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Vsevolod [243]
3 years ago
5

Valerie sold six tickets to the school play and Mark sold 16 tickets. What is the ratio of the number of tickets Valerie sold to

the number of tickets Mark sold? No I'm
Mathematics
1 answer:
Alenkasestr [34]3 years ago
4 0
The ratio would be 6:22 because they said whats the ratio for valire to the number of tickets mean inall so 6:22
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A mark up that is less than 1% give the original price and final price after the markdown
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3 years ago
M/PF Research, Inc. lists the average monthly apartment rent in some of the most expensive apartment rental locations in the Uni
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
3 years ago
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