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2 years ago

How many solutions does the nonlinear system of equations graphed below have?

Mathematics

2 answers:

k0ka [10]2 years ago

3 0

Answer:

B) 2

Step-by-step explanation:

There are 2 points of intersection

julsineya [31]2 years ago

3 0

Answer:2

Step-by-step explanation:

Looking at the 2 points of intersections, those are your answers. Good luck Apex kids :')

You might be interested in

A class with n kids lines up for recess. The order in which the kids line up is random with eachordering being equally likely. T

Elis [28]

Answer:

A) P(Betty is first in line and mary is last) = P(B₁) + P(Mₙ) - (P(B₁) × P(Mₙ/B₁))

B) The method used is Relative frequency approach.

Step-by-step explanation:

From the question, we are told a sample of n kids line up for recess.

Now, the order in which they line up is random with each ordering being equally likely. Thus, this means that the probability of each kid to take a position is n(total of kids/positions).

Since we are being asked about 3 kids from the class, let's assign a letter to each kid:

J: John

B: Betty

M: Mary

A) Now, we want to find the probability that Betty is first in line or Mary is last in line.

In this case, the events are not mutually exclusive, since it's possible that "Betty is first but Mary is not last" or "Mary is last but Betty is not first" or "Betty is the first in line and Mary is last". Thus, there is an intersection between them and the probability is symbolized as;

P(B₁ ∪ Mₙ) = P(B₁) + P (Mₙ) - P(B₁ ∩ Mₙ) = P(B₁) + P(Mₙ) - (P(B₁) × P(Mₙ/B₁))

Where;

The suffix 1 refers to the first position while the suffix n refers to the last position.

Also, P(B₁ ∩ Mₙ) = P(B₁) × P(Mₙ/B₁)

This is because the events "Betty" and "Mary" are not independent since every time a kid takes his place the probability of the next one is affected.

B) The method used is Relative frequency approach.

In this method, the probabilities are usually assigned on the basis of experimentation or historical data.

For example, If A is an event we are considering, and we assume that we have performed the same experiment n times so that n is the number of times A could have occurred.

Also, let n_A be the number of times that A did occur.

Now, the relative frequency would be written as (n_A)/n.

Thus, in this method, we will define P(A) as:

P(A) = lim:n→∞[(n_A)/n]

7 0

2 years ago

One year ago, you purchased 600 shares of stock for $14 a share. the stock pays $.41 a share in dividends each year. today, you

alukav5142 [94]

If you paid $14 for 600, you paid:

600 * 14 = $8,400.00

If you sold them for $15.30 a share, you made:

600 * 15.30 = $9,180

9,180 - 8,400 = $780

The return on the investment being $780

Very very safe investment and not worth it.

5 0

2 years ago

What is the true solution to In 20+In 5= 2 In x? #31

Neko [114]

Answer:

ln20+ln5=2lnx

ln(20x5)=lnx^2

ln100=1nx^2

100=x^2

square root

x=10

Hope This Helps!

5 0

2 years ago

A school librarian can buy books at a 20% discount from the list price. One month she spent $72 for books. What was the list pri

Nataly_w [17]

Answer:

$\boxed{\sf \ \ YES \ \ }$

Step-by-step explanation:

Hello

let's say that the price of the book was x

the price after a 20% discount is x - 20%*x = x*(1-20%)=x*(1-.20)=0.8*x

and this is $72 so we can write that

0.8*x=72

and then divide by 0.8 both parts

x = 72/0.8=90

So the list price value of the book is $90

and we can verify as 90 - 20%*90 = 90 - 18 = 72

Hope this helps

7 0

2 years ago

An annexation suit against a county subdivision of 1200 residences is being considered by a neighboring city. If the occupants o

Sunny_sXe [5.5K]

Answer:

Probability that in a random sample of 10 at least 3 favor the annexation suit is 0.9453.

Step-by-step explanation:

We are given that an annexation suit against a county subdivision of 1200 residences is being considered by a neighboring city. The occupants of half the residences object to being annexed.

Also, a random sample of 10 residents is taken.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 10 residents

r = number of success = at least 3

p = probability of success which in our question is probability that

residents favor the annexation suit, which is calculated as below;

p = $\frac{\text{Number of residents favoring the annexation suit }}{\text{Total number of residents considered } }$ = $\frac{600}{1200}$ = 0.50

LET X = Number of residents favoring the annexation suit

So, it means X ~ $Binom(n=10, p=0.50)$

Now, Probability that in a random sample of 10 at least 3 favor the annexation suit is given by = P(X $\geq$ 3)

P(X $\geq$ 3) = 1 - P(X < 3) = 1 - P(X $\leq$ 2)

= 1 - [P(X = 0) + P(X = 1) + P(X = 2)]

= $1- [\binom{10}{0}\times 0.50^{0} \times (1-0.50)^{10-0} + \binom{10}{1}\times 0.50^{1} \times (1-0.50)^{10-1} +\binom{10}{2}\times 0.50^{2} \times (1-0.50)^{10-2}]$

= $1-[ 1 \times 1 \times 0.50^{10}+10 \times 0.50^{1} \times 0.50^{9}+45 \times 0.50^{2} \times 0.50^{8}]$

= $1-[ 0.50^{10}+10 \times 0.50^{10}+45 \times 0.50^{10}]$

= $1-0.50^{10}[ 1+10 +45 ]$ = $1-0.50^{10} \times 56$

= 0.9453

Therefore, Probability that in a random sample of 10 at least 3 favor the annexation suit is 0.9453.

8 0

1 year ago