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3 years ago

Solve the equation x2 = 80. X =

Mathematics

1 answer:

Tasya [4]3 years ago

8 0

Answer:

x=40

Step-by-step explanation:

hope i helped

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PLZ HALP!!!!! the length of a rectange is 1 7/9 inches, and its width is 3/4 of its length. find the area of this rectangle.

11Alexandr11 [23.1K]

9514 1404 393

Answer:

2 10/27 square inches

Step-by-step explanation:

The length is 1 7/9 = 16/9 inches. The width is 3/4 of that, so is ...

(3/4)(16/9) = 4/3 inches

The area is the product of length and width, so is ...

A = (16/9 in)(4/3 in) = 64/27 in²

The area of the rectangle is 2 10/27 square inches.

7 0

3 years ago

Hey please sorry its three questions thank you.

Karo-lina-s [1.5K]

Bacd the ancestors

8 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 $\\ P(Z$ .

Therefore, the probability $\\ P(\overline{x}$ .

Part e

We can follow a similar way than the previous step.

$\\ P(\overline{x} > 23) = P(Z > 1.5)$

For $\\ P(Z > 1.5)$ using the cumulative standard normal table, we can find this probability knowing that

$\\ P(Z1.5) = 1$

$\\ P(Z>1.5) = 1 - P(Z$

Thus

$\\ P(Z>1.5) = 1 - 0.9332$

$\\ P(Z>1.5) = 0.0668$

Therefore, the probability $\\ P(\overline{x}>23) = 1 - P(Z$ .

Part f

This probability is $\\ P(\overline{x} > 16)$ and $\\ P(\overline{x} < 23)$ .

For finding this, we need to subtract the cumulative probabilities for $\\ P(\overline{x} < 16)$ and $\\ P(\overline{x} < 23)$

Using the previous standardized values for them, we have from Part d:

$\\ P(\overline{x}$

We know from Part e that

$\\ P(\overline{x} > 23) = P(Z>1.5) = 1 - P(Z$

$\\ P(\overline{x} < 23) = P(Z1.5)$

$\\ P(\overline{x} < 23) = P(Z$

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

5 0

2 years ago

Is -5 a solution to the equation 6x + 5 = 5x + 2x? Explain?

poizon [28]

Answer:

no it is 5

Step-by-step explanation:

Solve for x by simplifying both sides of the equation, then isolating the variable. x = 5

8 0

2 years ago

Read 2 more answers

Please solve each of the following by factoring!

Oxana [17]

Answer:

see explanation

Step-by-step explanation:

(a)

x² - 36 = 0 ← is a difference of squares and factors as

(x - 6)(x + 6) = 0

equate each factor to zero and solve for x

x + 6 = 0 ⇒ x = - 6

x - 6 = 0 ⇒ x = 6

(b)

x² - 5x + 4 = 0

consider the product of the factors of the constant term (+ 4) which sum to give the coefficient of the x- term (- 5)

the factors are - 1 and - 4 , since

- 1 × - 4 = + 4 and - 1 - 4 = - 5 , then

(x - 1)(x - 4) = 0 ← in factored form

equate each factor to zero and solve for x

x - 1 = 0 ⇒ x = 1

x - 4 = 0 ⇒ x = 4

(c)

x² - 2x = 3 ( subtract 3 from both sides )

x² - 2x - 3 = 0 ← in standard form

consider the product of the factors of the constant term (- 3) which sum to give the coefficient of the x- term (- 2)

the factors are + 1 and - 3 , since

1 × - 3 = - 3 and 1 - 3 = - 2 , then

(x + 1)(x - 3) = 0 ← in factored form

equate each factor to zero and solve for x

x + 1 = 0 ⇒ x = - 1

x - 3 = 0 ⇒ x = 3

(d)

6x² - 11x - 10 = 0

consider the product of the factors of the coefficient of the x² term and the constant term which sum to give the coefficient of the x- term.

product = 6 × - 10 = - 60 and sum = - 11

the factors are + 4 and - 15

use these factors to split the x- term

6x² + 4x - 15x - 10 = 0 ( factor the first/second and third/fourth terms )

2x(3x + 2) - 5(3x + 2) = 0 ← factor out (3x + 2) from each term

(3x + 2)(2x - 5) = 0

equate each factor to zero and solve for x

3x + 2 = 0 ⇒ 3x = - 2 ⇒ x = - $\frac{2}{3}$

2x - 5 = 0 ⇒ 2x = 5 ⇒ x = $\frac{5}{2}$

6 0

1 year ago