All categories

3 years ago

The equation of a straight line is 5y=7x+3 what is the gradient of the line?

Mathematics

1 answer:

Mrac [35]3 years ago

3 0

Answer:

7 over 5

Step-by-step explanation:

i believe this might be correct

You might be interested in

What is (64x^2y^3)-1/2 as radical expression?

oksano4ka [1.4K]

Answer:

$\frac{1}{8xy \sqrt{y} }$

Step-by-step explanation:

$(64 {x}^{2} {y}^{3} )^{ - \frac{ 1 }{2} } \\ \\ = \frac{1}{(64 {x}^{2} {y}^{3} )^{\frac{ 1 }{2} } } \\ \\ = \frac{1}{ \sqrt{64 {x}^{2} {y}^{3}} } \\ \\ = \frac{1}{ \sqrt{ {(8xy)}^{2} y} } \\ \\ = \frac{1}{8xy \sqrt{y} }$

3 0

3 years ago

Equation by elimination -4x-9y=20 -2x-9y=-24

storchak [24]

I'm not 100% sure this is right but I think this is how you do it

5 0

3 years ago

A system of linear inequalities is shown below:

Yuki888 [10]

The 4th selection is appropriate:
The first line is a dashed line which joins ordered pairs (-2, 1) and (3, 1). The second line is a dashed line and joins ordered pairs (-2, -2) and (3, 3). The portion common above the first line and above the second line is shaded

6 0

3 years ago

suppose you are trying to increase your coin collection to at least 500 coins. How many coins do you need if you already have a

Contact [7]

You will need to add 125 coins to get 500 coins from 375.

6 0

3 years ago

Read 2 more answers

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 $\\ P(Z$ .