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GaryK [48]
2 years ago
15

14. Janie Christopher lent $6000 to a friend for 90 days at 12%. After 30 days, she sold the note to a third party for $6000. Wh

at annual simple interest rate did the third party receive? Use 360 days in a year
Mathematics
1 answer:
Elenna [48]2 years ago
3 0

Can you do plsssssssss

help me plssssssssssss meee

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Write a fraction that is less then 5/6 and has a denominator of 8
Verdich [7]

So we want to write a fraction that is less than 5/6 and has a denominator of 8. So lets start from 1/1 that is greater then 5/6 and doesn't have 8 as a denominator. Lets take one half from 1/1 and well get 1/2.

8 0
3 years ago
Read 2 more answers
You are given a list of students names and their test scores. Design a pseudocode that does the following:(a) Calculates the ave
Rashid [163]

Answer:

List_of_students = {};    //Structure array that contains the scores of students and Names

Number_of_students = size(List_of_students);

Average_score ;   //Variable Stores average score

Below_Average = [];   //That is initially filled with zeros  

Highest_score ; //Initialize variable to store highest score

for ( i = 0 ; i < Number_of_students; i = i + 1)

{

      Average_score  = Average_score + List_of_students(i);    

     if ( i = 15)

      {

        Average_score  = Average_score / Number_of_students;

      }

}

for ( j = 0; j < Number_of_students ; i = i + 1)

{

    if (   List_of_students(j)  < Average_score)

     {

         print( "student  %name got %score , List_of_students(j),          List_of_students(j) )

     }

}

Compare_Score = 0;   //Variable to compare score

for ( k = 0; K < Number_of_students; k = k+1)

{

     if ( List_of_students(k) => Compare_Score)

    {

        Compare_Score = List_of_students(k);

    }

}

Highest_Score = Compare_Score;

Step-by-step explanation:

4 0
3 years ago
Manert’s grandfather put $4100 in the bank for him when he was born. The account has been earning 6.5% interest compounded annua
Travka [436]

Answer:

Ans: $19,793

Step-by-step explanation:

Compound interest:

A = P(1+6/100)²⁵

= 4100(1+0.065)²⁵

=19,793

3 0
2 years ago
Lauren divided a pile of paper into 5 stakes. The line plot shows the height of each stack of paper.
sergiy2304 [10]

Answer:

Step-by-step explanation:

4 0
3 years ago
Kids with cell phones: A marketing manager for a cell phone company claims that less than 58% of children aged 8-12 have cell ph
Vikentia [17]

Answer:

z=\frac{0.54 -0.58}{\sqrt{\frac{0.58(1-0.58)}{820}}}=-2.32  

p_v =P(z  

Since the p value is lower than the significance level \alpha=0.1 we have enough evidence to reject the null hypothesis and the claim for the manager makes sense.

For the Ti84 preocedure we need to do this:

STAT> TESTS> 1-Z prop Test

And then we need to input the following values:

po= 0.58

x = 443 , n= 820

prop <po

And then calculate and we will get the same results

Step-by-step explanation:

Data given and notation

n=820 represent the random sample taken

X=443 represent the people that had cell phones

\hat p=\frac{443}{820}=0.540 estimated proportion of people that had cell phones

p_o=0.58 is the value that we want to test

\alpha=0.1 represent the significance level

Confidence=90% or 0.90

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

System of hypothesis

We need to conduct a hypothesis in order to test the claim that the true proportion is less than 0.58.:  

Null hypothesis:p\geq 0.58  

Alternative hypothesis:p < 0.58  

The statistic is given by:

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

And replacing we got:

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

z=\frac{0.54 -0.58}{\sqrt{\frac{0.58(1-0.58)}{820}}}=-2.32  

Decision

The significance level provided \alpha=0.1. Now we can calculate the p value

Since is a left tailed test the p value would be:  

p_v =P(z  

Since the p value is lower than the significance level \alpha=0.1 we have enough evidence to reject the null hypothesis and the claim for the manager makes sense.

For the Ti84 preocedure we need to do this:

STAT> TESTS> 1-Z prop Test

And then we need to input the following values:

po= 0.58

x = 443 , n= 820

prop <po

And then calculate and we will get the same results

5 0
3 years ago
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