Christine wants to use the net below to create a three-dimensional design for art class.

On a coordinate plane, solid circles appear at the following points: (negative 4, 2), (negative 2, 2), (negative 1, 4), (1, 1),

lawyer [9240]

Answer:

(1, 3)

Step-by-step explanation:

Provided:

A circle containing these specified points-

(4, 2), (-2, 2), (-1, 4), (1, 1), (1, 3), (2, -3).

A function resembles a mechanism that produces an output for each unique input. Each input must yield a distinct output.

The function's input is characterized as the domain, and the output corresponds to its range.

The domain is indicated by the 'x' coordinate while the range is represented by the 'y' coordinate.

This means the domain serves as the independent variable, and the range is contingent upon the domain’s values.

In this case, a relation or set of points constitutes a function only if every 'x' in the ordered pairs is unique.

Within the described relation, the pairs (1, 1) and (1, 3) share the same input. Thus, removing either one will convert the relation into a function.

Therefore, choosing (1, 3) is the right answer.

5 0

1 month ago

Read 2 more answers

A tobacco company claims that the amount of nicotine in its cigarettes is a randomvariable with mean 2.2 mg and standard deviati

PIT_PIT [9117]

Response:

0.0000

Unusual

To explain step-by-step:

A tobacco company asserts that the nicotine content in its cigarettes behaves as a random variable with a mean of 2.2 mg and a standard deviation of.3 mg.

This means the population parameters are

$\mu =2.2 \\s = 0.3$

The likelihood that the sample average would equal or exceed 3.1

$=P(X\geq 3.1)\\=P(Z\geq \frac{3.1-2.2}{\frac{0.3}{\sqrt{100} } } )\\=P(Z\geq 30)\\$

equals 0.0000

8 0

8 days ago

If BCD ~ GEF, find BD

PIT_PIT [9117]

Response:

BD equals 19

Detailed breakdown:

4 0

26 days ago

A frog is sitting exactly in the middle of a board that is five feet long. Every ten seconds he jumps one foot to the left or on

PIT_PIT [9117]

Answer:

Given that the frog jumps every 10 seconds

(using digits from a random number table)

It requires 7 jumps with 2 in the reverse direction (either left or right) for the frog to get off the board in 60 seconds.
Alternatively, 3 jumps in the same direction will also lead to the frog being off the board.
Furthermore, it would take 5 jumps with one in the opposite direction within the time limit of 60 seconds to leave the board.

Step-by-step explanation:

A frog positioned right at the center of a 5ft long board is 2.5 ft away from either edge.

Every 10 seconds, the frog jumps left or right.

If the frog's jumps are LLRLRL, it will remain on the board at the leftmost square.

If it jumps as LLRLL, it will jump off the board after fifty seconds.

Given that the frog jumps every 10 seconds

(using digits from a random number table)

It requires 7 jumps with 2 in reverse direction (either left or right) for the frog to get off the board in 60 seconds.
Alternatively, 3 jumps in the same direction will also lead to the frog being off the board.
Furthermore, it would take 5 jumps with one in the opposite direction within the time limit of 60 seconds to leave the board.

4 0

1 month ago

A certain pen has been designed so that true average writing lifetime under controlled conditions (involving the use of awriting

tester [8842]

Answer:

a) Null hypothesis: $\mu \geq 10$

Alternative hypothesis: $\mu < 10$

b) $p_v =P(t_{17}$

Given that the p-value is lower than the significance threshold in this situation, we have sufficient grounds to reject the null hypothesis.

c) $p_v =P(t_{17}$

In this case, since the p-value exceeds the significance threshold, we have adequate evidence to FAIL to reject the null hypothesis.

d) $p_v =P(t_{17}$

Here again, with the p-value being less than the significance level, we can reject the null hypothesis.

Step-by-step explanation:

1) Provided data and references

$\bar X$ represents the average of the samples

$s$ denotes the standard deviation of the samples

$n=18$ indicates the number of samples

$\mu_o =10$ is the value we are examining

$\alpha$ defines the significance level for the test.

t represents the specific statistic of interest

$p_v$ indicates the p-value relevant to the test (the variable of concern)

Define the null and alternative hypotheses.

To assess if the true mean is at least 10 hours, we must set up a hypothesis:

Part a

Null hypothesis: $\mu \geq 10$

Alternative hypothesis: $\mu < 10$

If we consider the sample size being less than 30 and the population deviation unknown, it’s more appropriate to use a t-test to compare the actual mean with the reference value, calculated as:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Part b

In this scenario t=-2.3, $\alpha=0.05$

Initially, we need to calculate the degrees of freedom $df=n-1=18-1=17$

Since this is a left-tailed test, the p-value is determined by:

$p_v =P(t_{17}$

In this instance, with the p-value being less than the significance level, we have sufficient evidence to reject the null hypothesis.

Part c

For this situation t=-1.8, $\alpha=0.01$

We need to find the degrees of freedom $df=n-1=18-1=17$

For the left-tailed test, the p-value is given by:

$p_v =P(t_{17}$

In this case, since the p-value is above the significance level, we have enough grounds to FAIL to reject the null hypothesis.

Part d

For this case t=-3.6, $\alpha=0.05$

Firstly, we find the degrees of freedom $df=n-1=18-1=17$

Since we are conducting a left-tailed test, the p-value is calculated as:

$p_v =P(t_{17}$

Here, with the p-value being lower than the significance threshold, we can reject the null hypothesis.

5 0

29 days ago