write an inequality for the description of: ten times a number increased by four is no more than twenty five

Every day your friend commutes to school on the subway at 9 AM. If the subway is on time, she will stop for a $3 coffee on the w

Shtirlitz [24]

Answer:

1.02% probability of spending 0 dollars on coffee over the course of a five day week

7.68% probability of spending 3 dollars on coffee over the course of a five day week

23.04% probability of spending 6 dollars on coffee over the course of a five day week

34.56% probability of spending 9 dollars on coffee over the course of a five day week

25.92% probability of spending 12 dollars on coffee over the course of a five day week

7.78% probability of spending 12 dollars on coffee over the course of a five day week

Step-by-step explanation:

For each day, there are only two possible outcomes. Either the subway is on time, or it is not. Each day, the probability of the train being on time is independent from other days. So we use the binomial probability distribution to solve this problem.

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.

In this problem we have that:

The probability that the subway is delayed is 40%. 100-40 = 60% of the train being on time, so $p = 0.6$

The week has 5 days, so $n = 5$

She spends 3 dollars on coffee each day the train is on time.

Probabability that she spends 0 dollars on coffee:

This is the probability of the train being late all 5 days, so it is P(X = 0).

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

$P(X = 0) = C_{5,0}.(0.6)^{0}.(0.4)^{5} = 0.0102$

1.02% probability of spending 0 dollars on coffee over the course of a five day week

Probabability that she spends 3 dollars on coffee:

This is the probability of the train being late for 4 days and on time for 1, so it is P(X = 1).

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

$P(X = 1) = C_{5,1}.(0.6)^{1}.(0.4)^{4} = 0.0768$

7.68% probability of spending 3 dollars on coffee over the course of a five day week

Probabability that she spends 6 dollars on coffee:

This is the probability of the train being late for 3 days and on time for 2, so it is P(X = 2).

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

$P(X = 2) = C_{5,2}.(0.6)^{2}.(0.4)^{3} = 0.2304$

23.04% probability of spending 6 dollars on coffee over the course of a five day week

Probabability that she spends 9 dollars on coffee:

This is the probability of the train being late for 2 days and on time for 3, so it is P(X = 3).

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

$P(X = 3) = C_{5,3}.(0.6)^{3}.(0.4)^{2} = 0.3456$

34.56% probability of spending 9 dollars on coffee over the course of a five day week

Probabability that she spends 12 dollars on coffee:

This is the probability of the train being late for 1 day and on time for 4, so it is P(X = 4).

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

$P(X = 4) = C_{5,4}.(0.6)^{4}.(0.4)^{1} = 0.2592$

25.92% probability of spending 12 dollars on coffee over the course of a five day week

Probabability that she spends 15 dollars on coffee:

Probability that the subway is on time all days of the week, so $P(X = 5)$ .

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

$P(X = 5) = C_{5,5}.(0.6)^{5}.(0.4)^{0} = 0.0778$

7.78% probability of spending 12 dollars on coffee over the course of a five day week

8 0

3 years ago

At 2:00 p.m. two cars start toward each other from towns 240 miles apart. If the rate of one car is 10 mph faster than the other

Juli2301 [7.4K]

Let’s call the speed of the slower car S, then the speed of the other is S+10mph.
At 5pm they have been travelling for 3 hours. The slower car travels a distance 3S and the faster one 3(S+10).
But the two distances must add up to 240 miles so 3S+3(S+10)=240, 3S+3S+30=240, 6S=210, S=35 mph. The faster car’s speed is 45mph. We can see that 3S is the same distance as 3x, so x=S=35 mph, and the distance the faster car travels is 3×45=135 miles.

5 0

3 years ago

Read 2 more answers

Find all of the zeros of the function and write the polynomial as the product of linear factors.

Tomtit [17]

Answer:

(x+6i)(x-6i)(x+1)

So the zeroes are 6i, -6i, -1

Step-by-step explanation:

Factor by grouping

h(x) = x^2(x+1) + 36(x+1)

h(x) = (x^2+36)(x+1)

3 0

3 years ago

Suppose that we wish to assess whether more than 60 percent of all U.S. households in a particular income class bought life insu

IceJOKER [234]

Answer:

A. 0.0049

B. Yes

Step-by-step explanation:

Sample proportion = 0.64

N = 1000

Population proportion = 0.60

We solve for standard deviation

= √p(1-p)/n

= √0.60(1-0.60)/1000

= √0.60x0.40/1000

= √0.00024

= 0.0155

A.

The probability of sample >=0.64

Z>=0.64-0.60/0.0155

Z >= 0.04/0.0155

So z >= 2.5806

Using excel this equal to 0.0049

0.0049 is probability of sample proportion being 0.64 at least.

B.

This answer in a shows that than 60% of households in the united states income class purchased life insurance last year.

3 0

3 years ago

PLEASE HELP!!! Two trains simultaneously left points M and N and headed towards each other. The distance between point M and poi

iogann1982 [59]

Answer:

not complete question pls finish it

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers