All categories

1 year ago

The graph below shows the solution to which system of inequalities?

Mathematics

1 answer:

Brut [27]1 year ago

7 0

Answer: C
Explanation: Let's find each equation without any inequalities first. We have a dotted line at y = 1, so we can already remove A and B. Now, a dotted line represents exclusivity and hence; it should not include the = sign.

Thus, by elimination, the answer must be C.

You might be interested in

A construction company is considering submitting bids for contracts of three different projects. The company estimates that it h

julsineya [31]

Answer:

a. $P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}\\$

b. $E(x) = 0.3$

c. $S(x)=0.5196$

d. $E=5,000$

Step-by-step explanation:

The probability that the company won x bids follows a binomial distribution because we have n identical and independent experiments with a probability p of success and (1-p) of fail.

So, the PMF of X is equal to:

$P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}\\$

Where p is 0.1 and it is the chance of winning. Additionally, n is 3 and it is the number of bids. So the PMF of X is:

$P(x)=\frac{3!}{x!(3-x)!}*0.1^{x}*(1-0.1)^{n-x}\\$

For binomial distribution:

$E(x)=np\\S(x)=\sqrt{np(1-p)}$

Therefore, the company can expect to win 0.3 bids and it is calculated as:

$E(x) = np = 3*0.1 = 0.3$

Additionally, the standard deviation of the number of bids won is:

$S(x)=\sqrt{np(1-p)}=\sqrt{3(0.1)(1-0.1)}=0.5196$

Finally, the probability to won 1, 2 or 3 bids is equal to:

$P(1)=\frac{3!}{1!(3-1)!}*0.1^{1}*(1-0.1)^{3-1}=0.243\\P(2)=\frac{3!}{2!(3-2)!}*0.1^{2}*(1-0.1)^{3-2}=0.027\\P(3)=\frac{3!}{3!(3-3)!}*0.1^{3}*(1-0.1)^{3-3}=0.001$

So, the expected profit for the company is equal to:

$E=-10,000+50,000(0.243)+100,000(0.027)+150,000(0.001)\\E=5,000$

Because there is a probability of 0.243 to win one bid and it will produce 50,000 of income, there is a probability of 0.027 to win 2 bids and it will produce 100,000 of income and there is a probability of 0.001 to win 3 bids and it will produce 150,000 of income.

5 0

2 years ago

A school district has a student population of $6,734 students. If the maximum class size is 25 students. How many more teacher m

emmasim [6.3K]

Answer:

53 teachers

Step-by-step explanation:

Basically, what we need to do here is to find how many teachers there need to be, first. If there are 6,734 students in the school district and if maximum class size is 25, then the number of teachers needed is:

6,734 / 25 = 269.36

Of course, it's obvious that we can't have a decimal number of teachers, so we need to find integer (269 or 270).

If we take 269 teachers and 25 students per class, we get:

269 • 25 = 6,725 students, which is not enough, since there are 6,734 students.

That means that the number of teachers needed is 270.

It is given that there are already 217 teachers, meaning that 270-217=53 teachers have to be supplemented.

8 0

2 years ago

A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the l

Katena32 [7]

Answer:

(a) The fraction of the calls last between 4.50 and 5.30 minutes is 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is 0.1271.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is 0.745.

(e) The time is 5.65 minutes.

Step-by-step explanation:

We are given that the mean length of time per call was 4.5 minutes and the standard deviation was 0.70 minutes.

Let X = the length of the calls, in minutes.

So, X ~ Normal( $\mu=4.5,\sigma^{2} =0.70^{2}$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean time = 4.5 minutes

$\sigma$ = standard deviation = 0.7 minutes

(a) The fraction of the calls last between 4.50 and 5.30 minutes is given by = P(4.50 min < X < 5.30 min) = P(X < 5.30 min) - P(X $\leq$ 4.50 min)

P(X < 5.30 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{5.30-4.5}{0.7}$ ) = P(Z < 1.14) = 0.8729

P(X $\leq$ 4.50 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.5-4.5}{0.7}$ ) = P(Z $\leq$ 0) = 0.50

The above probability is calculated by looking at the value of x = 1.14 and x = 0 in the z table which has an area of 0.8729 and 0.50 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.8729 - 0.50 = 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is given by = P(X > 5.30 minutes)

P(X > 5.30 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{5.30-4.5}{0.7}$ ) = P(Z > 1.14) = 1 - P(Z $\leq$ 1.14)

= 1 - 0.8729 = 0.1271

The above probability is calculated by looking at the value of x = 1.14 in the z table which has an area of 0.8729.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is given by = P(5.30 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 5.30 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 5.30 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{5.30-4.5}{0.7}$ ) = P(Z $\leq$ 1.14) = 0.8729

The above probability is calculated by looking at the value of x = 2.14 and x = 1.14 in the z table which has an area of 0.9838 and 0.8729 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.8729 = 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is given by = P(4.00 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 4.00 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 4.00 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.0-4.5}{0.7}$ ) = P(Z $\leq$ -0.71) = 1 - P(Z < 0.71)

= 1 - 0.7612 = 0.2388

The above probability is calculated by looking at the value of x = 2.14 and x = 0.71 in the z table which has an area of 0.9838 and 0.7612 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.2388 = 0.745.

(e) We have to find the time that represents the length of the longest (in duration) 5 percent of the calls, that means;

P(X > x) = 0.05 {where x is the required time}

P( $\frac{X-\mu}{\sigma}$ > $\frac{x-4.5}{0.7}$ ) = 0.05

P(Z > $\frac{x-4.5}{0.7}$ ) = 0.05

Now, in the z table the critical value of x which represents the top 5% of the area is given as 1.645, that is;

$\frac{x-4.5}{0.7}=1.645$

${x-4.5}{}=1.645 \times 0.7$

x = 4.5 + 1.15 = 5.65 minutes.

SO, the time is 5.65 minutes.

7 0

2 years ago

If m(x) = x+5/x-1, and n(x) = x-3, which function has the same domain as m of n of x?

mamaluj [8]

Is that the exact question? because the last part doesn't really make sense to me

4 0

1 year ago

which of the following represents a chord, but not a diameter, of the circle? A. TS, B. SN, C.TR, D.MN

Georgia [21]

Answer: A diameter goes straight across the circle and passes through the center point (point M). A radius stops at the center point, making the line go halfway across the circle (basically half the diameter). A chord goes across the circle, but does not go through the center point. Therefore, the answer is C, because it does not go through the center point.

5 0

1 year ago