Three times the difference between t and y

Solve the equation <br><br> 4(x-2)=2(2x+6)

lesantik [10]

Answer:

No solution

Step-by-step explanation:

4(x-2) = 2(2x+6)

4x-8 = 4x+12 <-- Your answer

-8 = 12

No solution

7 0

3 years ago

Read 2 more answers

I’m confused on the type of angles these are and how to solve them . (Parallel lines and Transversals)

svp [43]

Answer:

7) 49°

8) 77°

9) 87°

10) 135°

Step-by-step explanation:

7) The angles are between the parallel lines, so are "interior." They are on opposite sides of the transversal, so are "opposite interior" angles. Such angles are congruent, so ...

... ? ≅ 49°

8) The angles are adjacent interior angles, so are supplementary.

... ? + 103° = 180°

... ? = 77°

9) The angles are outside the parallel lines, so are "exterior." They are on opposite sides of the transversal, so are "opposite exterior" angles. Such angles are congruent.

... ? ≅ 87°

10) These are vertical angles, so are congruent. (The other parallel line is irrelevant and doesn't need to be there for this relationship to be true.)

... ? ≅ 135°

4 0

3 years ago

With a short time remaining in the day a delivery driver has to make deliveries at 5 locations among the 6 locations remaining.

nika2105 [10]

The delivery driver has to make deliveries at 5 locations <span>among the 6 locations. </span>This means the order of the probability is important because the route he will take from A to B is different with A to C.
So, you need to use permutation for this problem. The calculation would be:
6P5= 6!/ (6-5)!= 720 different routes

6 0

3 years ago

5(y+2)=4(x-3) Can someone help me find the slope of this line pls

Oxana [17]

Answer:

4/5 is the slope.

6 0

3 years ago

23% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and as

zlopas [31]

Answer:

a) There is a 29.42% probability that the number of college students who say they use credit cards because of the rewards program is exactly two.

b) There is a 41.37% probability that the number of college students who say they use credit cards because of the rewards program is more than two.

c) There is a 69.49% probability that the number of college students who say they use credit cards because of the rewards program is between two and five, inclusive.

Step-by-step explanation:

There are only two possible outcomes. Either the student use credit cards because of the rewards program, or they use for other reason. So, we can solve this exercise using the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

In this problem, we have that:

10 students are randomly selected, so $n = 10$ .

23% of college students say they use credit cards because of the rewards program. This means that $\pi = 0.23$

(a) exactly two

This is P(X = 2).

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

$P(X = 2) = C_{10,2}.(0.23)^{2}.(0.77)^{8} = 0.2942$

There is a 29.42% probability that the number of college students who say they use credit cards because of the rewards program is exactly two.

(b) more than two

This is $P(X > 2)$ .

Either a value is larger than two, or it is smaller of equal. The sum of the decimal probabilities of these events must be 1. So:

$P(X \leq 2) + P(X > 2) = 1$

$P(X > 2) = 1 - P(X \leq 2)$

In which

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

So

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

$P(X = 0) = C_{10,0}.(0.23)^{0}.(0.77)^{10} = 0.0733$

$P(X = 1) = C_{10,1}.(0.23)^{1}.(0.77)^{9} = 0.2188$

$P(X = 2) = C_{10,2}.(0.23)^{2}.(0.77)^{8} = 0.2942$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0733 + 0.2188 + 0.2942 = 0.5863$

$P(X > 2) = 1 - P(X \leq 2) = 1 - 0.5863 = 0.4137$

There is a 41.37% probability that the number of college students who say they use credit cards because of the rewards program is more than two.

(c) between two and five inclusive.

This is

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

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

$P(X = 2) = C_{10,2}.(0.23)^{2}.(0.77)^{8} = 0.2942$

$P(X = 3) = C_{10,3}.(0.23)^{3}.(0.77)^{7} = 0.2343$

$P(X = 4) = C_{10,4}.(0.23)^{4}.(0.77)^{6} = 0.1225$

$P(X = 5) = C_{10,3}.(0.23)^{5}.(0.77)^{5} = 0.0439$

So

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.2942 + 0.2343 + 0.1225 + 0.0439 = 0.6949$

There is a 69.49% probability that the number of college students who say they use credit cards because of the rewards program is between two and five, inclusive.

8 0

3 years ago