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Tcecarenko [31]
2 years ago
10

Can someone please help me with math.

Mathematics
2 answers:
Rus_ich [418]2 years ago
7 0

Answer:

1. 10 pigs and 12 geese

2. 20 and 80

2. pretzel would be 3 dollars

Step-by-step explanation:

i had the same question and got it right

lorasvet [3.4K]2 years ago
5 0

Answer:

Step-by-step explanation:

3) Number of pigs = p

Number of geese = g

 p + g = 20  --------------(I)

       p = 20 - g --------------(II)

4p + 2g = 64 ----------------------(III)

Substitute  p  = 20 - g in (III)

4*(20 - g) + 2g = 64

4*20 - 4*g + 2g = 64

80 - 4g + 2g = 64   {Combine like terms}

80 - 2g = 64        {Subtract 80 form both sides}

      -2g = 64 - 80

      -2g = - 16  {Divide both sides by (-2)}

         g = -16/-2

g = 8

Plugin g = 8 in equation (II)

p = 20 - 8

P = 12

Number of pigs = 12

4) Let the number be x , y

x + y = 100       ---------------(I)

x = 4y  --------------(II)

Substitute x = 4y in equation (I)

4y + y = 100

      5y = 100

        y = 100/5

y = 20

Plugin y = 20 in equation (II)

x = 4*20 = 80

x = 80

Smaller number = 20

5) Cost of soda = s

Cost of pretzel = p

2s + p = 7 --------------(I)

       p = 7 - 2s -------------(II)

4s + 3p = 17     --------------(III)

Substitute p = 7 - 2s in (III)

4s + 3*(7 - 2s) = 17

4s + 3*7 - 3*2s = 17

4s + 21 - 6s = 17

21 - 2s = 17

      -2s = 17 -21

       -2s = -4

          s = -4/-2

s = 2

Plugin s = 2 in (II)

p = 7 - 2*2

  = 7 - 4

p = 3

Cost of soda = $ 2

Cost of pretzel = $ 3

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Step-by-step explanation:

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Let z denote a random variable that has a standard normal distribution. Determine each of the probabilities below. (Round all an
Gelneren [198K]

Answer:

(a) P (<em>Z</em> < 2.36) = 0.9909                    (b) P (<em>Z</em> > 2.36) = 0.0091

(c) P (<em>Z</em> < -1.22) = 0.1112                      (d) P (1.13 < <em>Z</em> > 3.35)  = 0.1288

(e) P (-0.77< <em>Z</em> > -0.55)  = 0.0705       (f) P (<em>Z</em> > 3) = 0.0014

(g) P (<em>Z</em> > -3.28) = 0.9995                   (h) P (<em>Z</em> < 4.98) = 0.9999.

Step-by-step explanation:

Let us consider a random variable, X \sim N (\mu, \sigma^{2}), then Z=\frac{X-\mu}{\sigma}, is a standard normal variate with mean, E (<em>Z</em>) = 0 and Var (<em>Z</em>) = 1. That is, Z \sim N (0, 1).

In statistics, a standardized score is the number of standard deviations an observation or data point is above the mean.  The <em>z</em>-scores are standardized scores.

The distribution of these <em>z</em>-scores is known as the standard normal distribution.

(a)

Compute the value of P (<em>Z</em> < 2.36) as follows:

P (<em>Z</em> < 2.36) = 0.99086

                   ≈ 0.9909

Thus, the value of P (<em>Z</em> < 2.36) is 0.9909.

(b)

Compute the value of P (<em>Z</em> > 2.36) as follows:

P (<em>Z</em> > 2.36) = 1 - P (<em>Z</em> < 2.36)

                   = 1 - 0.99086

                   = 0.00914

                   ≈ 0.0091

Thus, the value of P (<em>Z</em> > 2.36) is 0.0091.

(c)

Compute the value of P (<em>Z</em> < -1.22) as follows:

P (<em>Z</em> < -1.22) = 0.11123

                   ≈ 0.1112

Thus, the value of P (<em>Z</em> < -1.22) is 0.1112.

(d)

Compute the value of P (1.13 < <em>Z</em> > 3.35) as follows:

P (1.13 < <em>Z</em> > 3.35) = P (<em>Z</em> < 3.35) - P (<em>Z</em> < 1.13)

                            = 0.99960 - 0.87076

                            = 0.12884

                            ≈ 0.1288

Thus, the value of P (1.13 < <em>Z</em> > 3.35)  is 0.1288.

(e)

Compute the value of P (-0.77< <em>Z</em> > -0.55) as follows:

P (-0.77< <em>Z</em> > -0.55) = P (<em>Z</em> < -0.55) - P (<em>Z</em> < -0.77)

                                = 0.29116 - 0.22065

                                = 0.07051

                                ≈ 0.0705

Thus, the value of P (-0.77< <em>Z</em> > -0.55)  is 0.0705.

(f)

Compute the value of P (<em>Z</em> > 3) as follows:

P (<em>Z</em> > 3) = 1 - P (<em>Z</em> < 3)

             = 1 - 0.99865

             = 0.00135

             ≈ 0.0014

Thus, the value of P (<em>Z</em> > 3) is 0.0014.

(g)

Compute the value of P (<em>Z</em> > -3.28) as follows:

P (<em>Z</em> > -3.28) = P (<em>Z</em> < 3.28)

                    = 0.99948

                    ≈ 0.9995

Thus, the value of P (<em>Z</em> > -3.28) is 0.9995.

(h)

Compute the value of P (<em>Z</em> < 4.98) as follows:

P (<em>Z</em> < 4.98) = 0.99999

                   ≈ 0.9999

Thus, the value of P (<em>Z</em> < 4.98) is 0.9999.

**Use the <em>z</em>-table for the probabilities.

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Maya wrote the following matrix equation to represent a system of equations:
Ne4ueva [31]
I am somehow confused with the given matrix. Because usually, when the system of equations is given as:

ax + by = c
dx + ey = f

the matrix would be written in this manner:

\left[\begin{array}{ccc}a&b&c\\d&e&f\end{array}\right]

Let me just assume that the arrangement of the given corresponds to this system of equations:

4x + 8y = 0
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So you write it as 

\left[\begin{array}{ccc}4&8&0\\2&5&6\end{array}\right]

In order to solve this, the matrix should look like this where x and y are the answers

\left[\begin{array}{ccc}1&0&x\\0&1&y\end{array}\right]


So, the first thing to do is apply this pattern: Row 2 = Row 1 - 2*Row 2. The result will be

\left[\begin{array}{ccc}4&8&0\\0&-2&-12\end{array}\right]

Next, Row 1 = (Row 1/4) + R2:

\left[\begin{array}{ccc}1&0&-12\\0&-2&-12\end{array}\right]

Lastly, Row 2 = Row 2 / -2:

\left[\begin{array}{ccc}1&0&-12\\0&1&6\end{array}\right]

Thus, the answer is x=-12 and y=6.
7 0
3 years ago
Can i please get help with this question?
KATRIN_1 [288]

Answer:

Not proportional.

Step-by-step explanation:

A proportional relationship is a relationship which crosses through the origin (0,0) and which has a proportional constant. We can determine this either by finding (0,0) where x=0 and y=0 in the table or by dividing y/x. None of the tables contain (0,0) so we will divide y by x. We are looking for a table which when each y is divided by its x we have the same constant appearing.

\frac{2}{1} \neq \frac{3}{2} \neq \frac{4}{3}

Though it has a point at the origin, these fractions are not equal. This is not proportional.

4 0
3 years ago
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