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sladkih [1.3K]
3 years ago
10

The 6th term of an arithmetic progression is 35 and the 13th term is 77. find the 1st term

Mathematics
1 answer:
GenaCL600 [577]3 years ago
8 0

Answer:

5

Step-by-step explanation:

The 6th term of an arithmetic progression is 35 and the 13th term is 77. find the 1st term

We know we have an arithmetic sequence.

a_n = (n - 1)*d + a_1

given:

a_6 = 35 = (6-1)*d + a_1

a_13 = (13 -1)*d + a_1 = 77

find a_1

we have  35 = 5d + a_1

and  77 = 12d + a_1

2 equations in 2 unknowns.

We can solve this.

77 = 12d + a_1

35 = 5d + a_1

minus  the 2 equations from each other.

77 - 35 = 12d - 5d + a_1 - a_1

42 = 7d

d = 6

Find a_1

35 = 5*6 + a_1

35 = 30 + a_1

a_1 = 35 -30 = 5

a_1 = 5

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GarryVolchara [31]

Answer:

Michael and Carlos will have the same amount after 7 weeks

Step-by-step explanation:

Assume the total number of baseball cards that Carlos and Michael need to have to be equal is y, and the amount of time in terms of weeks that it will take for this to be achieved is x.

The following expressions can be derived

For Carlos to achieve y, he will need 85 and an addition of 10 per week (x):

y=85+(10x)..........equation 1

For Michael to achieve y, he will need 120 and an addition of 5 per week (x):

y=120+(5x)..........equation 2

Equating equation 1 and 2 to solve for x

85+10x=120+5x

10x-5x=120-85

5x/5=35/5

x=7

Number of weeks it will take Michael and Carlos to have the same amount=x=7 weeks

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The choir sells tickets to its performances for an assortment of prices and gives some tickets away. The choir
galina1969 [7]

The least square regression equation for the information provided is y = 9.72x - 40.776

<u>The general equation of a regression model can be expressed thus</u>:

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Calculating slope, b :

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<u>Plugging the values of b, y and x into the equation in other to calculate c</u> :

696 = 9.72(75.8) + c

696 = 736.776 + c

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Square root of 129 is greater than, less than or equal to 95/8?​
Tanzania [10]

√129 is less than \frac{95}{8}.

\large\mathfrak{{\pmb{\underline{\purple{Step-by-step\:explanation}}{\purple{:}}}}}

Let us first solve for √129.

➝\:\sqrt{129}

➝\:11.3578

Now,

➝\: \frac{95}{8}

➝\:11.875

Clearly,

➵\:11.3578 < 11.875

➵\: \sqrt{129}  <  \frac{95}{8}

Hence, √129 is less than \frac{95}{8}.

\red{\large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese \: \: Mystique35☂}}}}}

5 0
3 years ago
Each letter of the word "supercalifragilisticexpialidocious" is placed into a bag and drawn at 3 times, replacing the letter aft
Makovka662 [10]

Answer:

P(X ≥ 1) = 0.50

Step-by-step explanation:

Given that:

The word "supercalifragilisticexpialidocious" has 34 letters in which 'i' appears 7 times in the word.

Then; the probability of success = 7/34 = 0.20588

Using Binomial distribution to determine the probability; we have:

P(X = x)  = ^nC_x  \ \beta^x   \  (1 - \beta)^{n-x}

where;

x = 0,1,2,...n    and    0  <  β   <   1

and x represents the  number of successes.

However; since the letter is drawn thrice; the probability that the letter "i" is drawn at least once can be computed as:

P(X ≥ 1) = 1 - P(X< 1)

P(X ≥ 1) = 1 - P(X =0)

P(X \ge 1) =  1 - \bigg [ {^3C__0} (0.21)^0 (1-0.21)^{3-0} \bigg]

P(X \ge 1) =  1 - \bigg [ 1 \times 1 (0.79)^{3} \bigg]

P(X ≥ 1) = 1 - 0.50

P(X ≥ 1) = 0.50

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