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3 years ago

Which equation shows the quadratic formula used correctly to solve 5x2 + 3x – 4 = 0 for x?

Mathematics

1 answer:

enot [183]3 years ago

6 0

Answer:

[-3 ±√(89)]/10

Step-by-step explanation:

<u>Points to remember</u>

Quadratic formula for finding the solution of a quadratic equation ax² + bx + c = 0 is given by,

x = [-b ± √(b² - 4ac)]/2a

It is given a quadratic equation,

5x² + 3x - 4

<u>To find the solution using formula</u>

Here a = 5, b = 3 and c = -4

x = [-b ± √(b² - 4ac)]/2a

= [-3 ± √((-3)² - 4*5*(-4))]/2*5

= [-3 ±√(9 +80)]/10

= [-3 ±√(89)]/10

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Please help me answer this question. I will mark braineist to first answers no work nessacary.

andriy [413]

Plug in the given values and you'll see its option 2.

x=0 , f(x) = 0^2 + 1 = 1

x = 1 , f(x) = 1^2 + 1 = 2

x = 2, f(x) = 2^2 + 1 = 5

4 0

3 years ago

Read 2 more answers

Graph the solution. x/8>2÷(3+2x2+1)⋅2

Digiron [165]

X is the grater one becouse x/8>2 and evrethangs peced together at that point

7 0

3 years ago

7. Sky Castle Company is conducting research on 3 independent communities of 12 members each to see how they have been consuming

tresset_1 [31]

Answer:

1.98% probability that at least 11 members are categorized as low risk participants in two of three communities

Step-by-step explanation:

To solve this question, we need to use two separate binomial trials.

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 of at least 11 members being categorized as low risk participants.

12 members in the sample, so $n = 12$

30% chance to be categorized as high risk participants. So 100-30 = 70% probability of being categorized as low risk participants. So $p = 0.7$

This probability is

$P(X \leq 11) = P(X = 11) + P(X = 12)$

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

$P(X = 1) = C_{12,11}.(0.7)^{11}.(0.3)^{1} = 0.0712$

$P(X = 2) = C_{12,12}.(0.7)^{12}.(0.3)^{0} = 0.0138$

$P(X \leq 11) = P(X = 11) + P(X = 12) = 0.0712 + 0.0138 = 0.0850$

What is the probability that at least 11 members are categorized as low risk participants in two of three communities?

For each community, 8.50% probability of at least 11 members being categorized as low risk. So $p = 0.085$

Three comunities, so $n = 3$

This probability is P(X = 2).

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

$P(X = 2) = C_{3,2}.(0.085)^{2}.(0.915)^{1} = 0.0198$

1.98% probability that at least 11 members are categorized as low risk participants in two of three communities

6 0

4 years ago

A contractor cut 28 feet of wire into sections that were 4 2/3 feet long. How many pieces of wire did the contractor have?

Alisiya [41]

6

28 divided by 4 2/3=6
:)

3 0

3 years ago

30 points! would really appreciate the help!

katrin2010 [14]

Answer:

Step-by-step explanation:

cos43=60/AC

AC=60/cos43=82.04

cos27=60/BC

BC=60/cos27=67.34

sin(90-43)/BC=sin(43+27)/AB

AB=BCsin70/sin47

AB=67.34sin70/sin47

AB=86.522

P=86.522+67.34+82.04

P=235.9m

8 0

3 years ago