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2 years ago

If R is the midpoint of QS, RS = 2x-4, ST = tx-1, and RT = 8x-43

Mathematics

2 answers:

saul85 [17]2 years ago

6 0

The question is incomplete. Here is the complete question.

If R is the midpoint of QS, $RS=2x-4$ , ST = $4x-1$ and RT = $8x-43$ , find QS.

Answer: QS = 68 units

Step-by-step explanation: The figure below shows a line segment QT.

To determine QS, first, determine value of x:

RT = RS + ST

$8x-43=2x-4+4x-1$

$8x-43=6x-5$

2x = 38

x = 19

Now, we determine QS:

Midpoint is a point dividing a line segment in two equal parts.

Then, QR = RS.

QS = QR + RS

QS = 2RS

$QS=2(2x-4)$

$QS=4x-8$

Substituting x = 19:

$QS=4.19-8$

QS = 68

<u>The segment QS is 68 units.</u>

serious [3.7K]2 years ago

5 0

Answer: x = 1

Step-by-step explanation:

I have no idea what the actual answer is, as you didn't ask a question. So i'm assuming that you want the value of x.

Step 1: Draw a line containing all the points

R is in between point Q and point S, while T is last, based on alphabetical order.

Illustration:

2x - 4 Tx - 1

8x - 43

Step 2: Find T's value

To find t, create the substitution equation RS + ST = RT

2x - 4 + tx - 1 = 8x - 43

-2x -2x

-4 + tx - 1 = 6x - 43

tx - 5 = 6x - 43

tx = 6x - 38

tx/x = 6x/x - 38

t = 6 - 38

t = -32

This is the value of t

By the way, there is no correlation between point T and the variable t.

Step 3: Find x's value

Now that we know what t is, we can solve for x

2x - 4 -32x - 1 = 8x - 43

-30x - 5 = 8x - 43

+30x +30x

-5 = 38x - 43

+43 +43

38 = 38x

38/38 = 38x/38

1 = x

This is the value of x

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Answer:

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Step-by-step explanation:

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3 years ago

In Illinois, 9% of all drivers arrested for DUI (Driving Under the Influence) are repeat offenders; that is, they have been arre

Veseljchak [2.6K]

Answer:

a) 21.58% probability that exactly 3 people are repeat offenders

b) 97.91% probability that at least one person is a repeat offender

c) 3.69

d) 1.83

Step-by-step explanation:

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.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

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

9% of all drivers arrested for DUI (Driving Under the Influence) are repeat offenders

This means that $p = 0.09$

41 people arrested for DUI in Illinois are selected at random.

This means that $n = 41$

a. What is the probability that exactly 3 people are repeat offenders?

This is P(X = 3).

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

$P(X = 3) = C_{41,3}.(0.09)^{3}.(0.91)^{38} = 0.2158$

21.58% probability that exactly 3 people are repeat offenders

b. What is the probability that at least one person is a repeat offender?

Either none are repeat offenders, or at least one is. The sum of the probabilities of these outcomes is 1. So

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

We want $P(X \geq 1)$ .

Then

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

In which

$P(X = 0) = C_{41,0}.(0.09)^{0}.(0.91)^{41} = 0.0209$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0209 = 0.9791$

97.91% probability that at least one person is a repeat offender

c. What is the mean number of repeat offenders?

$E(X) = np = 41*0.09 = 3.69$

d. What is the standard deviation of the number of repeat offenders?

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{41*0.09*0.91} = 1.83$

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3 years ago

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3 years ago

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Answer:

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3 years ago

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