1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Nat2105 [25]
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
Mathematics
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.

You might be interested in
Simplify this expression 10b+7-3b-2
iren2701 [21]

Answer:

7b+5

Step-by-step explanation:

10b + 7 − 3b −2

=10b + 7 + −3b + −2

Combine Like Terms:

=10b + <u>7 </u>+ −3b + <u>−2 </u>

=(10b + −3b) + ( <u>7 </u>+ <u>−2</u>)

=7b + <u>5</u>

Answer:

7b + 5

Hope this helps : )

3 0
3 years ago
Read 2 more answers
Don operates a taxi business, and this year one of his taxis was damaged in a traffic accident. The taxi was originally purchase
scoray [572]
I believe the answer is c
3 0
3 years ago
Please help me to prove this!​
Ymorist [56]

Answer:  see proof below

<u>Step-by-step explanation:</u>

Given: A + B + C = π              → A + B = π - C

                                              → B + C = π - A

                                              → C + A = π - B

                                              → C = π - (B +  C)

Use Sum to Product Identity:  cos A + cos B = 2 cos [(A + B)/2] · cos [(A - B)/2]

Use the Sum/Difference Identity: cos (A - B) = cos A · cos B + sin A · sin B

Use the Double Angle Identity: sin 2A = 2 sin A · cos A

Use the Cofunction Identity: cos (π/2 - A) = sin A

<u>Proof LHS → Middle:</u>

\text{LHS:}\qquad \qquad \cos \bigg(\dfrac{A}{2}\bigg)+\cos \bigg(\dfrac{B}{2}\bigg)+\cos \bigg(\dfrac{C}{2}\bigg)

\text{Sum to Product:}\qquad 2\cos \bigg(\dfrac{\frac{A}{2}+\frac{B}{2}}{2}\bigg)\cdot \cos \bigg(\dfrac{\frac{A}{2}-\frac{B}{2}}{2}\bigg)+\cos \bigg(\dfrac{C}{2}\bigg)\\\\\\.\qquad \qquad \qquad \qquad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\cos \bigg(\dfrac{C}{2}\bigg)

\text{Given:}\qquad \quad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\cos \bigg(\dfrac{\pi -(A+B)}{2}\bigg)

\text{Sum/Difference:}\quad  =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\sin \bigg(\dfrac{A+B}{2}\bigg)

\text{Double Angle:}\quad  =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\sin \bigg(\dfrac{2(A+B)}{2(2)}\bigg)\\\\\\.\qquad \qquad  \qquad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+2\sin \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A+B}{4}\bigg)

\text{Factor:}\quad  =2\cos \bigg(\dfrac{A+B}{4}\bigg)\bigg[ \cos \bigg(\dfrac{A-B}{4}\bigg)+\sin \bigg(\dfrac{A+B}{4}\bigg)\bigg]

\text{Cofunction:}\quad  =2\cos \bigg(\dfrac{A+B}{4}\bigg)\bigg[ \cos \bigg(\dfrac{A-B}{4}\bigg)+\cos \bigg(\dfrac{\pi}{2}-\dfrac{A+B}{4}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\cos \bigg(\dfrac{2\pi-(A+B)}{4}\bigg)\bigg]

\text{Sum to Product:}\ 2\cos \bigg(\dfrac{A+B}{4}\bigg)\bigg[2 \cos \bigg(\dfrac{2\pi-2B}{2\cdot 4}\bigg)\cdot \cos \bigg(\dfrac{2A-2\pi}{2\cdot 4}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =4\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi-B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi -A}{4}\bigg)

\text{Given:}\qquad \qquad 4\cos \bigg(\dfrac{\pi -C}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi-B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi -A}{4}\bigg)\\\\\\.\qquad \qquad \qquad =4\cos \bigg(\dfrac{\pi -A}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi-B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi -C}{4}\bigg)

LHS = Middle \checkmark

<u>Proof Middle → RHS:</u>

\text{Middle:}\qquad 4\cos \bigg(\dfrac{\pi -A}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi-B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi -C}{4}\bigg)\\\\\\\text{Given:}\qquad \qquad 4\cos \bigg(\dfrac{B+C}{4}\bigg)\cdot \cos \bigg(\dfrac{C+A}{4}\bigg)\cdot \cos \bigg(\dfrac{A+B}{4}\bigg)\\\\\\.\qquad \qquad \qquad =4\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{B+C}{4}\bigg)\cdot \cos \bigg(\dfrac{C+A}{4}\bigg)

Middle = RHS \checkmark

3 0
3 years ago
Order the numbers from least to greatest: 0.8, 8% 18% 1/8 8/18
DiKsa [7]
From least to greatest: 8%, 1/8, 18%, 8/18, 0.8
4 0
3 years ago
Read 2 more answers
Environmentalists concerned about the impact of high-frequency radio transmissions on birds found that there was no evidence of
babunello [35]

Answer:

No, we cannot conclude anything about 0.10

Yes, they have made the same decision.

Step-by-step explanation:

Consider the provided information.

Part (A)

They based this conclusion on a test using α = 0.05. Would they have made the same decision at α = 0.10

A significance level of 0.05 indicates a 5% risk of concluding that a difference exists

From the normal table the critical value is Z(0.05) = 1.645 and we accept null hypothesis.

If α = 0.1, the critical value is Z(0.1)=1.28

1.28 is less than 1.645.

Therefore, we cannot conclude anything about 0.10

Part (B)

α = 0.01, the critical value is Z(0.01)=2.33

The critical value of α = 0.05 is 1.645 which is less than 2.33,

So we would again reject the null hypothesis.

Hence, they have made the same decision.

6 0
3 years ago
Other questions:
  • 3 solutions for equation <br> y=7x
    14·1 answer
  • If GE=42 and DH=16, find GF
    14·1 answer
  • What is AE ? <br><br> Enter your answer in the box. <br><br> ____units
    12·2 answers
  • A video game company is designing a game based on professional race cars. Professional race cars have a length of 570 cm and a w
    8·1 answer
  • What is 48% written as a fraction? (in simplest form) A) 1 48 B) 6 25 C) 12 25 D) 24 25
    10·1 answer
  • How many centimeters are in 2 inches?<br><br><br> Answer: There are 5.08 centimeters in 2 inches
    10·2 answers
  • Please help ASAP I NEED THIS TOMORROW!!!
    11·1 answer
  • A radio is giving away tickets to a play. They plan to give away tickets for seats that cost $10 and $20. They want to give away
    12·2 answers
  • Jill has quiz scores of 74, 72, 76, 80, and 73. To continue to receive her scholarship, Jill must maintain an average of 75. Wha
    11·1 answer
  • Someone please please help help me please please ASAP with my work <br><br> And NO LINKS or FILES
    10·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!