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Rufina [12.5K]
1 year ago
7

Find the 13th term of the arithmetic sequence whose common difference is d=7 and whose first term is a1=-3

Mathematics
1 answer:
timama [110]1 year ago
3 0

Answer:

81

Step-by-step explanation:

the formula for an arithmetic sequence:

aₙ = a₁ + (n-1)d

where aₙ = the nth term

a₁ = the 1st term = -3

n = the term number = 13

d = the common difference = 7

Plugging in those values into the formula to find a₁₃:

a₁₃ = -3 + (13-1)7

= -3 + (12)7

= -3 + 84

= 81

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kompoz [17]

Answer:

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Step-by-step explanation:

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3 years ago
Write this number in standard form. 300+80+0.9+0.06+0.001
Lina20 [59]

Hey there! I'm happy to help!

First, let's add the hundreds and the tens.

300+80=380

We see that there is nothing in the ones place, so we keep our ones place 0 and we move onto adding the tenths.

380+0.9=380.9

We add the hundredths.

380.9+0.06=380.96

And finally, we add the thousandths.

380.96+0.001=380.961

Therefore, this number in standard form is 380.961.

Have a wonderful day! :D

5 0
2 years ago
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Find the slope of the roof of a home that rises 8 feet for every horizontal change of 24 feet
Mrrafil [7]
Rise over run 8\24 = 1/3
6 0
3 years ago
A and B are points on a line that has the slope 2. The abscissa of B exceeds the abscissa of A by 3.5. State the relationship of
Troyanec [42]

Answer:

The ordinate of B exceeds the ordinate of A by 7

Step-by-step explanation:

Let

A(x1,y1),B(x2,y2)

where

x1 is the abscissa of A

x2 is the abscissa of B

y1 is the ordinate of A

y2 is the ordinate of B

we know that

m=\frac{y2-y1}{x2-x1}

m=2

so

2=\frac{y2-y1}{x2-x1}  -----> equation A

x2=x1+3.5 ----> equation B

substitute equation B in equation A

2=\frac{y2-y1}{x1+3.5-x1}  

2=\frac{y2-y1}{3.5}  

7=y2-y1  

y2=y1+7  

therefore

The ordinate of B exceeds the ordinate of A by 7

6 0
3 years ago
If t follows a t7 distribution, find t0 such that (a) p(|t | < t0) = .9 and (b) p(t > t0) = .05.
Thepotemich [5.8K]

Answer:

a) P(-t_o < t_7

Using the symmetrical property we can write this like this:

1-2P(t_7

We can solve for the probability like this:

2P(t_7

P(t_7

And we can find the value using the following excel code: "=T.INV(0.05,7)"

So on this case the answer would be t_o =\pm 1.895

b) For this case we can use the complement rule and we got:

1-P(t_7

We can solve for the probability and we got:

P(t_7

And we can use the following excel code to find the value"=T.INV(0.95;7)"

And the answer would be t_o = 1.895

Step-by-step explanation:

Previous concepts

The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".

The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.  

The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."

Solution to the problem

For this case we know that t \sim t(df=7)

And we want to find:

Part a

P(|t|

We can rewrite the expression using properties for the absolute value like this:

P(-t_o < t_7

Using the symmetrical property we can write this like this:

1-2P(t_7

We can solve for the probability like this:

2P(t_7

P(t_7

And we can find the value using the following excel code: "=T.INV(0.05,7)"

So on this case the answer would be t_o =\pm 1.895

Part b

P(t>t_o) =0.05

For this case we can use the complement rule and we got:

1-P(t_7

We can solve for the probability and we got:

P(t_7

And we can use the following excel code to find the value"=T.INV(0.95;7)"

And the answer would be t_o = 1.895

5 0
3 years ago
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