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3 years ago

8

In a mid-size company in Philadelphia, the distribution of the number of phone calls answered each day by each of the 12 recepti

onists is bell-shaped and has a mean of 57 and a standard deviation of 7. Using the empirical (68-95-99.7) rule, what is the approximate percentage of daily phone calls numbering between 50 and 64

1 answer:

alexandr1967 [171]3 years ago

8 0

Answer:

Approximately 68% of daily phone calls will be in this interval.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 57

Standard deviation = 7

What is the approximate percentage of daily phone calls numbering between 50 and 64

50 = 57 - 7

So 50 is one standard deviation below the mean

64 = 57 + 7

So 64 is one standard deviation above the mean

By the Empirical Rule, approximately 68% of daily phone calls will be in this interval.

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On the first day it was posted online, a music video got 390 views. The number of

aleksley [76]

Answer:

The answer is 7212 views

Step-by-step explanation:

<h3><u>Given</u>;</h3>

A music video got 390 views.
The number of views that the video got each day increased by 30% per day.

<h3><u>To Find</u>;</h3>

Total views did the video get over the course of the first 13 days

So, For the first day the number of views is 390

Second Day

30% increased of total views

390 × 30 ÷ 100 = 11.7 = 12

390 + 12 = 402

Third Day

30% increased of total views

402 × 30 ÷ 100 = 120.6 = 121

402 + 121 = 523

Fourth Day

30% increased of total views

523 × 30 ÷ 100 = 156.9 = 157

523 + 157 = 680

Fifth Day

30% increased of total views

680 × 30 ÷ 100 = 204

680 + 204 = 884

Sixth Day

30% increased of total views

884 × 30 ÷ 100 = 265.2 = 265

884 + 265 = 1149

Seventh Day

30% increased of total views

1149 × 30 ÷ 100 = 344.7 = 345

1149 + 345 = 1494

Eight Day

30% increased of total views

1494 × 30 ÷ 100 = 448.2 = 448

1494 + 448 = 1942

Ninth Day

30% increased of total views

1942 × 30 ÷ 100 = 582.6 = 583

1942 + 583 = 2525

Tenth Day

30% increased of total views

2525 × 30 ÷ 100 = 757.5 = 758

2525 + 758 = 3283

Eleventh Day

30% increased of total views

3283 × 30 ÷ 100 = 984.9 = 985

3283 + 985 = 4268

Twelth Day

30% increased of total views

4268 × 30 ÷ 100 = 1280.4 = 1280

4268 + 1280 = 5548

Thirteenth Day

30% increased of total views

5548 × 30 ÷ 100 = 1664.4 = 1664

5548 + 1664 = 7212

Thus, Total views did the video get over the course of the first 13 days is 7212 Views.

<u>-TheUnknownScientist 72</u>

6 0

3 years ago

Hershey’s is making pumpkin-shaped chocolates for Halloween. These are spherically formed with a radius of 2 cm. What is the min

Lady_Fox [76]

Answer:

Sa = 1.5*V

Step-by-step explanation:

We have that the surface area of a sphere has the following formula:

Sa = 4*pi*r^2

we know that r = 2, so if we replace it, we're left:

Sa = 4*pi*(2^2)= = 4*pi*4

Sa = 16*pi

Now, the volume has the following formula:

V = 4/3*pi* r^3

Replacing it is our job:

V = 4/3*pi* (2^3) = 4/3*pi*8

V = 32/3*pi

Now, Sa/V, it would remain:

Sa/V = 16*pi / 32/3*pi

Sa/V = 48/32 = 1.5

Therefore, Sa = 1.5*V

4 0

3 years ago

Pleasantburg has a population growth model of P(t)=at2+bt+P0 where P0 is the initial population. Suppose that the future populat

yulyashka [42]

Answer:

The population will reach 34,200 in February of 2146.

Step-by-step explanation:

Population in t years after 2012 is given by:

$P(t) = 0.8t^{2} + 6t + 19000$

In what month and year will the population reach 34,200?

We have to find t for which P(t) = 34200. So

$P(t) = 0.8t^{2} + 6t + 19000$

$0.8t^{2} + 6t + 19000 = 34200$

$0.8t^{2} + 6t - 15200 = 0$

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

In this question:

$0.8t^{2} + 6t - 15200 = 0$

So $a = 0.8, b = 6, c = -15200$

Then

$\bigtriangleup = 6^{2} - 4*0.8*(-15100) = 48356$

$t_{1} = \frac{-6 + \sqrt{48356}}{2*0.8} = 134.14$

$t_{2} = \frac{-6 - \sqrt{48356}}{2*0.8} = -141.64$

We only take the positive value.

134 years after 2012.

.14 of an year is 0.14*365 = 51.1. The 51st day of a year happens in February.

So the population will reach 34,200 in February of 2146.

6 0

3 years ago

You want to put a fence around your field. The cost of fencing is $60 per meter. The

Darina [25.2K]

Cost per meter(C) = $60/m
Length of rectangular field(L) = 50m
L = 2W
-> W = L/2
-> W = 50/2
-> Width of rectangular field = 25m
Cost of one field length(l) = L x C
-> l = 50 x 60
-> l = $3000
Two of the lengths of the field = 2 x l
-> 2 x $3000
-> $6000

Cost of one field width(w) = W x C
-> w = 25 x 60
-> w = $1500
Two of the widths of the field = 2 x w
-> 2 x $1500
-> $3000

Cost of fencing entire field = $6000+$3000
Hence, total field cost = $9000

3 0

3 years ago

Tesla Battery Recharge Time.The electric vehicle manufacturing company Tesla estimates that a driver who commutes 50 miles per

madam [21]

Answer:

a.) f(x) = $\frac{1}{30}$ where 90 < x < 120

b.) $\frac{2}{3}$

c.) $\frac{2}{3}$

d.) $\frac{1}{2}$

Step-by-step explanation:

Let

X be a uniform random variable that denotes the actual charging time of battery.

Given that, the actual recharging time required is uniformly distributed between 90 and 120 minutes.

⇒X ≈ ∪ ( 90, 120 )

a.)

Probability density function , f (x) = $\frac{1}{120 - 90} = \frac{1}{30}$ where 90 < x < 120

b.)

P(x < 110) = $\int\limits^{110}_{90} {\frac{1}{30} } \, dx$

= $\frac{1}{30}[x]\limits^{110}_{90} = \frac{1}{30} [ 110 - 90 ] = \frac{1}{30} [ 20] = \frac{2}{3}$

c.)

P(x > 100 ) = $\int\limits^{120}_{100} {\frac{1}{30} } \, dx$

= $\frac{1}{30}[x]\limits^{120}_{100} = \frac{1}{30} [ 120 - 100 ] = \frac{1}{30} [ 20] = \frac{2}{3}$

d.)

P(95 < x< 110) = $\int\limits^{110}_{95} {\frac{1}{30} } \, dx$

= $\frac{1}{30}[x]\limits^{110}_{95} = \frac{1}{30} [ 110 - 95 ] = \frac{1}{30} [ 15] = \frac{1}{2}$

7 0

3 years ago