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3 years ago

Write a recursive formula and an explicit formula for the following arithmetic sequence.

Mathematics

1 answer:

Serjik [45]3 years ago

3 0

Answer:

Explicit: a₁=61 aₙ=a₁ -3ₓ(n-1)

Recursive: a₁=61 aₙ=aₙ₋₁ - 3

Step-by-step explanation:

Explicit: a₁=61 aₙ=a₁ -3ₓ(n-1)

Recursive: a₁=61 aₙ=aₙ₋₁ - 3

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If the surface area of the box of cereal is 272 inches squared, find the length of the box.

LekaFEV [45]

Answer:

x= 8 inches

Step-by-step explanation:

you left out the other lengths but since i had this same question once, the other lengths are 12 and 2-

dimensions of the box:

12 in x 2 in x ?

=272 square inches

and we know that there are 6 faces

area= 2 *(12 * 2) + 2 *(2*x) + 2 *(12*x)

area = 272 sq. in. = 48 + 4x + 24x

272 = 48 + 28x

272 - 48 = 28x

224 = 28x

x = 8 inches

good luck :)

i hope this helps

brainliest would be highly appreciated

have a nice day!

5 0

2 years ago

Help please!!!!!!!!!!!!!!

OLga [1]

Answer:

the distance is 3/4

Step-by-step explanation:

If you have any questions about the way I solved it, don't hesitate to ask me in the comments below :)

8 0

2 years ago

Graph the line with slope - 1 passing through the point (2,-5).

alexira [117]

Answer:

there you go

Step-by-step explanation:

7 0

3 years ago

When we reject the null hypothesis, we would expect the variance ratio, over the long run, to be:_________

FinnZ [79.3K]

Answer:

Cuando rechazamos la hipótesis nula, esperaríamos que la razón de varianza, a largo plazo, sea: <u>mayor al valor crítico.</u>

Step-by-step explanation:

5 0

2 years ago

An article reports that 1 in 500 people carry the defective gene that causes inherited colon cancer. In a sample of 2000 individ

Tema [17]

Answer:

a) P=0.558

b) P=0.021

Step-by-step explanation:

We can model this random variable as a Poisson distribution with parameter λ=1/500*2000=4.

The approximate distribution of the number who carry this gene in a sample of 2000 individuals is:

$P(x=k)=\frac{\lambda^ke^{-\lambda}}{k!} =\frac{4^ke^{-4}}{k!}$

a) We can calculate that the approximate probability that between 4 and 9 (inclusive) as:

$P(4\leq x\leq 9)=\sum_{k=4}^9P(k)\\\\\\ P(4)=4^{4} \cdot e^{-4}/4!=256*0.0183/24=0.195\\\\P(5)=4^{5} \cdot e^{-4}/5!=1024*0.0183/120=0.156\\\\P(6)=4^{6} \cdot e^{-4}/6!=4096*0.0183/720=0.104\\\\P(7)=4^{7} \cdot e^{-4}/7!=16384*0.0183/5040=0.06\\\\P(8)=4^{8} \cdot e^{-4}/8!=65536*0.0183/40320=0.03\\\\P(9)=4^{9} \cdot e^{-4}/9!=262144*0.0183/362880=0.013\\\\\\$

$P(4\leq x\leq 9)=\sum_{k=4}^9P(k)=0.195+0.156+0.104+0.060+0.030+0.013=0.558$

b) The approximate probability that at least 9 carry the gene is:

$P(x\geq9)=1-P(x\leq 8)\\\\\\$

$P(0)=4^{0} \cdot e^{-4}/0!=1*0.0183/1=0.018\\\\P(1)=4^{1} \cdot e^{-4}/1!=4*0.0183/1=0.073\\\\P(2)=4^{2} \cdot e^{-4}/2!=16*0.0183/2=0.147\\\\P(3)=4^{3} \cdot e^{-4}/3!=64*0.0183/6=0.195\\\\P(4)=4^{4} \cdot e^{-4}/4!=256*0.0183/24=0.195\\\\P(5)=4^{5} \cdot e^{-4}/5!=1024*0.0183/120=0.156\\\\P(6)=4^{6} \cdot e^{-4}/6!=4096*0.0183/720=0.104\\\\P(7)=4^{7} \cdot e^{-4}/7!=16384*0.0183/5040=0.06\\\\P(8)=4^{8} \cdot e^{-4}/8!=65536*0.0183/40320=0.03\\\\$

$P(x\geq9)=1-P(x\leq 8)\\\\P(x\geq9)=1-(0.018+0.073+0.147+0.195+0.195+0.156+0.104+0.060+0.030)\\\\P(x\geq9)=1-0.979=0.021$

8 0

2 years ago