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Masteriza [31]
3 years ago
14

Leo buy 5 video games for $60. at this rat, how much would he pay for 3 video games?

Mathematics
2 answers:
matrenka [14]3 years ago
7 0

Answer:36

Step-by-step explanation:

60÷5×3=36

Vadim26 [7]3 years ago
4 0
If 5 games equals $60. Divide those two and you get 12. Then you multiply 12 and 3. Your answer would be $36

60/5 =12 12x3=36
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A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the l
Katena32 [7]

Answer:

(a) The fraction of the calls last between 4.50 and 5.30 minutes is 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is 0.1271.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is 0.745.

(e) The time is 5.65 minutes.

Step-by-step explanation:

We are given that the mean length of time per call was 4.5 minutes and the standard deviation was 0.70 minutes.

Let X = <u><em>the length of the calls, in minutes.</em></u>

So, X ~ Normal(\mu=4.5,\sigma^{2} =0.70^{2})

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

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

where, \mu = population mean time = 4.5 minutes

           \sigma = standard deviation = 0.7 minutes

(a) The fraction of the calls last between 4.50 and 5.30 minutes is given by = P(4.50 min < X < 5.30 min) = P(X < 5.30 min) - P(X \leq 4.50 min)

    P(X < 5.30 min) = P( \frac{X-\mu}{\sigma} < \frac{5.30-4.5}{0.7} ) = P(Z < 1.14) = 0.8729

    P(X \leq 4.50 min) = P( \frac{X-\mu}{\sigma} \leq \frac{4.5-4.5}{0.7} ) = P(Z \leq 0) = 0.50

The above probability is calculated by looking at the value of x = 1.14 and x = 0 in the z table which has an area of 0.8729 and 0.50 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.8729 - 0.50 = <u>0.3729</u>.

(b) The fraction of the calls last more than 5.30 minutes is given by = P(X > 5.30 minutes)

    P(X > 5.30 min) = P( \frac{X-\mu}{\sigma} > \frac{5.30-4.5}{0.7} ) = P(Z > 1.14) = 1 - P(Z \leq 1.14)

                                                              = 1 - 0.8729 = <u>0.1271</u>

The above probability is calculated by looking at the value of x = 1.14 in the z table which has an area of 0.8729.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is given by = P(5.30 min < X < 6.00 min) = P(X < 6.00 min) - P(X \leq 5.30 min)

    P(X < 6.00 min) = P( \frac{X-\mu}{\sigma} < \frac{6-4.5}{0.7} ) = P(Z < 2.14) = 0.9838

    P(X \leq 5.30 min) = P( \frac{X-\mu}{\sigma} \leq \frac{5.30-4.5}{0.7} ) = P(Z \leq 1.14) = 0.8729

The above probability is calculated by looking at the value of x = 2.14 and x = 1.14 in the z table which has an area of 0.9838 and 0.8729 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.8729 = <u>0.1109</u>.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is given by = P(4.00 min < X < 6.00 min) = P(X < 6.00 min) - P(X \leq 4.00 min)

    P(X < 6.00 min) = P( \frac{X-\mu}{\sigma} < \frac{6-4.5}{0.7} ) = P(Z < 2.14) = 0.9838

    P(X \leq 4.00 min) = P( \frac{X-\mu}{\sigma} \leq \frac{4.0-4.5}{0.7} ) = P(Z \leq -0.71) = 1 - P(Z < 0.71)

                                                              = 1 - 0.7612 = 0.2388

The above probability is calculated by looking at the value of x = 2.14 and x = 0.71 in the z table which has an area of 0.9838 and 0.7612 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.2388 = <u>0.745</u>.

(e) We have to find the time that represents the length of the longest (in duration) 5 percent of the calls, that means;

            P(X > x) = 0.05            {where x is the required time}

            P( \frac{X-\mu}{\sigma} > \frac{x-4.5}{0.7} ) = 0.05

            P(Z > \frac{x-4.5}{0.7} ) = 0.05

Now, in the z table the critical value of x which represents the top 5% of the area is given as 1.645, that is;

                      \frac{x-4.5}{0.7}=1.645

                      {x-4.5}{}=1.645 \times 0.7

                       x = 4.5 + 1.15 = 5.65 minutes.

SO, the time is 5.65 minutes.

7 0
3 years ago
a rectangler mirror has a length of 9.5 inches and a width of 7.5 inches. what's is c the area of v the mirror
Alexeev081 [22]
The area of the mirror (v) is 9.5x7.5=71.25 inches
6 0
3 years ago
The cubic crystal company has a new crystal cube they want to sell. The packaging manager insists that the cubes be arranged to
Nat2105 [25]

Answer:

Following are the responses to the given question:

Step-by-step explanation:

height  \ \ \ \ \ \ \ Width    \ \ \ \ \ \ \ Length\\\\

   1 \ \ \ \ \ \ \  \ \ \ \ \ \ \ 1 \ \ \ \ \ \ \  \ \ \ \ \ \ \  12\\\\ 1   \ \ \ \ \ \ \ \ \ \ \ \ \ \  2 \ \ \ \ \ \ \  \ \ \ \ \ \ \  6\\\\ 1  \ \ \ \ \ \ \ \ \ \ \ \ \ \  3  \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4\\\\2  \ \ \ \ \ \ \ \ \ \ \ \ \ \  2   \ \ \ \ \ \ \ \ \ \ \ \ \ \  3\\\\

7 0
3 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bp%7D%7B5%7D" id="TexFormula1" title="\frac{p}{5}" alt="\frac{p}{5}" align="absmiddle
tatyana61 [14]

Answer:

p ≥ -10

Step-by-step explanation:

first, subtract one from each side to get:

p/5 ≥ -2

lastly, multiply each side by 5

p ≥ -10

6 0
2 years ago
Help me please..!!!!
wariber [46]
I think it is D but i am not too sure
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