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3 years ago

14

Find the 25th term of the arithmetic sequence.

2 answers:

dybincka [34]3 years ago

8 0

The answer is -80, i have done all the math and -80 is the answer

Phantasy [73]3 years ago

3 0

Answer:

C. -80

Step-by-step explanation:

i did all the math for this problem and -80 is the correct term.

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Please help freshman math much-needed

Damm [24]

You can make the equatin: y=15x+22
X is the number of weeks
Y is the total amount of money needed

Plug in 120 for y and solve for x
120=15x+22
98=15x
x=6.53

But the answer is 7 weeks because we have to round up so he gets paid for the whole week

5 0

3 years ago

Read 2 more answers

There are 10 balls in an urn, numbered from 1 to 10. If 5 balls are selected at random and their numbers are added, what is the

Kisachek [45]

Let $B_i$ denote the value on the $i$ -th drawn ball. We want to find the expectation of $S=B_1+B_2+B_3+B_4+B_5$ , which by linearity of expectation is

$E[S]=E\left[\displaystyle\sum_{i=1}^5B_i\right]=\sum_{i=1}^5E[B_i]$

(which is true regardless of whether the $X_i$ are independent!)

At any point, the value on any drawn ball is uniformly distributed between the integers from 1 to 10, so that each value has a 1/10 probability of getting drawn, i.e.

$P(X_i=x)=\begin{cases}\frac1{10}&\text{for }x\in\{1,2,\ldots,10\}\\0&\text{otherwise}\end{cases}$

and so

$E[X_i]=\displaystyle\sum_{i=1}^{10}x\,P(X_i=x)=\frac1{10}\frac{10(10+1)}2=5.5$

Then the expected value of the total is

$E[S]=5(5.5)=\boxed{27.5}$

8 0

3 years ago

Eitan randomly selected volcanoes to travel to and study. After traveling, he had seen 13 cinder cone volcanoes, 18 shield volca

Crazy boy [7]

Answer:

6 STRATOVOLCANOES

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

I need help with this one.

Mandarinka [93]

Answer:

$T_{n}$ = $\frac{1}{5}$ $T_{n-1}$ with T₁ = 5000

Step-by-step explanation:

There is a common ratio r between consecutive terms in the sequence, that is

r = $\frac{1000}{5000}$ = $\frac{200}{1000}$ = $\frac{40}{200}$ = $\frac{1}{5}$

This indicates the sequence is geometric with recursive rule

$T_{n}$ = r $T_{n-1}$

Here tr = $\frac{1}{5}$ , thus

$T_{n}$ = $\frac{1}{5}$ $T_{n-1}$ , T₁ = 5000

3 0

3 years ago

A six-sided die is loaded in such a way that the probability of each face turning up is proportional to the number of dots on th

agasfer [191]

Answer:

P(2U5) = 7/21 = 1/3

the probability of getting either a 5 or a 2 in one throw is 1/3

Step-by-step explanation:

Given that; the probability of each face turning up is proportional to the number of dots on that face

P(1) = 1×P(1)

P(2) = 2×P(1)

P(3) = 3×P(1)

P(4) = 4×P(1)

P(5) = 5×P(1)

P(6) = 6×P(1)

P(T) = 21×P(1)

Where;

P(x) is the probability of getting number x on the dice.

P(T) is the total probability of obtaining any number

N(x) is the number of possible number x in terms of the distribution function.

P(x) = N(x)/N(T) ....1

And since P(T) is constant, and P(T) is proportional to N(T) then,

P(x) is directly proportional to N(x)

So, equation 1 becomes;

P(x) = N(x)/N(T) = P(x)/P(T) ....2

The probability of getting either a 5 or a 2 in one throw

P(2U5) = (P(2) + P(5))/P(T)

Substituting the values of each probability;

P(2U5) = (2P(1) + 5P(1))/21P(1)

P(2U5) = 7P(1)/21P(1)

P(1) cancel out, to give;

P(2U5) = 7/21 = 1/3

the probability of getting either a 5 or a 2 in one throw is 1/3

8 0

3 years ago