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3 years ago

hich function has a range of {y|y ≤ 5}? f(x) = (x – 4)2 + 5 f(x) = –(x – 4)2 + 5 f(x) = (x – 5)2 + 4 f(x) = –(x – 5)2 + 4

Mathematics

1 answer:

Brums [2.3K]3 years ago

6 0

Answer:

$f(x)=-(x-4)^2+5$

Step-by-step explanation:

The function $f(x)=-(x-4)^2+5$ has the vertex at $(4,5)$ .

Since $a=-1\:$ , the graph of the function is a maximum graph.

The highest y-value is the y-coordinate of the vertex which is 5.

Therefore the range is { $y|y\le 5$ }

Therefore the correct answer is B.

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The probability that a randomly selected 2 2-year-old male garter snake garter snake will live to be 3 3 years old is 0.98861 0

Mnenie [13.5K]

Answer:

a. $Probability = 0.97735$

b. $Probability = 0.92294$

c. $P(At\ Least\ One) = 1$

No, it is not unusual if at least 1 lives up to 3.

Step-by-step explanation:

Given

Represent the probability that a 2 year old snake will live to 3 with P(Live);

$P(Live) = 0.98861$

Solving (a): Probability that two selected will live to 3 years.

Both snakes have a chance of 0.98861 to live up to 3 years.

So, the required probability is:

$Probability = P(Live)\ and\ P(Live)$

$Probability = 0.98861 * 0.98861$

$Probability = 0.9773497321$

$Probability = 0.97735$ <em>--- Approximated</em>

Solving (b): Probability that seven selected will live to 3 years.

All 7 snakes have a chance of 0.98861 to live up to 3 years.

So, the required probability is:

$Probability = P(Live)^n$

Where $n = 7$

$Probability = 0.98861^7$

$Probability = 0.92294324145$

$Probability = 0.92294$ <em>--- Approximated</em>

Solving (c): Probability that at least one of seven selected will not live to 3 years.

In probabilities, the following relationship exist:

$P(At\ Least\ One) = 1 - P(None).$

So, first we need to calculate the probability that none of the 7 lived up to 3.

If the probability that one lived up to 3 years is 0.98861, then the probability than one do not live up to 3 years is 1 - 0.98861

This gives:

$P(Not\ Live) = 0.01139$

The probability that none of the 7 lives up to 3 is:

$P(None) = P(Not\ Live)^7$

$P(None) = 0.01139^7$

Substitute this value for P(None) in

$P(At\ Least\ One) = 1 - P(None).$

$P(At\ Least\ One) = 1 - 0.01139^7$

$P(At\ Least\ One) = 0.99999999999997513055642436060443621$

$P(At\ Least\ One) = 1$ ---- Approximated

No, it is not unusual if at least 1 lives up to 3.

This is so because the above results, which is 1 shows that it is very likely for at least one of the seven to live up to 3 years

7 0

3 years ago

Estigation -Grade 12

zloy xaker [14]

Answer:

i don't know this type of work.

Step-by-step explanation:

i'm only in 10th grade

7 0

3 years ago

Which graph represents the function y = 2/3x-2

Alina [70]

Answer:

Step-by-step explanation:

so the y-int is where the point is on the y-axis (0,X). Or, in this situation-2.

for slope, you just need to do rise over run, or up and down first then, left and right. If I do that for C, I go 2 up and 3 right. So the slope os 2/3 (P.S. up and right is positive, and down and left is negative)

7 0

3 years ago

Suppose that you are given a bag containing n unbiased coins. You are told that n-1 of these coins are normal, with heads on one

gladu [14]

Answer:

The (conditional) probability that the coin you chose is the fake coin is 2/(1 + n)

Step-by-step explanation:

Given

Total unbiased coin = n

Normal coins =n - 1

Fake = 1

The (conditional) probability that the coin you chose is the fake coin is represented by

P(Fake | Head)

And it's calculated as follows;

P(Fake | Head) = P(Fake, Head) ÷ P(Head) ----- (1)

Where P(Fake, Head) = P(Fake) * P(Head | Fake)

P(Fake) = 1/n --- because only one is fake

P(Head | Fake) = n/n because all coins (including the fake) have head

So, P(Fake, Head) = P(Fake) * P(Head | Fake) becomes

P(Fake, Head) = 1/n * n/n

P(Fake, Head) = 1/n

P(Head) is calculated by

P(Fake) * P(Head | Fake) + P(Normal) * P(Head | Normal)

P(Fake) * P(Head | Fake) = P(Fake, Head) = 1/n (as calculated above)

P(Normal) * P(Head | Normal) = ½ * (n - 1)/n ----- considering that the coin also has a tail with equal probability as that of the head.

Going back to (1)

P(Fake | Head) = P(Fake, Head) ÷ P(Head) becomes

P(Fake | Head) = (1/n) ÷ ((1/n) + (½(n-1)/n))

= (1/n) ÷ ((1/n) + (½(n-1)/n))

= (1/n) ÷ (1/n + (n - 1)/2n)

= (1/n) ÷ (2 + n - 1)/(2n)

= (1/n) ÷ (1 + n)/(2n)

= (1/n) * (2n)/(1 + n)

= 2/(1 + n)

Hence, the (conditional) probability that the coin you chose is the fake coin is 2/(1 + n)

5 0

3 years ago

Which statement is not true for a binomial distribution with n = 15 and p = 1/20 ? a) The standard deviation is 0.8441 b) The nu

malfutka [58]

Answer:

d) The highest probability occurs when x equals 0.7500

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability, given by the following formula:

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

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.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

In this problem, we have that:

$n = 15, p = \frac{1}{20}$

a) The standard deviation is 0.8441

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{15*0.05*0.95} = 0.8441$

This is correct

b) The number of trials is equal to 15

n is the number of trials and $n = 15$ . So this option is correct.

c) The probability that x equals 1 is 0.3658

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

$P(X = 1) = C_{15,1}.(0.05)^{1}.(0.95)^{14} = 0.3658$

This option is correct.

d) The highest probability occurs when x equals 0.7500

False. The number of sucesses is a discrete number, that is, 0, 1, 2,...,15. P(X = 0.75), for example, does not exist.

e) The mean equals 0.7500

$E(X) = np = 15*0.05 = 0.75$

This option is correct.

f) None of the above

d is false

3 0

3 years ago