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Lady bird [3.3K]
3 years ago
11

a country club offers membership to it's customers. there is a one-time application fee of $75 and the dues are $25 per month. w

rite an equation to represent the total cost of membership at this country club​
Mathematics
1 answer:
Lilit [14]3 years ago
3 0

Answer: 25x + 75

Step-by-step explanation:

The 25 is the monthly fee, x stands for the number of months and the 75 is the set fee

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Josh attempted 56 free throws during the basketball season. He made 57 of them. How many free throws did he make?
Scorpion4ik [409]

Answer:

He made approximately 32 free throws.

Step-by-step explanation:

Assuming that there's a typo in the question and that he made 57% of his attempted free throws, then we can solve it as shown below:

We can apply a rule of three in order to calculate the number of free throws he made. This is done as follows:

56 free throws -> 100%

x free throws -> 57%

56/x = 100/57

100*x = 57*56

x = 57*56/100 = 31.92

It can also be solved by transforming the percentage in a fraction such as 57% = 57/100 and then multiplying it by the total attempts.

free throws made = 56*57/100 = 31.92

He made approximately 32 free throws.

7 0
3 years ago
Solve for y:<br> y+5x=6;x= -1,0,1
AlekseyPX
When x = -1

y + 5x = 6

y = 6 - 5x

y = 6 - (5)(-1)

y = 6 - (-5)

y = 11

when x = 0

y = 6 - 5x

y = 6 - (5)(0)

y = 6 - 0

y = 6

when x = 1

y = 6 - 5x

y = 6 - (5)(1)

y = 6 - 5

y = 1
7 0
3 years ago
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Find three consecutive even numbers whose sum is 42.<br> ​
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Answer:chicken banana ckae

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2 years ago
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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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