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goldfiish [28.3K]
3 years ago
15

Find the 6th partial sum of summation of 7 open parentheses 4 close parentheses to the I minus 1 power from 1 to infinity

Mathematics
1 answer:
kozerog [31]3 years ago
4 0
Answer: option C. 9,555.

Justification:

1) The summation is:

6
∑  7(4)^ (i - 1)
i=1

2) Substitute with the values of i from 1 to 6 (note that 7 is a common factor of all the terms) =>

7 { 4^(1-1) + 4^(2-1) + 4^(3-1) + 4^(4-1) + 4^(5-1) + 4^(6-1) } =

= 7 { 4^0 + 4^1 + 4^2 + 4^3 + 4^4 + 4^5 } = 7 { 1 + 4 + 16 + 64 + 256 + 1024} =

7{1365} = 9,555.

Answer: 9,555


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Say youre going to conduct a t procedure for the difference between the means of two populations. You decide the samples of size
stiv31 [10]

Answer:

all 3 are correct.

I

II

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you're welcome

6 0
3 years ago
A certain genetic condition affects 8% of the population in a city of 10,000. Suppose there is a test for the condition that has
enot [183]

Solution:

Population in the city= 10,000

As genetic condition affects 8% of the population.

8 % of 10,000

=\frac{8}{100}\times 10,000=800

As, it is also given that, there is an error rate of 1% for condition (i.e., 1% false negatives and 1% false positives).

So, 1% false negatives means out of 800 tested who are found affected , means there are chances that 1% who was found affected are not affected at all.

So, 1% of 800 =\frac{1}{100}\times 800=8

Also,  1% false positives means out of 10,000 tested,[10,000-800= 9200] who are found not affected , means there are chances that 1% who was found not affected can be affected also.

So, 1% of 9200 =\frac{1}{100}\times 9200=92

1. Has condition Does not have condition totals  = 800

2. Test positive =92

3. Test negative =8

4. Total =800 +92 +8=900

5. Probability (as a percentage) that a person has the condition if he or she tests positive= As 8% are found positive among 10,000 means 9200 are not found affected.But there are chances that out of 9200 , 1% may be affected

=\frac{\text{1 percent of 9200}}{9200}\\\\ \frac{\frac{1}{100}\times 9200}{9200}=\frac{92}{9200}\\\\ =0.01

that is Probability equal to 0.01 or 1%.

7 0
3 years ago
Help me PLEASEEEEE I need help
V125BC [204]

Answer: maybe 18.6 °F?

7 0
2 years ago
Read 2 more answers
Last year 190 student attended the spring dance and the total cost was 1250. Two years ago 175 people attended and the total cos
kondor19780726 [428]
Unit cost of a ticket = Income from ticket sales / number of tickets sold:

$1250
--------------- = $6.58 per ticket
190 tickets

Again:

$1175
--------------- = $6.71
175 tickets

While ticket prices do change (usually increase) from year to year, it's unusual to see such a situation here.

Don't have any guidelines by which to determine the "fixed cost of a ticket".

If we use the cost of a ticket of 2 years ago ($6.58/ticket), then the income from the sale of 225 tickets this year would be ($6.58/ticket)(225 tickets), or $1480.50.
6 0
3 years ago
The table shows a company’s profit based on the number of pounds of food produced.
Bingel [31]

Answer:

$5,300

Step-by-step explanation:

Formulae used,

a=\dfrac{(\sum x^2y\sum xx)-(\sum xy\sum xx^2)}{(\sum xx\sum x^2x^2)-({\sum xx^2)}^2}

b=\dfrac{(\sum xy\sum x^2x^2)-(\sum x^2y\sum xx^2)}{(\sum xx\sum x^2x^2)-({\sum xx^2)}^2}

c=\dfrac{\sum y}{n}-b\frac{\sum x}{n}-a\frac{\sum x^2}{n}

Where,

\sum xx=\sum x^2-\dfrac{(\sum x)^2}{n}

\sum xy=\sum xy-\dfrac{\sum x\sum y}{n}

\sum xx^2=\sum x^3-\dfrac{\sum x\sum x^2}{n}

\sum x^2y=\sum x^2y-\dfrac{\sum x^2\sum y}{n}

\sum x^2x^2=\sum x^4-\dfrac{(\sum x^2)^2}{n}

Putting the values from the table, we get the best fit line as,

y= -0.0817x^2 + 102.24x - 20421

As we want to calculate the profit at 350 pounds, so putting x=350, we get

y= -0.0817(350)^2 + 102.24(350) - 20421=\$5354.75


5 0
3 years ago
Read 2 more answers
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