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jek_recluse [69]
4 years ago
11

$3,300 principal earning 4% compounded annually after three years you need to find the balance in the account. Thanks:)

Mathematics
2 answers:
jasenka [17]4 years ago
7 0
Total = 3,300 * (1.04)^3
Total = <span> <span> <span> 3,712.05

</span></span></span>
galina1969 [7]4 years ago
3 0

Answer:

$3,712.05

Step-by-step explanation:

Using compounding:  

result= 3300*1.04^3= $3,712.05

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Mr.Williams physical education class lasts 7/8 hour if instruciton is 1/5 how many minutes are not spent on instruction?
g100num [7]
= 7/8 − 1/5

= ((7 × 5) − (1 × 8)) / (8 × 5)

= (35 - 8) / 40

= 27/40


SO, 27/40 minutes are NOT spend on instructions.
Hope I was able to help! :)
6 0
3 years ago
Read 2 more answers
The following prices, in dollars, of 7.5-cubic-foot refrigerators were recorded from a random sample. 314 305 344 283 285 310​ 3
Mariana [72]

Answer:

t=\frac{310.9-300}{\frac{31.09}{\sqrt{10}}}=1.109      

The degrees of freedom are given by:

df=n-1=10-1=9  

And the p value would be given by:

p_v =P(t_{9}>1.109)=0.148  

Since the p value is higher than the significance level of 0.05 we don't have enough evidence to conclude that the true mean is significantly higher than $300 because we FAIL to reject the null hypothesis.

Step-by-step explanation:

Information given

314 305 344 283 285 310​ 383​ 285​ 300​ 300

We can calculate the sample mean and deviation with the following formula:

\bar X =\frac{\sum_{i=1}^n X_i}{n}

s =\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}

\bar X=310.9 represent the sample mean      

s=31.09 represent the standard deviation for the sample      

n=10 sample size      

\mu_o =300 represent the value to test

\alpha=0.05 represent the significance level

t would represent the statistic

p_v represent the p value

Hypothesis to test

We want to verify if the true mean is greater than 300, the system of hypothesis would be:      

Null hypothesis:\mu \leq 300      

Alternative hypothesis:\mu > 300      

The statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)      

Replacing the info given we got:

t=\frac{310.9-300}{\frac{31.09}{\sqrt{10}}}=1.109      

The degrees of freedom are given by:

df=n-1=10-1=9  

And the p value would be given by:

p_v =P(t_{9}>1.109)=0.148  

Since the p value is higher than the significance level of 0.05 we don't have enough evidence to conclude that the true mean is significantly higher than $300 because we FAIL to reject the null hypothesis.

6 0
3 years ago
I need help now assssssap
Andreyy89

Answer:

assssssssssssssap

Step-by-step explanation:

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  °   •  .°•    ✯   ★ *     °      °·                             .   • ° ★ •  ☄▁▂▃▄▅▆▇▇▆▅▄▃▁▂

7 0
3 years ago
Read 2 more answers
I need help with this questions
Keith_Richards [23]
QS=7x-1
BC=5x-19
I think so...

8 0
4 years ago
7 1/6- 4 2/3<br><br><br><br><br> Please help meee
Pavlova-9 [17]

Answer:

2\frac{1}{2}

Step-by-step explanation:

7-4=3

now let's change the denominator by finding the lcd (lowest common denominator)

<em>factors of 3:</em> 3, 6, 9, 12...

<em>factors of 6:</em> 6, 12, 18, 24...

both 3 and 6 has 6 as their lcd

3 x 2= 6

1/6 -4/6= -3/6

REDUCE

-1/2

3-1/2

2 1/2

3 0
1 year ago
Read 2 more answers
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