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devlian [24]
2 years ago
15

Choose all selections that represent the domain of the relation, {(2,5)(3, 6)(2,8)}?

Mathematics
1 answer:
Murrr4er [49]2 years ago
8 0

Answer:

{2,3}

Step-by-step explanation:

Domain are the x-axis numbers ordered from least to greatest

Range is the y-axis numbers ordered from least to greatest

This is the symbol when you are ordering the range and domain { }

If a range or domain number is repeated, you only put it once

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Which quadratic function shows a y-intercept of 4 and a maximum value of 5
Elena L [17]

Answer:

The 4th graph

Step-by-step explanation:

To have a maximum value, your parabola would have to open downwards. So the first 2 are wrong. Since your y-intercept is positive 4, your graph would have to touch the y axis at 4. The 3rd graph is wrong since it touches the y-axis at a negative point. The correct answer would be the 4th answer choice.

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2 years ago
If the angles of a triangle are 10 45 and 23 what is the perimerter of the triangle
djverab [1.8K]

Answer:

No Solutions

Step-by-step explanation:

In a triangle, the sum of the angles has to be 180 degrees. It also is impossible to find the length of sides without at least one side, since the range of lengths is practically infinite. There are no solutions to this problem.

3 0
2 years ago
Suppose that you are in charge of evaluating teacher performance at a large elementary school. One tool you have for this evalua
Strike441 [17]

Answer:

a) Standard error = 2

b) Range = (76.08, 83.92)

c) P=0.69

d) Smaller

e) Greater

Step-by-step explanation:

a) When we have a sample taken out of the population, the standard error of the mean is calculated as:

\sigma_m=\dfrac{\sigma}{\sqrt{n}}=\dfrac{10}{\sqrt{25}}=\dfrac{10}{5}=2

where n is te sample size (n=25) and σ is the population standard deviation (σ=10).

Then, the standard error of the classroom average score is 2.

b) The calculations for this range are the same that for the confidence interval, with the difference that we know the population mean.

The population standard deviation is know and is σ=10.

The population mean is M=80.

The sample size is N=25.

The standard error of the mean is σM=2.

The z-value for a 95% confidence interval is z=1.96.

The margin of error (MOE) can be calculated as:

MOE=z\cdot \sigma_M=1.96 \cdot 2=3.92

Then, the lower and upper bounds of the confidence interval are:

LL=M-t \cdot s_M = 80-3.92=76.08\\\\UL=M+t \cdot s_M = 80+3.92=83.92

The range that we expect the average classroom test score to fall 95% of the time is (76.08, 83.92).

c) We can calculate this by calculating the z-score of X=79.

z=\dfrac{X-\mu}{\sigma}=\dfrac{79-80}{2}=\dfrac{-1}{2}=-0.5

Then, the probability of getting a average score of 79 or higher is:

P(X>79)=P(z>-0.5)=0.69146

The approximate probability that a classroom will have an average test score of 79 or higher is 0.69.

d) If the sample is smaller, the standard error is bigger (as the square root of the sample size is in the denominator), so the spread of the probability distribution is more. This results then in a smaller probability for any range.

e) If the population standard deviation is smaller, the standard error for the sample (the classroom) become smaller too. This means that the values are more concentrated around the mean (less spread). This results in a higher probability for every range that include the mean.

6 0
3 years ago
A rose garden Is formed by jolning a rectangle and a semicircle, as shown below. The rectangle Is 23 ft long and 14 ft wide.Find
Ratling [72]

Answer:

Area of the garden:

\begin{equation*} 398.93\text{ ft}^2 \end{equation*}

Explanation:

Given the below parameters;

Length of the rectangle(l) = 23 ft

Width of the rectangle(w) = 14 ft

Value of pi = 3.14

Since the width of the rectangle is 14 ft, so the diameter(d) of the semicircle is also 14 ft.

The radius(r) of the semicircle will now be;

r=\frac{d}{2}=\frac{14}{2}=7\text{ ft}

Let's now go ahead and determine the area of the semicircle using the below formula;

A_{sc}=\frac{\pi r^2}{2}=\frac{3.14*\left(7\right)^2}{2}=\frac{3.14*49}{2}=\frac{153.86}{2}=76.93\text{ ft}^2

Let's also determine the area of the rectangle;

A_r=l*w=23*14=322\text{ ft}^2

We can now determine the area of the garden by adding the area of the semicircle and that of the rectangle together;

\begin{gathered} Area\text{ of the garden = Area of semi circle + Area of rectangle } \\ =76.93+322 \\ =398.93\text{ ft}^2 \end{gathered}

Therefore, the area of the garden is 398.93 ft^2

8 0
1 year ago
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