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Vladimir79 [104]
3 years ago
15

0.25 + 1 + 4 + 16 + 64 Enter the values used in finding a partial sum.

Mathematics
2 answers:
Deffense [45]3 years ago
6 0
This is a geometric series

r = 4
a1 = 0.25
n = 5
Alexeev081 [22]3 years ago
4 0

Answer:  The required values are

r=4,~~a_1=0.25~~\textup{and}~~n=5.

Step-by-step explanation:  We are given a partial sum as follows :

0.25 + 1 + 4 + 16 + 64.

We are to find the values of r,~~a_1~~\textup{and}~~n.

From the given partial series, we note that

the series is a geometric one with first term 0.25 and common ratio calculated as follows :

r=\dfrac{1}{0.25}=\dfrac{4}{1}=\dfrac{16}{4}=\dfrac{64}{16}=4.

Also, since there are 5 terms in the given sum, so the value of n is 5.

Thus, the required values are

r=4,~~a_1=0.25~~\textup{and}~~n=5.

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3 years ago
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Step-by-step explanation:

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7 0
4 years ago
If a line has a slope of -4, what is the y-intercept if it also passed through the point (-2, 7)?
slamgirl [31]

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3 0
3 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
(5a-2)(5a+2)-3(1-a)-3a=
attashe74 [19]
So first we need to know the order of operations
P - Parenthesis
E - Exponents
M - Multiplication
D - Division
A - Addition
S - Subtraction

Then the special cases:
(I have attached a picture below)

Lets get into action!
(5a - 2)(5a + 2) - 3( 1 - a ) - 3a 
25a² - 4 - 3 + 3a - 3a 
25a² - 7 
25a² - 7 is the answer


8 0
3 years ago
Read 2 more answers
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