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KengaRu [80]
2 years ago
6

In melanie's styling salon, the time to complete a simple haircut is normally distributed with a mean of 25 minutes and a standa

rd deviation of 4 minutes. for a simple haircut, the middle 90 percent of the customers will require:
Mathematics
1 answer:
vovikov84 [41]2 years ago
6 0
They will require between 18.42 and 31.58 minutes.

We want the middle 90%.  The two probability values associated with this would be 0.95 and 0.05; this would leave 5% below and 5% above, giving us the middle 90%.

Using a z-table (http://www.z-table.com) we see that the z-score associated with an area to the left of 0.05 is between -1.64 and -1.65; since it is equally distant from both we will use -1.645.

The z-score associated with an area of the left of 0.95 is between 1.64 and 1.65; since it is equally distant from both we will use 1.645.

The formula for a z-score is

z = (X-μ)/σ
-1.645 = (X-25)/4

Multiplying by 4 on both sides,
-1.645(4) = X-25
-6.58 = X-25

Adding 25 to both sides,
-6.58+25 = X
18.42 = X

For the upper bound,
1.645 = (X-25)/4

Multiplying both sides by 4,
1.645(4) = X-25
6.58 = X-25

Adding 25 to both sides,
6.58+25 = X
31.58 = X

The times are between 18.42 and 31.58 minutes.
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Candy draws a square design with a side length of x inches for the window at the pet shop. She takes the design to the printer a
Inga [223]

Answer:

This is late, but for anyone else that needs it, it's B. 4x-5

8 0
2 years ago
Read 2 more answers
a(0.3−y)+1.1+2.4x (y−1.2) ​ =0 =−1.2(x−0.5) ​ Consider the system of equations above, where aaa is a constant. For which value o
Veseljchak [2.6K]

Answer:

For a = 1.22 there is one solution where y = 1.3

Step-by-step explanation:

Hi there!

Let´s write the system of equations:

a(0.3 - y) + 1.1 +2.4x(y-1.2) = 0

-1.2(x-0.5) = 0

Let´s solve the second equation for x:

-1.2(x-0.5) = 0

x- 0.5 = 0

x = 0.5

Now let´s repalce x = 0.5 and y = 1.3 in the first equation and solve it for a:

a(0.3 - y) + 1.1 +2.4x(y-1.2) = 0

a(0.3 - 1.3) + 1.1 + 2.4(0.5)(1.3 -1.2) = 0

a(-1) + 1.1 + 1.2(0.1) = 0

-a + 1.22 = 0

-a = -1.22

a = 1.22

Let´s check the solution and solve the system of equations with a = 1.22. Let´s solve the first equation for y:

1.22(0.3 - y) + 1.1 +2.4(0.5)(y-1.2) = 0

0.366 - 1.22y + 1.1 + 1.2 y - 1.44 = 0

-0.02y +0.026 = 0

-0.02y = -0.026

y = -0.026 / -0.02

y = 1.3

Then, the answer is correct.

Have a nice day!

4 0
2 years ago
Suppose you roll a six-sided 50 times and calculate the mean roll, x .
Vitek1552 [10]

Answer:

0.07

a. the distribution will be a normal distribution.

c. we would suspicious ins there is a 2 % chance of getting the required value.

Step-by-step explanation:

Let the number of times, t be = 50

Assuming that the die is fair

standard deviation = 1.71

mean = 3.5

suppose we want tp find the probability of a 2 showing. The solution becomes:

probability = \frac{2}{6}  = \frac{1}{3}

c. the  mean of rolls will be 0.07

8 0
2 years ago
Jenny’s mother is 5 years older than twice Jenny's age. The sum of their ages is 62 years. This is represented by the equation x
Salsk061 [2.6K]

Answer:

Jenny´s age is 19 and her mother is 43.

Step-by-step explanation:

1. Plug in the first value of the set

41 is different from 62, so Jenny isn´t 12 years old.

2. Plug in the second value of the set

50 is different from 62, so Jenny isn´t 15 years old.

3. Plug in the third value of the set

59 is different from 62, so Jenny isn´t 18 years old.

4. Plug in the fourth value of the set

Therefore Jenny´s 19 years old.

5. Calculate Jenny´s mother age:

As the sum of their ages is 62 ages, we have the following equation

where

x=Jenny´s age

y=Jenny´s mother age

Therefore Jenny's mother age is 43 years old.

pls mark me brainliest

8 0
2 years ago
among a group of students 50 played cricket 50 played hockey and 40 played volleyball. 15 played both cricket and hockey 20 play
kondaur [170]

Answer:

Cricket only= 30

Volleyball only = 15

Hockey only = 25

Explanation:

Number of students that play cricket= n(C)

Number of students that play hockey= n(H)

Number of students that play volleyball = n(V)

From the question, we have that;

n(C) = 50, n(H) = 50, n(V) = 40

Number of students that play cricket and hockey= n(C∩H)

Number of students that play hockey and volleyball= n(H∩V)

Number of students that play cricket and volleyball = n(C∩V)

Number of students that play all three games= n(C∩H∩V)

From the question; we have,

n(C∩H) = 15

n(H∩V) = 20

n(C∩V) = 15

n(C∩H∩V) = 10

Therefore, number of students that play at least one game

n(CᴜHᴜV) = n(C) + n(H) + n(V) – n(C∩H) – n(H∩V) – n(C∩V) + n(C∩H∩V)

= 50 + 50 + 40 – 15 – 20 – 15 + 10

Thus, total number of students n(U)= 100.

Note;n(U)= the universal set

Let a = number of people who played cricket and volleyball only.

Let b = number of people who played cricket and hockey only.

Let c = number of people who played hockey and volleyball only.

Let d = number of people who played all three games.

This implies that,

d = n (CnHnV) = 10

n(CnV) = a + d = 15

n(CnH) = b + d = 15

n(HnV) = c + d = 20

Hence,

a = 15 – 10 = 5

b = 15 – 10 = 5

c = 20 – 10 = 10

Therefore;

For number of students that play cricket only;

n(C) – [a + b + d] = 50 – (5 + 5 + 10) = 30

For number of students that play hockey only

n(H) – [b + c + d] = 50 – ( 5 + 10 + 10) = 25

For number of students that play volleyball only

n(V) – [a + c + d] = 40 – (10 + 5 + 10) = 15

3 0
2 years ago
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