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3 years ago

Given that f(x) = x2 + 2x + 3 and g(x) = quantity of x plus four, over three , solve for f(g(x)) when x = 2.

Mathematics

2 answers:

mihalych1998 [28]3 years ago

6 0

G(2)=(x+4)/3=2 F(2)=x^2+2x+3 4+4+3=11

Sidana [21]3 years ago

4 0

$f(x)=x^2+2x+3$
$g(x)=\frac{x+4}{3}$
$f(g(x))=(\frac{x+4}{3})^2+2(\frac{x+4}{3})+3$
$f(g(x))=\frac{x^2+8x+16}{9}+\frac{2x+8}{3}+3$
$f(g(2))=\frac{2^2+8(2)+16}{9}+\frac{2(2)+8}{3}+3$
$f(g(2))+\frac{4+16+16}{9}+\frac{4+8}{3}+3$
$f(g(2))=\frac{36}{9}+\frac{12}{3}+3$
$f(g(2))=4+4+3$
$f(g(2)=11$

Check:
g(2)=(2+4)/3=2
f(2)=2^2+2(2)+3=4+4+3=11 Checks.

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A computer can be classified as either cutting dash edge or ancient. Suppose that 94% of computers are classified as ancient.

taurus [48]

Answer:

(a) 0.8836

(b) 0.6096

(c) 0.3904

Step-by-step explanation:

We are given that a computer can be classified as either cutting dash edge or ancient. Suppose that 94% of computers are classified as ancient.

(a) Two computers are chosen at random.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 2 computers

r = number of success = both 2

p = probability of success which in our question is % of computers

that are classified as ancient, i.e; 0.94

LET X = Number of computers that are classified as ancient

So, it means X ~ $Binom(n=2, p=0.94)$

Now, Probability that both computers are ancient is given by = P(X = 2)

P(X = 2) = $\binom{2}{2}\times 0.94^{2} \times (1-0.94)^{2-2}$

= $1 \times 0.94^{2} \times 1$

= 0.8836

(b) Eight computers are chosen at random.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 8 computers

r = number of success = all 8

p = probability of success which in our question is % of computers

that are classified as ancient, i.e; 0.94

LET X = Number of computers that are classified as ancient

So, it means X ~ $Binom(n=8, p=0.94)$

Now, Probability that all eight computers are ancient is given by = P(X = 8)

P(X = 8) = $\binom{8}{8}\times 0.94^{8} \times (1-0.94)^{8-8}$

= $1 \times 0.94^{8} \times 1$

= 0.6096

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 8 computers

r = number of success = at least one

p = probability of success which is now the % of computers

that are classified as cutting dash edge, i.e; p = (1 - 0.94) = 0.06

LET X = Number of computers classified as cutting dash edge

So, it means X ~ $Binom(n=8, p=0.06)$

Now, Probability that at least one of eight randomly selected computers is cutting dash edge is given by = P(X $\geq$ 1)

P(X $\geq$ 1) = 1 - P(X = 0)

= $1 - \binom{8}{0}\times 0.06^{0} \times (1-0.06)^{8-0}$

= $1 - [1 \times 1 \times 0.94^{8}]$

= 1 - $0.94^{8}$ = 0.3904

Here, the probability that at least one of eight randomly selected computers is cutting dash edge is 0.3904 or 39.04%.

For any event to be unusual it's probability is very less such that of less than 5%. Since here the probability is 39.04% which is way higher than 5%.

So, it is not unusual that at least one of eight randomly selected computers is cutting dash edge.

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Snezhnost [94]

Answer:

Step-by-step explanation:

(19-5)/3.5

14/3.5

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