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3 years ago

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If that is = a, find the value of a + 1 over a

1" title="3 - \sqrt{5} " alt="3 - \sqrt{5} " align="absmiddle" class="latex-formula">

1 answer:

Doss [256]3 years ago

6 0

Answer:

7+√5/4

Step-by-step explanation:

if that is = a, find the value of a + 1 over a

Given that a = 3 - √5

a+1/1 = (3-√5)+1/3 - √5

4-√5/3 - √5

Rationalize

4-√5/3 - √5 * 3+√5/3 + √5

= 12 +4√5-3√5-√25/9-5

= 12+√5-5/4

= 7+√5/4

Hence the requred answer is 7+√5/4

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If a cylinder has a height of 21 cm and a volume of 11,144 cm^3.

nataly862011 [7]

$\bf \textit{volume of a cylinder}\\\\
V=\pi r^2 h\qquad 
\begin{cases}
h=height\\
r=radius\\
----------\\
h=21\\
V=11,144
\end{cases}\implies 11,144=\pi \cdot r^2\cdot 21$

solve for "r"

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2 years ago

Use the formula speed distance/time to calculate each speed A train travels 120miles in 2hours speed= Mph

antoniya [11.8K]

Answer:

60

Step-by-step explanation:

120 miles divided by 2 hours equals 60 mph

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2 years ago

Math how many models of 100 do u need to to model 3,200 explain

Keith_Richards [23]

32 because 32 x 100 = 3,200

4 0

2 years ago

Complete the square to put () in vertex form. () = 3<br> 2 + 30 − 12. What is the vertex?

gulaghasi [49]

Answer:

A and A

the equation of a parabola in vertex form is

y = a(x - h)² + k

where ( h, k ) are the coordinates of the vertex and a is a multiplier

y = - 2(x + 3)² + 2 is in this form

with vertex = ( - 3, 2)

To find the y-intercept let x = 0

y = - 2(3)² + 2 = - 18 + 2 = - 16

Similarly

y = - 2(x + 2)² + 2 is in vertex form

vertex = ( - 2 , 2)

x = 0 : y = - 2(2)² + 2 = - 8 + 2 = - 6 ← y- intercept

hope this helped

8 0

2 years ago

A particular telephone number is used to receive both voice calls and fax messages. Suppose that 25% of the incoming calls invol

bagirrra123 [75]

Answer:

a) 0.214 = 21.4% probability that at most 4 of the calls involve a fax message

b) 0.118 = 11.8% probability that exactly 4 of the calls involve a fax message

c) 0.904 = 90.4% probability that at least 4 of the calls involve a fax message

d) 0.786 = 78.6% probability that more than 4 of the calls involve a fax message

Step-by-step explanation:

For each call, there are only two possible outcomes. Either it involves a fax message, or it does not. The probability of a call involving a fax message is independent of other calls. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

25% of the incoming calls involve fax messages

This means that $p = 0.25$

25 incoming calls.

This means that $n = 25$

a. What is the probability that at most 4 of the calls involve a fax message?

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$ .

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001$

$P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006$

$P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025$

$P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064$

$P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118$

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.001 + 0.006 + 0.025 + 0.064 + 0.118 = 0.214$

0.214 = 21.4% probability that at most 4 of the calls involve a fax message

b. What is the probability that exactly 4 of the calls involve a fax message?

$P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118$

0.118 = 11.8% probability that exactly 4 of the calls involve a fax message.

c. What is the probability that at least 4 of the calls involve a fax message?

Either less than 4 calls involve fax messages, or at least 4 do. The sum of the probabilities of these events is 1. So

$P(X < 4) + P(X \geq 4) = 1$

We want $P(X \geq 4)$ . Then

$P(X \geq 4) = 1 - P(X < 4)$

In which

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001$

$P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006$

$P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025$

$P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064$

$P(X$

$P(X \geq 4) = 1 - P(X < 4) = 1 - 0.096 = 0.904$

0.904 = 90.4% probability that at least 4 of the calls involve a fax message.

d. What is the probability that more than 4 of the calls involve a fax message?

Very similar to c.

$P(X \leq 4) + P(X > 4) = 1$

From a), $P(X \leq 4) = 0.214)$

Then

$P(X > 4) = 1 - 0.214 = 0.786$

0.786 = 78.6% probability that more than 4 of the calls involve a fax message

8 0

2 years ago