Solve the equation 2/5x - 1 = 9. Complete the solution and check your solution.

If the rectangular screen of a 40-inch monitor is 35 inches wide, what is the height of the monitor? Round to the nearest tenth.

zloy xaker [14]

Answer:

19.4

Step-by-step explanation:

We can solve this question using Pythagorean theorem

40^2 = h^2 + 35^2

1600= h^2+ 1225

1600-1225= h^2

375= h^2

h= √375

h = 19.4

Hence the height is 19.4

7 0

3 years ago

Need geometry help. ASAP pls & thanks

Sladkaya [172]

Answer:

1008.0 yd square

Step-by-step explanation:

6 0

3 years ago

Matthew made a fruit salad, in which the ratio of blueberries to red grapes is 8:6. If Matthew used 28 blueberries, how many red

Inga [223]

Answer:

Step-by-step explanation:

Let's look at some other ratios that equal 8:6 We could increase them or decrease them. For instance if we cut one in half we cut the other one in half, so 8:6 would be the same ratio as 4:3. With the same thinking if we double one we double the other so 8:6 s the same ratio as 16:12. Now, we want to go from 8 to 28. What do we multiply 8 by to get to it? Or, it might be easier to use 4:3 and see what you multiply 4 by. Then you multiply 3 by that same number to get the other number.

Let me know if that doesn't make sense.

4 0

3 years ago

What is the percent of change from 35 to 62? round to the nearest percent.

Leona [35]

In order to change from 35 to 62, we have to add 27. So, the question becomes: which percentage of 35 is 27?

To answer this question, we set this simple equation

$27 = 35\cdot\dfrac{x}{100}$

And solving for x we have

$x = \dfrac{2700}{35} \approx 77.14$

So, if you change from 35 to 62, you have an increase of about 77%

7 0

3 years ago

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