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BlackZzzverrR [31]
4 years ago
9

Mrs. Jackson put a total of $10,000 into two accounts one account earns 6% simple annual interest the other account earns 6.5% s

imple interest after one year the two accounts earn $632 and 50 interest find how much money was invested in each account
Mathematics
1 answer:
kozerog [31]4 years ago
7 0

Answer:

The money invested in account at 6 % rate is $ 9230.77

The money invested in account at 6.5 % rate is $ 769.23

Step-by-step explanation:

Given as :

The total amount saves in two account = $ 10,000

The two rate of interest are 6 % and 6.5 % at simple interest

The interest earn at 6 % rate = $ 632

The interest earn at 6.5 % ate = $ 50

Now, Let The amount deposit at 6 % rate = $ x

And The amount deposit at 6.5 % rate = $ 10,000 - $ x

<u>From Simple Interest method</u>

Simple Interest = \dfrac{\textrm principal \times \textrm Rate\times \textrm time}{100}

or, At 6.5 % rate

$ 50 =  \dfrac{\textrm (10,000 - x ) \times \textrm 6.5\times \textrm 1}{100}

or, 10000 - x =  \frac{5000}{6.5}

Or, 10000 - x = 769.23

∴  x = 10000 - 769.23

I.e x = $ 9230.77

So , At 6.5 % the amount deposit = 10,000 - 9230.77 = $ 769.23

So, At 6 % The amount deposit  = $ 9230.77

Hence The money invested in account at 6 % rate is $ 9230.77

And  The money invested in account at 6.5 % rate is $ 769.23 Answer

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Now that you know this information, you simply subtract the days he missed from that total.

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And

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In a recent Super Bowl, a TV network predicted that 50 % of the audience would express an interest in seeing one of its forthcom
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Answer:

z= -0.968

We can conclude that we fail to reject the null hypothesis, and we can said that at 5% of significance the proportion of people who says that  they would watch one of the television shows not differs from 0.5 or 50% .  

Step-by-step explanation:

1) Data given and notation n  

n=106 represent the random sample taken

X=48 represent the people who says that  they would watch one of the television shows.

\hat p=\frac{48}{106}=0.453 estimated proportion of people who says that  they would watch one of the television shows.

p_o=0.5 is the value that we want to test

\alpha represent the significance level  

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that 50% of people who says that  they would watch one of the television shows.:  

Null hypothesis:p=0.5  

Alternative hypothesis:p \neq 0.5  

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.

3) Calculate the statistic  

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

z=\frac{0.453 -0.5}{\sqrt{\frac{0.5(1-0.5)}{106}}}=-0.968  

4) Statistical decision  

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The significance level is not provided, but we can assume \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:  

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