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jeka57 [31]
2 years ago
11

A box of 3 picture hangers sells for $2.22. What is the unit rate

Mathematics
1 answer:
Anestetic [448]2 years ago
7 0

Answer:

0.74 cents each unit piece

Step-by-step explanation:

2.22 ÷ 3 = 0.74

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What is 4to the power of 12
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-1 3/4+ 1 1/6 what is the answer I say -7/12
Bingel [31]

Answer:

No, the correct answer of -1\frac{3}{4}  + 1\frac{1}{6} = \frac{11}{12}

Step-by-step explanation:

The given expression is -1\frac{3}{4}  + 1\frac{1}{6}

As we know that a mixed fractiona\frac{b}{c} can be converted into proper or improper fraction as a\frac{b}{c}  = \frac{ac + b}{c}

⇒ -1\frac{3}{4}  = \frac{(-1) (4)+ 3}{c}  = \frac{-4 + 3}{4}  =  \frac{-1}{4}

and, 1\frac{1}{6}  = \frac{(1) (6)+ 1}{6}  = \frac{6 + 1}{6}  =  \frac{7}{6}

Hence, the simplified expression = \frac{-1}{4}  + \frac{7}{6}  = \frac{-1(3)  + 7(2)}{12}   = \frac{11}{12}

or, -1\frac{3}{4}  + 1\frac{1}{6} = \frac{11}{12}

6 0
2 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
2 years ago
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