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3 years ago

Given a1=4, find the term of the recursivr sequence using: a=2an-1

1 answer:

7 0

So, basically substitute as many times as they tell you (I don;t see it):

a1=4, a2=2*4-1=7, a3=2*7-1=13, ... As you see it grows up pretty quickly so that's why I think your statement may be missing how many times.

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Find the value of x such that sin (90-x) = cos 50

Arte-miy333 [17]

Sin (90-x) = cos x

so x = 40

3 0

3 years ago

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What is the best estimate for the value of x

Vladimir [108]

C
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6 0

3 years ago

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How many solutions are there to the equation below?

Elza [17]

Answer:

no solution

Step-by-step explanation:

heres why:

Let's solve your equation step-by-step.

4x−9(x+1)=20−5x

Step 1: Simplify both sides of the equation.

4x−9(x+1)=20−5x

4x+(−9)(x)+(−9)(1)=20+−5x(Distribute)

4x+−9x+−9=20+−5x

(4x+−9x)+(−9)=−5x+20(Combine Like Terms)

−5x+−9=−5x+20

−5x−9=−5x+20

Step 2: Add 5x to both sides.

−5x−9+5x=−5x+20+5x

−9=20

Step 3: Add 9 to both sides.

−9+9=20+9

0=29

so there are no solution

4 0

3 years ago

If f(1) = 4 and f(n) = f(n − 1) + 5 then find the value of f(5).

notka56 [123]

Answer:

Step-by-step explanation:

f(5)=5(5-1)+5

f(5)=5(4)+5

f(5)=20+5

f(5)=25

5 0

3 years ago

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Suppose a biological cell contains 400 genes. When treated radioactively the probability that

Anton [14]

Let $X\sim\mathrm{Bin}(400,0.006)$ be the random variable representing the number of genes that do get mutated. Here $\mathrm{Bin}(n,p)$ denotes a binomial distribution with parameters $n$ (total number of genes) and $p$ (probability of mutation).

Then the probability that *at most* 4 genes get mutated is

$\mathbb P(X\le4)=\displaystyle\sum_{x=0}^4\mathbb P(X=x)$

where

$\mathbb P(X=x)=\begin{cases}\dbinom{400}x0.006^x(1-0.006)^{400-x}&\text{for }x\in\{0,1,2,\ldots,400\}\\\\0&\text{otherwise}\end{cases}$

You should find that

$\mathbb P(X\le4)\approx0.9047$

8 0

3 years ago