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nadezda [96]
2 years ago
12

What is the difference of the fractions? Use the number line to help find the answer. 1 3 5 4 -1 ok -12 -10 8 20 20 000 0 10 12

14 16 18 20 O 19 20 о O 11 20 O 20 19 20​

Mathematics
1 answer:
Nookie1986 [14]2 years ago
3 0

Answer: -11/20

Step-by-step explanation:

1/5-3/4= 4/20-15/20= -11/20

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A box was 3/4 full of marbles. The box fell on the floor. Thirty marbles fell out. This was 20% of the marbles in the box. How m
Hunter-Best [27]

Answer: -5.7

Step-by-step explanation:

3/4- 30= -29.25 and that was only 20% of the box that was dropped out of the box. -29.25 of 20% equals -5.85. 3/4 of 20% equals 0.15 subtract the 2 decimals and you will get -5.7

7 0
3 years ago
Determine the equation of the circle graphed below. <br><br> ( please help me )
Brums [2.3K]

Answer:

Equation : \\(x+3)^2 + ( y -2)^2 = 36

Step-by-step explanation:

For standard form the circle's equation we need the centre of the circle and the radius.

Step 1: <em><u>Find the centre</u></em>

If the centre is not given find the end points of the diameter

and then find the mid point.

Let the end points of the diameter be : ( - 3 , 8 ) and ( -3 , -4 )

The mid-point of the diameter is :

                        Mid-point = (\frac{-3 + - 3}{2}, \frac{-4+8}{2}) =  (-3, 2)

Therefore, centre of the circle = ( -3 , 2 )

Step 2 : <u>Find radius</u>

<u></u>Radius = \frac{Diameter }{2}<u></u>

Diameter is the distance between the end points ( -3 , 8) and ( -3 , -4 )

That is ,

 Diameter = \sqrt{(-3-(-3))^2 + ( -4 -8)^2}\\

                 = \sqrt{(-3 + 3)^2 + (-12)^2}\\\\=\sqrt{0 + 144}\\\\=12

Therefore ,

       Radius = \frac{12}{2} = 6

Step 3 : <u>Equation of the circle</u>

Standard equation of the circle with centre ( h ,k )

and radius ,r is :

       (x - h)^2+(y -k)^2 = r^2

Therefore, the equation of the circle with centre ( -3, 2)

and radius = 6 is :

    (x - (-3))^2 +  (y - 2)^2 = 6^2\\\\(x + 3)^2 + (y - 2)^2 = 36

5 0
2 years ago
Given point A at(3, 2) and B at (4,1). Find the slope of line segment AB.
Vladimir79 [104]

Answer:

-1

Step-by-step explanation:

4-3/1-2=slope

1/-1

-1

8 0
3 years ago
What is 4x² / 4x² ? Is the answer still a polynomial?
seraphim [82]

Answer:

4x² / 4x² = 1

Step-by-step explanation:

Anything divided by itself is always 1

4 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 <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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