0.09} + \frac{(60 - 425*0.20)^2}{425*0.20}]\\\\X^2 = [ 0.06590 + 0.03132 + 4.37044 + 2.48529 + 7.35294]\\\\X^2 = 14.30589$

- Then we determine the chi-square critical value ( X^2- critical ). The two parameters for evaluating the X^2- critical are:

Significance Level ( α ) = 0.10

Degree of freedom ( v ) = Data points - 1 = 5 - 1 = 4

Therefore,

X^2-critical = X^2_α,v = X^2_0.1,4

X^2-critical = 7.779

- We see that X^2 test value = 14.30589 is greater than the X^2-critical value = 7.779. The test statistics value lies in the rejection region. Hence, the Null hypothesis is rejected.

Conclusion:-

This provides us enough evidence to conclude that there as been a change in the claimed/expected distribution of the motivations of college students to use credit cards.

6 0

4 years ago