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lyudmila [28]
3 years ago
11

What is the solution of the system of equations? Enter the solution as an ordered pair (use decimal rounded to the nearest hundr

edth if necessary)
y=\frac{3}{2} x+\frac{5}{4} \\2x+4y=25
I may give Brainleist
Mathematics
1 answer:
Ulleksa [173]3 years ago
4 0

Answer:

(2.5,5)

Step-by-step explanation:

y=3x/2 +5/4

4y=6x+5

-6x+4y=5

6x+12y=75

16y=80

y=5

x=5/2

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sdas [7]
I think it would be 3
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Sketch the graph of a 4th degree polynomial function f(x) such that f(- 3) = 0, f(- 1) = 0, f(1) = 0 , and f(3) = 0 and f(x) is
fenix001 [56]

Answer:

  see attached

Step-by-step explanation:

For each listed root x=p as signified by f(p) = 0, the function has a factor (x-p). The given roots and the given end behavior tell us the factored form is ...

  f(x) = -(x +3)(x +1)(x -1)(x -3)

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The graph is attached. It shows the function increasing at extreme left, crossing the x-axis at x=-3, and again at x=-1, x=1 and x=3. Since it crosses an even number of times, the right-side end behavior is "decreasing" toward negative infinity.

4 0
2 years ago
What does "interpret the meaning of the quotient in terms of the two fractions given" mean in math? 6th grade
Olenka [21]

Answer:

The quotient is the answer to a division problem, the result you get when you divide one number by another. ... If you are dividing by a fraction, you'll first divide your unit boxes into the number given in the denominator of the fraction you are dividing by.

Step-by-step explanation:

Done

4 0
3 years ago
Read 2 more answers
"What is the probability that in 10 dice throws, you will throw AT LEAST two ‘3’s on the dice? Assume you’re throwing a single d
dem82 [27]

Answer:

There is a 51.61% probability that in 10 dice throws, you will throw AT LEAST two ‘3’s on the dice.

Step-by-step explanation:

For each throw, there are only two possible outcomes. Either it is a '3', or it is not. This means that we solve this problem 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}.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:

There are 6 possible outcomes for the dice. This means that the probability that it is a '3' is \frac{1}{6} = 0.167

There are 10 throws, so n = 10.

Probability of throwing AT LEAST two ‘3’s on the dice?

Either you throw less than two, or you throw at least two. The sum of the probabilities of these events is 1. So

P(X < 2) + P(X \geq 2) = 1

P(X \geq 2) = 1 - P(X < 2).

In which

P(X < 2) = P(X = 0) + P(X = 1).

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

P(X = 0) = C_{10,0}.(0.167)^{0}.(0.833)^{10} = 0.1609

P(X = 1) = C_{10,1}.(0.167)^{1}.(0.833)^{9} = 0.3225

So

P(X < 2) = P(X = 0) + P(X = 1) = 0.1609 + 0.3225 = 0.4834.

Finally:

P(X \geq 2) = 1 - P(X < 2) = 1 - 0.4839 = 0.5161.

There is a 51.61% probability that in 10 dice throws, you will throw AT LEAST two ‘3’s on the dice.

5 0
3 years ago
Okay...last time i ask. 4ths times a charm i suppose. please help. will mark brainliest. ill even write you a frickisagfbsdfn po
barxatty [35]

Ok, so when it says "3x, x\leq2," it means that when the x value is less than or equal to 3, you graph 3x. You would have the following points on the line: (-3,-9), (-2,-6), (-1,-3), (0,0), (1,3), and (2,6). The endpoint of the ray on the left side (the points I just said) would be (2,6)

When it says, "1/2 x +5, 2\leqx," it means that when the value of x is greater than or equal to 2, the y value is .5x + 5. You would graph the following points: (2,6), (4,7), (6,8), and (8,9). The endpoint of th ray on the right (the points I just said) would be (2,6).

Keep in mind that though the graph is continuous, it isn't one line, but 2 rays (which you probably already know).

6 0
3 years ago
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