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2 years ago

- Plot two points to graph the second equation.

Mathematics

1 answer:

denis-greek [22]2 years ago

6 0

Answer:

heyyyyyyy

Step-by-step explanation:

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Graph y = -½ x -3 mathematics

Dominik [7]

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3 years ago

Solve graphically please send me by solving and clicking photo plzzzz

Delicious77 [7]

Answer:

Subtract 5x from both sides of the equation.

7y=1−5x

x+4y=−5

Divide each term by 7 and simplify.

y=17−5x/7

x+4y=−5

Subtract x from both sides of the equation.

y=1/7−5x/7

4y=−5−x

y=1/7−5x/7

Divide each term by 4 and simplify.

y=1/7−5x/7

y=−5/4−x/4

y=1/7−5x/7

Create a graph to locate the intersection of the equations. The intersection of the system of equations is the solution. (3,−2)

Step-by-step explanation:

8 0

2 years ago

⦁ The data set shows the number of practice throws players in a basketball competition made and the number of free throws they m

barxatty [35]

Answer:

Kindly check explanation

Step-by-step explanation:

Given the data:

Practice throws(X) : 4 10 6 15 0 7 11

Free throws(Y) : 8 23 9 34 5 11 27

Using the online linear regression calculator :

ŷ = 2.1559X + 0.3912

2.1559 = slope

0.3912 = intercept

X = independent variable ; y = dependent variable

The Coefficient of determination (R²) = 0.9506² = 0.9036

Quadratic model : y = 0.0961x²+0.7138x+3.8031

Coefficient of determination(R²) = 0.9739² = 0.9485

Exponential model:

a*b^x

Coefficient of determination (R²) = 0.9753² = 0.9512

The exponential model fits the data best.

7 0

3 years ago

A random sample of n = 64 observations is drawn from a population with a mean equal to 20 and standard deviation equal to 16. (G

dezoksy [38]

Answer:

a) The mean of a sampling distribution of $\\ \overline{x}$ is $\\ \mu_{\overline{x}} = \mu = 20$ . The standard deviation is $\\ \frac{\sigma}{\sqrt{n}} = \frac{16}{\sqrt{64}}=2$ .

b) The standard normal z-score corresponding to a value of $\\ \overline{x} = 16$ is $\\ Z = -2$ .

c) The standard normal z-score corresponding to a value of $\\ \overline{x} = 23$ is $\\ Z = 1.5$ .

d) The probability $\\ P(\overline{x}$ .

e) The probability $\\ P(\overline{x}>23) = 1 - P(Z$ .

f) $\\ P(16 < \overline{x} < 23) = P(-2 < Z < 1.5) = P(Z$ .

Step-by-step explanation:

We are dealing here with the concept of a sampling distribution, that is, the distribution of the sample means $\\ \overline{x}$ .

We know that for this kind of distribution we need, at least, that the sample size must be $\\ n \geq 30$ observations, to establish that:

$\\ \overline{x} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

In words, the distribution of the sample means follows, approximately, a normal distribution with mean, $\mu$ , and standard deviation (called standard error), $\\ \frac{\sigma}{\sqrt{n}}$ .

The number of observations is n = 64.

We need also to remember that the random variable Z follows a standard normal distribution with $\\ \mu = 0$ and $\\ \sigma = 1$ .

$\\ Z \sim N(0, 1)$

The variable Z is

$\\ Z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$ [1]

With all this information, we can solve the questions.

Part a

The mean of a sampling distribution of $\\ \overline{x}$ is the population mean $\\ \mu = 20$ or $\\ \mu_{\overline{x}} = \mu = 20$ .

The standard deviation is the population standard deviation $\\ \sigma = 16$ divided by the root square of n, that is, the number of observations of the sample. Thus, $\\ \frac{\sigma}{\sqrt{n}} = \frac{16}{\sqrt{64}}=2$ .

Part b

We are dealing here with a random sample. The z-score for the sampling distribution of $\\ \overline{x}$ is given by [1]. Then

$\\ Z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ Z = \frac{16 - 20}{\frac{16}{\sqrt{64}}}$

$\\ Z = \frac{-4}{\frac{16}{8}}$

$\\ Z = \frac{-4}{2}$

$\\ Z = -2$

Then, the standard normal z-score corresponding to a value of $\\ \overline{x} = 16$ is $\\ Z = -2$ .

Part c

We can follow the same procedure as before. Then

$\\ Z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ Z = \frac{23 - 20}{\frac{16}{\sqrt{64}}}$

$\\ Z = \frac{3}{\frac{16}{8}}$

$\\ Z = \frac{3}{2}$

$\\ Z = 1.5$

As a result, the standard normal z-score corresponding to a value of $\\ \overline{x} = 23$ is $\\ Z = 1.5$ .

Part d

Since we know from [1] that the random variable follows a standard normal distribution, we can consult the cumulative standard normal table for the corresponding $\\ \overline{x}$ already calculated. This table is available in Statistics textbooks and on the Internet. We can also use statistical packages and even spreadsheets or calculators to find this probability.

The corresponding value is Z = -2, that is, it is two standard units below the mean (because of the negative value). Then, consulting the mentioned table, the corresponding cumulative probability for Z = -2 is $\\ P(Z$ .