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3 years ago

(c) x2 - 7x + 10 = 0

Mathematics

2 answers:

Sphinxa [80]3 years ago

8 0

Answer:

I love algebra anyways

The ans is in the picture with the steps how i got it

(hope this helps can i plz have brainlist :D hehe)

Step-by-step explanation:

DedPeter [7]3 years ago

6 0

Answer:

x = 2

Step-by-step explanation:

U mean 2x? & i don’t know if (c) is parta the equation

2x - 7x + 10 = 0

-5x + 10 = 0

-10 -10

-5x/-5 = -10/-5

X = 2

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What is the best estimate of the sum of the fraction? 31/4 +17/3 +8/18

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The best estimate of the sum of the expression given is $\rm 13 \dfrac {31}{26}$

What is a Fraction ?

A fraction is written as p/q , where q $\rm \ne$ 0

It is asked to determine the sum of the fraction

31/4 + 17/3 + 8/18

Taking the LCM of 4 , 3 , 18

LCM of 4 , 3 ,18 is 36

Therefore

$\rm \dfrac{ 31 *9 + 17 *12 + 8 *2}{36 }$

(279+204+16) /36

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$\rm 13 \dfrac {31}{26}$

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3 years ago

What is the area of this figure

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2 years ago

A certain medical test is known to detect 73% of the people who are afflicted with the disease Y. If 10 people with the disease

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The probability of an event is the measurement of the chance of that event's occurrence. The probabilities of considered events are:

P(At least 8 have the disease) ≈ 0.4378
P(At most 4 have the disease) ≈ 0.0342

<h3 /><h3>How to find that a given condition can be modeled by binomial distribution?</h3>

Binomial distributions consist of n independent Bernoulli trials. Bernoulli trials are those trials that end up randomly either on success (with probability p) or on failures( with probability 1- p = q (say))

Suppose we have random variable X pertaining to a binomial distribution with parameters n and p, then it is written as

X = B(n,p)

The probability that out of n trials, there'd be x successes is given by

P(X=x) = $^nC_xp^x(1-p)^{n-x}$

Since 10 people can be either diseased or not and they be so independent of each other (assuming them to be selected randomly) , thus, we can take them being diseased or not as outputs of 10 independent Bernoulli trials.

Let we say

Success= Probability of a diseased person tagged as diseased by the clinic

Failure = Probability of a diseased person tagged as not diseased by the clinic.

Then,

P(Success) = p = 72% = 0.72 (of a single person)

P(Failure) = q = 1-p = 0.28

Let X be the number of people diagnosed diseased by the clinic out of 10 diseased people. Then we have: X ≈ B(n+10,P=0.73)

Calculating the needed probabilities, we get:

a) P(At leased 8 have disease) = P(X≥8) =P(X=8) + P(X=9) + P(X=10)

P(X≥8) = $^{10}C_8(0.73)^8(0.28)^2+^{10}C_9(0.73)^9(0.27)^1+^{10}C_{10}(0.73)^{10}(0.27)^0$

P(X≥8) ≈ 0.2548 + 0.1456 + 0.0374 ≈ 0.4378

b) P(At most 4 have the disease) = P(X≤4) = P(X=0) + P(X=1)+P(X=2)+P(X=3)+P(X=4)

P (X ≤ 4) =

$^{10}C_0(0.73)^0(0.27)^{10}+^{10}{C_1(0.73)^1(0.27)^9+^{10}{C_2(0.73)^2(0.27)^8+^{10}C_3(0.73)&^3(0.27)^7 \\$

$+^{10}C_4(0.73)^4(0.27)^6$

P (X ≤ 4) = 0.000003 + 0.000076+0.00088+0.00604+0.02719

P (X ≤ 4) = 0.0342

Thus,

The probabilities of considered events are:

P(At leased 8 have disease) = 0.4378 approx
P(At most 4 have the disease) = 0.0342 approx

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4 0

2 years ago

(c) x2 - 7x + 10 = 0​

(c) x2 - 7x + 10 = 0