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kow [346]
4 years ago
7

Determine the direction that this parabola opens y=x^2+2x+8up or down

Mathematics
1 answer:
Paha777 [63]4 years ago
6 0

Answer:

up

Step-by-step explanation:

The standard equation for a parabola is y = ax^2 + bx + c.  If the coefficient a is positive, then the curve (graph) opens up.

In y=x^2+2x+8  this coefficient is +1, so the curve opens up.

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What is the value of x in the equation 3x-1/9y=18 when y=27
IrinaK [193]
Yes it is true 
When we do the sum with the value as x -27 then we will get the answer 
6 0
3 years ago
Read 2 more answers
X =<br> y =<br> (3y + 2)<br> 4x°<br> 52°
antiseptic1488 [7]

TO GET Y I WILL USE THE REASON SUM OF ANGLES IN A TRIANGLE:

52 + 90 +( 3y + 2) = 180  \\ 52 + 90 + 2 +3 y = 180 \\ 3y = 180 - 90 - 52 - 2 \\ 3y = 36 \\  \frac{3y}{3}  =  \frac{36}{3}  \\ y = 12

y=12°

TO FIND X I WILL USE THE REQSON ALTERNATE ANGLES

4x = 52 \\  \frac{4x}{4}  =  \frac{52}{4}  \\ x = 13

x=13°

THEE ANGLES ARE INSIDE THE PARALLEL LINES :NOTE.

7 0
1 year ago
A cage holds two litter of rats. One litter comprises one female and five males. The other litter comprises seven females and tw
timurjin [86]

Answer:

The probability that the two rats are from the first litter is 14.28%, and the probability that the two rats are from the second litter is 34.28%.

Step-by-step explanation:

Since a cage holds two litter of rats, and one litter comprises one female and five males, while the other litter comprises seven females and two males, and a random selection of two rats is done, to find the probability that the two rats are from the same litter the following calculation must be performed:

6/15 x 5/14 = 0.1428

9/15 x 8/14 = 0.3428

Therefore, the probability that the two rats are from the first litter is 14.28%, and the probability that the two rats are from the second litter is 34.28%.

4 0
3 years ago
If KM = 2x - 4 and LM = 12, what is the<br>value of x?​
Sedaia [141]
X = 8.

2 (8) -4 = 12. LM=12
8 0
3 years ago
Two corporate baseball teams are scheduled to play a game together. They agree that if both teams attend or if neither team atte
sp2606 [1]

Answer:

2.99% probability that the cost will be paid by only one team

Step-by-step explanation:

The binomial probability distribution is used to solve this question.

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.

Probability that a team plays:

A team plays if it has at most 2 injured players out of 11.

11 players, so n = 11

Each player with a 5% probability of injury, so p = 0.05

Then

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

In which

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

P(X = 0) = C_{11,0}.(0.05)^{0}.(0.95)^{11} = 0.5688

P(X = 1) = C_{11,1}.(0.05)^{1}.(0.95)^{10} = 0.3293

P(X = 2) = C_{11,2}.(0.05)^{2}.(0.95)^{9} = 0.0867

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.5688 + 0.3293 + 0.0867 = 0.9848

Each team has a 0.9848 probability of showing up to play.

What is the probability that the cost will be paid by only one team?

This happens if one team shows up and the other do not.

2 teams, so n = 2

Each team has a 0.9848 probability of showing up to play, so p = 0.9848.

This probability is P(X = 1).

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

P(X = 1) = C_{2,1}.(0.9848)^{1}.(0.0152)^{1} = 0.0299

2.99% probability that the cost will be paid by only one team

7 0
4 years ago
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