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3 years ago

15

What Are The Points?

1 answer:

solong [7]3 years ago

7 0

-4,-4 - 9,-1 is the answer, i think

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A football quarterback enjoys practicing his long passes over 40 yards. He misses the first pass 40% of the time. When he misses

laila [671]

Answer:

Probability of missing two passes in a row is 0.08.

Step-by-step explanation:

Event E = A football player misses twice in a row.

P(E) = ?

Event X = Football player misses the first pass

P(X) = 0.4

Event Y = Football player misses just after he first miss

P(Y) = 0.2

Both the events are exclusive so the probability of occuring of these two events can be calculated by the formula:

P(E) = P(X).P(Y)

P(E) = 0.4*0.2

P(E) = 0.08

4 0

3 years ago

Seven years ago, lan purchased a $265,000 home with a 30-year mortgage at 3.5%. Having recently lost his job, he can no longer a

MariettaO [177]

Answer:

980.82

Step-by-step explanation:

5 0

3 years ago

Write the slope intercept form of the equation of the line through the given points (-5,3) and (3,0)

grigory [225]

Answer:Y=8,-5

Step-by-step explanation:

3 0

3 years ago

Determine formula of the nth term 2, 6, 12 20 30,42

nalin [4]

Check the forward differences of the sequence.

If $\{a_n\} = \{2,6,12,20,30,42,\ldots\}$ , then let $\{b_n\}$ be the sequence of first-order differences of $\{a_n\}$ . That is, for n ≥ 1,

$b_n = a_{n+1} - a_n$

so that $\{b_n\} = \{4, 6, 8, 10, 12, \ldots\}$ .

Let $\{c_n\}$ be the sequence of differences of $\{b_n\}$ ,

$c_n = b_{n+1} - b_n$

and we see that this is a constant sequence, $\{c_n\} = \{2, 2, 2, 2, \ldots\}$ . In other words, $\{b_n\}$ is an arithmetic sequence with common difference between terms of 2. That is,

$2 = b_{n+1} - b_n \implies b_{n+1} = b_n + 2$

and we can solve for $b_n$ in terms of $b_1=4$ :

$b_{n+1} = b_n + 2$

$b_{n+1} = (b_{n-1}+2) + 2 = b_{n-1} + 2\times2$

$b_{n+1} = (b_{n-2}+2) + 2\times2 = b_{n-2} + 3\times2$

and so on down to

$b_{n+1} = b_1 + 2n \implies b_{n+1} = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)$

We solve for $a_n$ in the same way.

$2(n+1) = a_{n+1} - a_n \implies a_{n+1} = a_n + 2(n + 1)$

Then

$a_{n+1} = (a_{n-1} + 2n) + 2(n+1) \\ ~~~~~~~= a_{n-1} + 2 ((n+1) + n)$

$a_{n+1} = (a_{n-2} + 2(n-1)) + 2((n+1)+n) \\ ~~~~~~~ = a_{n-2} + 2 ((n+1) + n + (n-1))$

$a_{n+1} = (a_{n-3} + 2(n-2)) + 2((n+1)+n+(n-1)) \\ ~~~~~~~= a_{n-3} + 2 ((n+1) + n + (n-1) + (n-2))$

and so on down to

$a_{n+1} = a_1 + 2 \displaystyle \sum_{k=2}^{n+1} k = 2 + 2 \times \frac{n(n+3)}2$

$\implies a_{n+1} = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}$

6 0

2 years ago

1.6÷_=4 a deciaml division promblem what is the missing divisor

Sladkaya [172]

1.6 divided by x =4

we can exchange the 4 and the x

1.6 divided by 4 =x

.4 =x

8 0

3 years ago

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