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3 years ago

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Mathematics

1 answer:

Doss [256]3 years ago

3 0

Answer:

First point: (1,0)

Second Point: (2,1)

Third Point: (3,2)

Fourth Point: (8,7)

Step-by-step explanation:

For the first two points, you have to plot the two pairs of coordinates that are given. After this, you have to plug the x values into the equation to get the missing y values.

For example: The third point

X= 3

Y= X - 1

Y= 3 - 1

Y=2

So this means that the third coordinate is (3,2)

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Which equations represent linear functions? Select all that apply.

liraira [26]

Answer: x-3y=3x

8y=6x8

y=6x

Step-by-step explanation:

<h2>Meaning of linear equation</h2><h3>A linear equation is a equation that goes at a constant pace and won't change overtime, so functions like x squared and functions that multiply x and y together doesn't go at a constant pace, there not linear equations, but functions like y=mx, M representing any constant, that is a linear equation, therefore x-3y=3x</h3>

8y=6x8

y=6x

<h3>Are the answers to your Question,</h3>

4 0

3 years ago

In T-ball, the distance to each successive base is 50 feet. If the distance from home plate to the pitcher’s mound is 38 feet, h

JulijaS [17]

Answer:

<h2>The distance from the pitcher's mound and to second base is 37.99 approximately.</h2>

Step-by-step explanation:

The diamond is a square, which in this case has 50 feet long each side, and from home to pitcher is 38 feet. Notice that home is a vertex of the square and the pitcher's mound is the intersection of the diagonals, where they cut half.

We can find the distance from the pitcher to first base using Pythagorean's Theorem, where 50 feet is the hypothenuse.

$50^{2} =38^{2}+x^{2}\\x^{2}=50^{2}-38^{2}\\x=\sqrt{2500-1444}\\ x=\sqrt{1056}\\ x \approx 32.5 \ ft$

Therefore, the distance from the pitcher to first base is 32.5 feet, approximately.

Now, we can use again Pythagorean's Theorem to find the distance from pitcher to second base, where the hypothenuse is 50 feet.

$50^{2}=32.5^{2}+y^{2}\\y^{2}=50^{2}-32.5^{2}\\y=\sqrt{2500-1056.25}\\ y =\sqrt{1443} \approx 37.99$

Therefore, the distance from the pitcher's mound and to second base is 37.99 approximately.

<em>(this results make sense, because the diagonals of a square intersect at half, that means all bases have the same distance from pitcher's mound, so the second way to find the distance asked in the question is just using theory)</em>

8 0

3 years ago

About % of the area under the curve of the standard normal distribution is between z = − 0.9 z = - 0.9 and z = 0.9 z = 0.9 (or w

prisoha [69]

Using the normal distribution, it is found that 63.18% of the area under the curve of the standard normal distribution is between z = − 0.9 z = - 0.9.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The area within 0.9 standard deviations of the mean is the <u>p-value of Z = 0.9(0.8159) subtracted by the p-value of Z = -0.9(0.1841)</u>, hence:

0.8159 - 0.1841 = 0.6318 = 63.18%.

More can be learned about the normal distribution at brainly.com/question/4079902

#SPJ1

3 0

2 years ago

A water wheel has a radius measuring between 13 feet and 24 feet. The wheel is able to turn 7π9 radians from its starting positi

madam [21]

Answer:

31.76 ft and 58.64 ft

Step-by-step explanation:

The radius measures between 13 feet and 24 feet.

The wheel is able to turn 7π/9 radians before getting stuck.

We need to find the range of distances that the wheel could spin before getting stuck. That is, the length of arc.

Length of an arc is given as:

$L = \frac{\theta}{2\pi} * 2\pi r$

where θ = central angle = 7π/9 radians

r = radius of the circle

Therefore, for 13 feet:

$L = \frac{7\pi}{18 \pi} * 2 * \pi * 13\\\\L = 31.76 ft$

For 24 feet:

$L = \frac{7\pi}{18 \pi} * 2 * \pi * 24\\\\L = 58.64 ft$

The wheel could spin between 31.76 ft and 58.64 ft before getting stuck.

4 0

3 years ago

The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e

Viefleur [7K]

Answer:

The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.2087, 0.2507).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

Suppose a sample of 1537 tenth graders is drawn. Of the students sampled, 1184 read above the eighth grade level. So 1537 - 1184 = 353 read at or below this level. Then

$n = 1537, \pi = \frac{353}{1537} = 0.2297$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2297 - 1.96\sqrt{\frac{0.2297*0.7703}{1537}} = 0.2087$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2297 + 1.96\sqrt{\frac{0.2297*0.7703}{1537}} = 0.2507$

The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.2087, 0.2507).

7 0

3 years ago