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Salsk061 [2.6K]

2 years ago

5

The solution to a quadric equation are x=2 and x=7. If k is a nonzero constant which of the following must be equal to the quadr

atic equation

1 answer:

nikklg [1K]2 years ago

7 0

Answer:

The quadratic equation form is x² - 9 x + 14 = 0

Step-by-step explanation:

Given in the question as ,

The solution of quadratic equation are x = 2 and x = 7

The constant K is a non-zero number

Now , let the quadratic equation be , ax² + bx + c = 0

And sum of roots = $\frac{ - b}{a}$

product of roots = $\frac{c}{a}$

So, 2 + 7 = $\frac{ - b}{a}$

Or, 2 × 7 = $\frac{c}{a}$

I.e $\frac{ - b}{a}$ = 9 , $\frac{c}{a}$ = 14

Or, b = - 9 a And c = 14 a

Put this value of b and c in standard form of quadratic equation

I.e ax² + bx + c = 0

Or, ax² - 9 ax + 14 a = 0

Or , a ( x² - 9 x + 14 ) = 0

∴ a = 0 And x² - 9 x + 14 = 0

Hence , The quadratic equation form is x² - 9 x + 14 = 0 Answer

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If i leave my house at 6:52 am and run 1 mile at an easy pace (8:15 per mile), then 3 miles at tempo (7:12 per mile) and 1 mile

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2 years ago

Past studies indicate that about 60 percent of the trees in a forested region are classified as softwood. A botanist studying th

alukav5142 [94]

Answer:

1. If it is true that 60 percent of the trees in a forested region are classified as softwood, 0.015 is the probability of obtaining a population proportion greater than 0.6.

Step-by-step explanation:

Hello!

The historical information indicates that 60% of the forest trees are classified as softwood.

A botanist thinks that the proportion might be greater than 60%, so he tested his belief obtaining:

H₀: p = 0.60

H₁: p > 0.60

p-value: 0.015

You need to interpret this p-value. Little reminder:

The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis. It represents the % of size n samples from a population with proportion p=p₀, which will produce a measure that provides evidence as (or stronger) than the current sample that p is not equal to p₀.

The correct answer is:

1. If it is true that 60 percent of the trees in a forested region are classified as softwood, 0.015 is the probability of obtaining a population proportion greater than 0.6.

I hope this helps!

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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