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inn [45]
2 years ago
9

Claudia claims that there is one operation in the algebraic equation above. Joanna claims that there are two. Who is correct? Ex

plain.
17 - r4= 80
Mathematics
2 answers:
andrey2020 [161]2 years ago
8 0
The answer is NOT 672727
aliina [53]2 years ago
4 0

Answer:

There are only 4 operations. Joanna is incorrect and Claudia is correct.

Hope this helps :)

-jp524

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False.
y= 3(1)+1= 4
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what TRANSFORMATIONS take place to the parent graph f(x)= log(x) to achieve the graph g(x)=log(-2x-4)+5?
Leona [35]

The -4 will translate the graph of f(x) 4 units to the right.
Then the -2 before the x will stretch it vertically with factor 2, then reflect it in the y -axis.
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Explain how to use the distributive property to find an expression that is equivalent to 20 + 16​
IceJOKER [234]

Answer:

What you need to do is to find a multiple of 20 and 16 in other words a number that can multiply into 20 and 16. You have 2 and you have 4. So now this is how it looks like.

Either:

4(5+4) Or

2(10+8)

Step-by-step explanation:

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3 years ago
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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Step-by-step explanation:

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