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Anna35 [415]
3 years ago
7

If (a, –5) is a solution to the equation 3a = –2b – 7, what is a?

Mathematics
1 answer:
sukhopar [10]3 years ago
6 0

Answer:

d)  1

Step-by-step explanation:

3a = –2b – 7

We have the point (a, -5) so b = -5

Substituting in

3a = -2(-5) -7

3a = 10-7

3a = 3

Divide each side by 3

3a/3 = 3/3

a =1

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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
Simplify-2(y+4) + 3y
Pie

Answer: y-8

Step-by-step explanation:

8 0
2 years ago
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Se tiene la sucesión 3, 8, 13, 18, 23 … ¿Cuáles son los siguientes dos términos y cuál es la regla general?
V125BC [204]

Respuesta:

28; 33

A (n) = 5n - 2

Explicación paso a paso:

A partir de los datos dados, 3, 8, 13, 18, 23…, podemos ver que cada valor sucesivo de la serie aumenta en 5;

Por lo tanto, los siguientes dos términos de la serie deberían ser:

23 + 5 = 28

28 + 5 = 33

La serie es una progresión aritmética:

Recuerde la fórmula general:

A (norte) = a + (norte - 1) d

Donde, a = Primer término = 3; d = diferencia común = 5

n = enésimo término

A (n) = 3 + (n-1) 5

A (n) = 3 + 5n - 5

A (n) = 5n - 2

6 0
3 years ago
How much 2 1/4×7=? Explain​
Law Incorporation [45]

Answer:

3.5

Step-by-step explanation:

2x1/4=0.5

0.5x7=3.5

=3.5

7 0
3 years ago
Read 2 more answers
Over the course of the school year, you keep track of how much snow falls on a given day and whether it was a snow day. Your dat
Free_Kalibri [48]

Answer:

Probability of the day to be a snow day is 0.34

Step-by-step explanation:

We have the data,

Out of 21 which had less than 3 inches of snow, 5 were snow days.

Out of 8 days which had more than 3 inches of snow, 6 were snow days.

So, we get,

Total number of days = 21 + 8 = 29

Total number of snow days = 5 + 6 = 11

Thus, the probability that the day will be a snow day is \frac{11}{29} i.e. 0.34

Hence, the probability of the day to be a snow day is 0.34.

5 0
2 years ago
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