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3 years ago

Consider the equation |(x-3)^2-4|=b. Determine the value of b so that the equation has no solution.

Mathematics

2 answers:

oee [108]3 years ago

3 0

As an absolute value makes any value positive, any negative value of b will cause the equation to be unsolvable.

saul85 [17]3 years ago

3 0

<span>|(x-3)^2-4|=b since this is an absolute value equation

b can have any value less than zero so that the above is never true, in interval notation:

b=(-oo,0)</span>

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Which graph shows a system of equations with no solutions?

podryga [215]

Graph of Parallel lines shows a system of equations with no solutions

Step-by-step explanation:

Consider a set of equations

$7x - 2y = 16\\21x + 6y =24$

If we solve this both equations using any one of the solving method, (Substitution method) then we will get

$7x-2y=16\\7x=16+2y\\x=\frac{16+2y}{7}$

substituting the following x in 2nd equation (21x + 6y = 24) We get

$21(\frac{16+2y}{7} )+6y=24\\3(16+2y)+6y=24\\48+6y+6y=24\\12y=24-48\\y=-\frac{24}{12} \\y=-2$

Put y= -2 in x equation

$x=\frac{16+2(-2)}{7}\\ x=\frac{16-4}{7}\\\x=\frac{12}{7} \\x=1.71$

Comparing these (x,y) values we can understand that they never meet at a point

4 0

3 years ago

I'm not sure what AAS,SAS,SSS,ASA mean .how do I solve this?

notka56 [123]

Answer:

AAS is an acronym for Angle-Angle-Side. It basically means that if two angles and a nonincluded side of one triangle are congruent to two angles and the corresponding nonincluded side of a second triangle, then the two triangles are congruent. SAS is an acronym for Side-Angle-Side. It means that if two sides and the included angle of one triangle are congruent to two sides and the included angle of a second triangle, then the two triangles are congruent. SSS is an acronym for Side-Side-Side. It means that if three sides of one triangle are congruent to three sides of a second triangle, then the two triangles are congruent. ASA is an acronym for Angle-Side-Angle. It means that if two angles and the included side of one triangle are congruent to two angles and the included side of a second triangle, then the two triangles are congruent.

In the problem, we know that the corresponding sides of both triangles are congruent to each other, so those would be given. The third side of each triangle would also be congruent because of reflexive property. Reflexive property means that the two triangles share a line segment. So, the answer would be SSS.

8 0

3 years ago

I need help with this question

timofeeve [1]

Answer:

67 people i believe. I am not 100% sure..

Step-by-step explanation:

3 0

3 years ago

34% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and a

finlep [7]

Answer:

a) There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

b) There is a 71.62% probability that more than two students use credit cards because of the rewards program.

c) There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

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 problem by the binomial distribution.

Binomial probability

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 student are sampled, so $n = 10$

34% of college students say they use credit cards because of the rewards program, so $\pi = 0.34$

(a) exactly two

This is P(X = 2).

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

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

(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 must be 1. So:

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

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

In which

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

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

$P(X = 0) = C_{10,0}.(0.34)^{0}.(0.66)^{10} = 0.0157$

$P(X = 1) = C_{10,1}.(0.34)^{1}.(0.66)^{9} = 0.0808$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0157 + 0.0808 + 0.1873 = 0.2838$

$P(X > 2) = 1 - P(X \leq 2) = 1 - 0.2838 = 0.7162$

There is a 71.62% probability that more than two students use credit cards because of the rewards program.

(c) between two and five inclusive

This is:

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

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

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X = 3) = C_{10,3}.(0.34)^{3}.(0.66)^{7} = 0.2573$

$P(X = 4) = C_{10,4}.(0.34)^{4}.(0.66)^{6} = 0.2320$

$P(X = 5) = C_{10,5}.(0.34)^{5}.(0.66)^{5} = 0.1434$

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1873 + 0.2573 + 0.2320 + 0.1434 = 0.82$

There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

6 0

3 years ago

Please help i will mark branliest

kotegsom [21]

Answer:

Step-by-step explanation:

5 0

3 years ago