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almond37 [142]
3 years ago
8

Jasmin, Quinn, and Maddy are rock climbing. Jasmin is 42 ½ feet below Quinn. Maddy is 67 ¾ feet above Jasmin. What is Maddy's po

sition compared to Quinn?

Mathematics
1 answer:
ladessa [460]3 years ago
8 0

Answer:

25\frac{1}{4} feet

<em />

Step-by-step explanation:

Given

Maddy = 67\frac{3}{4} above Jasmin

Jasmin = 42\frac{1}{2} below Quinn

Required

Determine Maddy's position relative to Quinn's

From the given parameter, one can easily deduce that Quinn is in between Maddy and Jasmin (See attachment)

Using the attachment as a point of reference;

z = 67\frac{3}{4}

x = 42\frac{1}{2}

The relationship between x, y and z is

z = y + x

Make y the subject of formula

y = z - x

Substitute the values of z and x

y = 67\frac{3}{4} - 42\frac{1}{2}

Convert to improper fractions

y = \frac{271}{4} - \frac{85}{2}

Take LCM

y = \frac{271 - 170}{4}

y = \frac{101}{4}

y = 25\frac{1}{4}

<em>Hence; Maddy's position relative to Quinn's is </em>25\frac{1}{4}<em> feet</em>

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xz_007 [3.2K]

Answer:

(a) $1,020 or more = 0.2358

(b) Between $880 and $1,130 = 0.7389

(c) Between $830 and $940 = 0.3524

(d) Less than $770 = 0.0294

Step-by-step explanation:

We are given that According to M/PF Research, Inc. report, the average cost of renting an apartment in Minneapolis is $951.

Suppose that the standard deviation of the cost of renting an apartment in Minneapolis is $96 and that apartment rents in Minneapolis are normally distributed.

<em>Let X = apartment rents in Minneapolis</em>

So, X ~ Normal(\mu=$951,\sigma^{2} =$96^{2})

The z score probability distribution for normal distribution is given by;

<em>                  </em>    Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where,  \mu = average cost of renting an apartment = $951

            \sigma = standard deviation = $96

(a) Probability that the price is $1,020 or more is given by = P(X \geq $1,020)

    P(X \geq $1,020) = P( \frac{X-\mu}{\sigma} \geq \frac{1,020-951}{96} ) = P(Z \geq 0.72) = 1 - P(Z < 0.72)

                                                               = 1 - 0.76424 = 0.2358

<em>The above probability is calculated by looking at the value of x = 0.72 in the z table which gives an area of 0.76424.</em>

<em />

(b) Probability that the price is between $880 and $1,130 is given by = P($880 < X < $1,130) = P(X < $1,130) - P(X \leq 880)

    P(X < $1,130) = P( \frac{X-\mu}{\sigma} < \frac{1,130-951}{96} ) = P(Z < 1.86) = 0.96856

     P(X \leq $880) = P( \frac{X-\mu}{\sigma} \leq \frac{880-951}{96} ) = P(Z \leq -0.74) = 1 - P(Z < 0.74)

                                                         = 1 - 0.77035 = 0.22965

<em>The above probability is calculated by looking at the value of x = 1.86 and x = 0.74 in the z table which gives an area of 0.96856 and 0.77035 respectively.</em>

<em />

Therefore, P($880 < X < $1,130) = 0.96856 - 0.22965 = 0.7389

<em />

(c) Probability that the price is between $830 and $940 is given by = P($830 < X < $940) = P(X < $940) - P(X \leq 830)

    P(X < $940) = P( \frac{X-\mu}{\sigma} < \frac{940-951}{96} ) = P(Z < -0.11) = 1 - P(Z \leq 0.11)

                                                          = 1 - 0.5438 = 0.4562

     P(X \leq $830) = P( \frac{X-\mu}{\sigma} \leq \frac{830-951}{96} ) = P(Z \leq -1.26) = 1 - P(Z < 1.26)

                                                         = 1 - 0.89617 = 0.10383

<em>The above probability is calculated by looking at the value of x = 0.11 and x = 1.26 in the z table which gives an area of 0.5438 and 0.89617 respectively.</em>

<em />

Therefore, P($830 < X < $940) = 0.4562 - 0.10383 = 0.3524

(d) Probability that the price is Less than $770 is given by = P(X < $770)

    P(X < $770) = P( \frac{X-\mu}{\sigma} < \frac{770-951}{96} ) = P(Z < -1.89) = 1 - P(Z \leq 1.89)

                                                         = 1 - 0.97062 = 0.0294

<em>The above probability is calculated by looking at the value of x = 1.89 in the z table which gives an area of 0.97062.</em>

8 0
2 years ago
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