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Sladkaya [172]
3 years ago
10

CAN SOMEONE HELP ME WITH THIS PLEASE AND THANK YOU

Mathematics
2 answers:
sammy [17]3 years ago
5 0
For this do 5/8-1/2!! in order to do this, find the common denominator. this one is 16, so do: 10/16-8/16, and simplify!
Komok [63]3 years ago
4 0

Answer:

She used 4/6 more fabric

Step-by-step explanation:

The equation asked how many more so you add how many more you need but you don't know what it is yet, so your next solution is to subtract 5/8 from 1/2 which is 4/6. In conclusion Salma used 4/6 more red fabric to complete the Shirt.

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Paha777 [63]
This is where you give up
5 0
3 years ago
Express 160 as product of their prime factor
MariettaO [177]

Answer: 5x2^5

Step-by-step explanation: Divide 160 by 2 as many times as you can to get 80x2, 40x2x2, 20x2x2x2, 10x2x2x2x2, 5x2x2x2x2x2. Then re-write your answer so that it is 5x2^5.

4 0
3 years ago
Which expression could help you find the distance between (10, 4) and (–6, 4)? A. |10| + |–4| B. |10| + |4| C. |–6| + |4| D. |10
bazaltina [42]
It would be C because |-6| + |4| is basically 6+4 because the lines make the number positive. 6+4 is 10, which is the x value for the point. That 4 unites us the distance between the points.
5 0
3 years ago
X and y are positive integers. explain why incorrect to claim that
dedylja [7]

You can find counterexamples to disprove this claim. We have positive integers that are perfect square numbers; when we take the square root of those numbers, we get an integer.

For example, the square root of 1 is 1, which is an integer. So if y = 1, then the denominator becomes an integer and thus we get a quotient of two integers (since x is also defined to be an integer), the definition of a rational number.

Example: x = 2, y = 1 ends up with \frac{2}{\sqrt{1}} = \frac{2}{1} which is rational. This goes against the claim that x/\sqrt{y} is always irrational for positive integers x and y.

Any integer y that is a perfect square will work to disprove this claim, e.g. y = 1, y = 4, y= 9, y = 16. So it is not always irrational.

4 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 <em>a sampling distribution</em>, 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 <em>normal distribution</em> with mean, \mu, and standard deviation (called <em>standard error</em>), \\ \frac{\sigma}{\sqrt{n}}.

The number of observations is n = 64.

We need also to remember that the random variable Z follows a <em>standard normal distribution</em> 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 <em>random sample</em>. 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 <em>standard normal z-score</em> 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 <em>standard normal z-score</em> 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 <em>standard normal distribution</em>, we can consult the <em>cumulative standard normal table</em> 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 <em>two standard units</em> <em>below</em> the mean (because of the <em>negative</em> 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 <em>cumulative standard normal table</em>, 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 <em>standardized values</em> for them, we have from <em>Part d</em>:

\\ P(\overline{x}

We know from <em>Part e</em> 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

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

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

5 0
3 years ago
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