I only got this partially correct please tell me the correct answer

Two water balloons were launched into the air at different moments and collided. the water balloons were modeled by the quadrati

DerKrebs [107]

Two water balloons were launched into the air at different moments and collided. the water balloons were modeled by the quadratic functions: y = −7x2 26x 3 and y = −6x2 23x 5, where y represents the height in meters and x represents the time in seconds after the launch. 2.0 seconds is the time, in seconds, that the balloons collided at the highest point

The time in seconds that the balloons collided at the highest point is; B: 2.0 seconds

We are told that the water balloons were modeled by the quadratic functions:

y = -7x² + 26x + 6 and y = -6x² + 25x + 4

where;

y = the height in meters

x = the time in seconds after the launch

The time at which they will collide has to be at the same height.

Thus;

-7x² + 26x + 6 = -6x² + 25x + 4

Rearranging this gives;

x² - x - 2 = 0.

Using an online quadratic equation solver gives us x = 2 or -1. We will take the positive one as x = 2.0 seconds

Learn more about quadratic equations at

brainly.com/question/1214333

#SPJ4

5 0

1 year ago

For numbers 7 and 8, determined whether the triangles are congruent. explain your reasoning.

raketka [301]

Answer:

#7 is congruent

#8 is not congruent

Step-by-step explanation:

#7 is congruent because of the angle angle side triangle congruency theorem.

#8 is not congruent because it’s properties don’t match with any of the triangle congruency theorems.

3 0

3 years ago

-3(x - 2) - 2x + 4x^2

Fantom [35]

Answer:

-6x+6+4x^2

Step-by-step explanation:

Distribute the -3

-3(x-2)= -3x+6

Combine like terms

<u>-3x</u>+6<u>-2x</u>+4x^2

-6x+6+4x^2

8 0

3 years ago

Among persons donating blood to a clinic, 85% have Rh+ blood (that is, the Rhesus factor is present in their blood.) Six people

Leona [35]

Answer:

a) There is a 62.29% probability that at least one of the five does not have the Rh factor.

b) There is a 22.36% probability that at most four of the six have Rh+ blood.

c) There need to be at least 8 people to have the probability of obtaining blood from at least six Rh+ donors over 0.95.

Step-by-step explanation:

For each person donating blood, there are only two possible outcomes. Either they have Rh+ blood, or they do not. This means that 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 combinatios 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:

$p = 0.85, n = 6$ .

a) fine the probability that at least one of the five does not have the Rh factor.

Either all six have the factor, or at least one of them do not. The sum of the probabilities of these events is decimal 1. So:

$P(X < 6) + P(X = 6) = 1$

$P(X < 6) = 1 - P(X = 6)$

In which:

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

$P(X = 6) = C_{6,6}.(0.85)^{6}.(0.15)^{0} = 0.3771$

So

$P(X < 6) = 1 - P(X = 6) = 1 - 0.3771 = 0.6229$

There is a 62.29% probability that at least one of the five does not have the Rh factor.

b) find the probability that at most four of the six have Rh+ blood.

Either more than four have Rh+ blood, or at most four have. So

$P(X \leq 4) + P(X > 4) = 1$

$P(X \leq 4) = 1 - P(X > 4)$

In which

$P(X > 4) = P(X = 5) + P(X = 6)$

$P(X = 5) = C_{6,5}.(0.85)^{5}.(0.15)^{1} = 0.3993$

$P(X = 6) = C_{6,6}.(0.85)^{6}.(0.15)^{0} = 0.3771$

$P(X > 4) = P(X = 5) + P(X = 6) = 0.3993 + 0.3771 = 0.7764$

$P(X \leq 4) = 1 - P(X > 4) = 1 - 0.7764 = 0.2236$

There is a 22.36% probability that at most four of the six have Rh+ blood.

c) The clinic needs six Rh+ donors on a certain day. How many people must donate blood to have the probability of obtaining blood from at least six Rh+ donors over 0.95?

With 6 donors:

$P(X = 6) = C_{6,6}.(0.85)^{6}.(0.15)^{0} = 0.3771$

37.71% probability of obtaining blood from at least six Rh+ donors over 0.95.

With 7 donors:

$P(X = 6) = C_{7,6}.(0.85)^{6}.(0.15)^{1} = 0.3960$