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2 years ago

PLEASE HELP WITH THIS ONE QUESTION

Mathematics

1 answer:

Snezhnost [94]2 years ago

8 0

Answer:

The fourth option.

Step-by-step explanation:

In the given equation x^2 + 4x + 9 = 0, a=1, b=4, and c=9. The quadratic formula is in the attachment below so if you plug in a, b, and c you get the fourth option.

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A national survey indicated that 30% of adults conduct their banking online. It also found that 40% are under the age of 50. and

Svet_ta [14]

Answer:

Step-by-step explanation:

Hello!

I'll express all the given percentages as probabilities:

Given the events:

Banking online (Bo)

Under the age of 50 (<50)

P(Bo)= 0.30

P(<50)= 0.40

P(Bo ∩ <50)= 0.25

1) What percentage of adults do not conduct their banking online?

The event "adults that do not conduct their baking online" is the complement of the event "adults that conduct their baking online" Symbolically $Bo^c$

P( $Bo^c$ )= 1 - P(Bo)= 1 - 0.30 = 0.70

2) What type of probability is 25%?

The probability P(Bo ∩ <50)= 0.25 is a joint probability, it indicates the intersection between both events.

3) Construct a contingency table showing all joint and marginal probabilities.

Check attachment.

4) What is the probability that an individual conducts banking online given that the individual is under the age of 50?

Symbolically:

P(Bo/<50)= P(Bo ∩ <50) = 0.25 = 0.625

P(<50) 0.40

I hope it helps!

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3 years ago

What value of b will cause the system to have an infinite number of solutions?

irga5000 [103]

b must be equal to -6 for infinitely many solutions for system of equations $y = 6x + b$ and $-3 x+\frac{1}{2} y=-3$

Solution:

Need to calculate value of b so that given system of equations have an infinite number of solutions

$\begin{array}{l}{y=6 x+b} \\\\ {-3 x+\frac{1}{2} y=-3}\end{array}$

Let us bring the equations in same form for sake of simplicity in comparison

$\begin{array}{l}{y=6 x+b} \\\\ {\Rightarrow-6 x+y-b=0 \Rightarrow (1)} \\\\ {\Rightarrow-3 x+\frac{1}{2} y=-3} \\\\ {\Rightarrow -6 x+y=-6} \\\\ {\Rightarrow -6 x+y+6=0 \Rightarrow(2)}\end{array}$

Now we have two equations

$\begin{array}{l}{-6 x+y-b=0\Rightarrow(1)} \\\\ {-6 x+y+6=0\Rightarrow(2)}\end{array}$

Let us first see what is requirement for system of equations have an infinite number of solutions

If $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y+c_{2}=0$ are two equation

$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ then the given system of equation has no infinitely many solutions.

In our case,

$\begin{array}{l}{a_{1}=-6, \mathrm{b}_{1}=1 \text { and } c_{1}=-\mathrm{b}} \\\\ {a_{2}=-6, \mathrm{b}_{2}=1 \text { and } c_{2}=6} \\\\ {\frac{a_{1}}{a_{2}}=\frac{-6}{-6}=1} \\\\ {\frac{b_{1}}{b_{2}}=\frac{1}{1}=1} \\\\ {\frac{c_{1}}{c_{2}}=\frac{-b}{6}}\end{array}$

As for infinitely many solutions $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

$\begin{array}{l}{\Rightarrow 1=1=\frac{-b}{6}} \\\\ {\Rightarrow6=-b} \\\\ {\Rightarrow b=-6}\end{array}$

Hence b must be equal to -6 for infinitely many solutions for system of equations $y = 6x + b$ and $-3 x+\frac{1}{2} y=-3$

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2 years ago