Is 4.47 a rational number

Six houses are located along the same road; the distances are shown. Where should the school bus stop to make the sum of distanc

sukhopar [10]

Hey there :)

Now lets name each house

A B C D E F
500ft <-> 500ft <-> 500ft <-> 500ft <-> 500ft <-> 2000ft

Let's say the bus stops at A
= 0 ( from A ) + 500 ( from B ) + 500 + 500 ( from C ) + 500 + 500 + 500 ( from D ) + 500 + 500 + 500 + 500 ( from E ) + 500 + 500 + 500 + 500 + 2000 ( from F )
= 0 + 500 + 1000 + 1500 + 2000 + 4000
= 9000 ft

at B
= 500 ( from A ) + 0 ( from B ) + 500 ( from C ) + 500 + 500 ( from D ) + 500 + 500 + 500 ( from E ) + 500 + 500 + 500 + 2000 ( from F )
= 500 + 0 + 500 + 1000 + 1500 + 3500
= 7000 ft

at C
= 500 ( from A ) + 500 + 500 ( from B ) + 0 ( from C ) + 500 ( from D ) + 500 + 500 ( from E ) + 500 + 500 + 2000 ( from F )
= 500 + 1000 + 0 + 500 + 1000 + 3000
= 6000 ft

Do the same for D , E and F

at D
= 6000 ft

at E
= 7000 ft

at F
= 15000 ft

You will find that the bus should stop at either C or D to make the sum of distances from every house to the stop as small as possible.

8 0

3 years ago

Read 2 more answers

Some parts of California are particularly earthquake-prone. Suppose that in one metropolitan area, 33% of all homeowners are ins

Lina20 [59]

Answer:

(a) The probability mass function of X is:

$P(X=x)={4\choose x}\ (0.33)^{x}\ (1-0.33)^{4-x};\ x=0,1,2,3...$

(b) The most likely value for X is 1.32.

(c) The probability that at least two of the four selected have earthquake insurance is 0.4015.

Step-by-step explanation:

The random variable X is defined as the number among the four homeowners who have earthquake insurance.

The probability that a homeowner has earthquake insurance is, p = 0.33.

The random sample of homeowners selected is, n = 4.

The event of a homeowner having an earthquake insurance is independent of the other three homeowners.

(a)

All the statements above clearly indicate that the random variable X follows a Binomial distribution with parameters n = 4 and p = 0.33.

The probability mass function of X is:

$P(X=x)={4\choose x}\ (0.33)^{x}\ (1-0.33)^{4-x};\ x=0,1,2,3...$

(b)

The most likely value of a random variable is the expected value.

The expected value of a Binomial random variable is:

$E(X)=np$

Compute the expected value of X as follows:

$E(X)=np$

$=4\times 0.33\\=1.32$

Thus, the most likely value for X is 1.32.

(c)

Compute the probability that at least two of the four selected have earthquake insurance as follows:

P (X ≥ 2) = 1 - P (X < 2)

= 1 - P (X = 0) - P (X = 1)

$=1-{4\choose 0}\ (0.33)^{0}\ (1-0.33)^{4-0}-{4\choose 1}\ (0.33)^{1}\ (1-0.33)^{4-1}\\\\=1-0.20151121-0.39700716\\\\=0.40148163\\\\\approx 0.4015$

Thus, the probability that at least two of the four selected have earthquake insurance is 0.4015.

3 0

3 years ago

Can someone help me ASAP plsssss

AlladinOne [14]

that is literally gibberish

4 0

3 years ago

Solve this problem -282 - (+1,017)=2

77julia77 [94]

The answer is not true.

6 0

3 years ago

The graph of y = ax 2 + bx + c is a parabola that opens up and has a vertex at (-2, 5). What is the solution set of the related

loris [4]

Answer: Assume c = 0

Step-by-step explanation:

Step-by-step explanation:

Assume c = 0

Using the formula for the x-coordinate of the vertex, b can be calculated in terms of a:

B can then be substituted into the quadratic equation, along with the coordinates of the vertex, to solve a:

AND

Substituting into the quadratic equation:

Because a is negative, the parabola opens up.

8 0

3 years ago