A store employs 11 women and 12 men.

A pack of paper costs $3.75 including tax. Mr. Cooper wants to purchase packs of paper for his class and has a $20 budget. Write

STALIN [3.7K]

If a pack of paper costs 3.75 including tax, and there is a $20 budget, then the equation to find out the answer would be 3.75p is greater or equal to $20. So the answer would be B. 3.75 would be multiplied by the tax and then applied up to $20.

7 0

2 years ago

A country's population in 1993 was 94 million. in 1999 in was 99 million. estimate the population in 2005 using the exponential

Jet001 [13]

The initial population is
P₀ = 94 million in 1993

The growth formula is
$P(t) = P_{0}e^{kt}$
where P(t) is the population (in millions) after t years, measured from 1993.
k = constant.

Because P(5) = 99 million (in 1999),
$94e^{5k} = 99 \\e^{5k}=1.0532 \\ 5k = ln(1.0532) \\ k = 0.010367$

In the year 2005, t = 12 years, and
$P(12)=94e^{0.010367*12} = 106.45$

Answer: 106 million (nearest million)

7 0

2 years ago

In high-school 135 freshmen were interviewed.

timama [110]

Answer:

a) n(none) = 25

b) n(PE but not Bio) = 25

c) n(ENG but not both BIO and PE) = 55

d) n(students that did not take Eng or Bio) = 40

e) P( Students did not take exactly two subjects) = 0.65

Step-by-step explanation:

From the Venn diagram drawn:

a) Number of students that took none

n(Freshmen) = 135

n(all three) = 5

n (PE and Bio) = 10

n(PE and Eng) = 15

n(Bio and Eng) = 7

n (PE and Bio only) = 10 - 5 = 5

n(PE and Eng only) = 15 - 5 = 10

n(Bio and Eng only) = 7 - 5 = 2

n(PE only) = 35 - 5 - 5 - 10 = 15

n(Bio only) = 42 - 5 - 5 - 2 = 30

n(Eng only) = 60 - 10 - 5 -2 = 43

n(Freshmen) = n(PE only) + n(Bio only) + n(Eng only) + n(PE and Bio only) + n(PE and Eng only) + n(Bio and Eng only) + n(all three) + n(none)

135 = 15 + 30 + 43 + 5 + 10 + 2 + 5 + n(none)

135 = 110 + n(none)

n(none) = 135 - 110

n(none) = 25

b)Number of students that too PE but not Bio

n(PE but not bio)= n(PE only) + n(PE and Eng only)

n(PE but not Bio) = 15 + 10

n(PE but not Bio) = 25

c) Number of students that took ENG but not both BIO and PE

n(ENG but not both BIO and PE) = n(Eng only) + n(Eng and Bio only) + n(Eng and PE only) = 43 + 2 + 10

n(ENG but not both BIO and PE) = 55

d) Number of students that did not take ENG or BIO

n( students that did not take Eng or Bio) = n(PE only) + n(none)

n(students that did not take Eng or Bio) = 15 + 25

n(students that did not take Eng or Bio) = 40

e) Probability that a randomly-chosen student from this group did not take exactly two subjects

n( Students that did not take exactly two subjects) = n(PE only) + n(Bio only) + n(Eng only)

n( Students that did not take exactly two subjects) = 15 + 30 + 43

n( Students that did not take exactly two subjects) = 88

P( Students did not take exactly two subjects) = 88/135

P( Students did not take exactly two subjects) = 0.65

3 0

2 years ago

Read 2 more answers

If e = 9 cm and f = 40 cm, what is the length of g?

rewona [7]

(9,40,41) is a Pythagorean Triple, farther down the list than teachers usually venture.

Answer: D. 41 cm

There's a subset of Pythagorean Triples where the long leg is one less than the hypotenuse,

a^2+b^2 = (b+1)^2

a^2 + b^2 = b^2 + 2b +1

a^2=2b+1

So we get one for every odd number, since the square of an odd number is odd and the square of an even number is even.

b = (a^2 - 1)/2

a=3, b=(3^2-1)/2=4, c=b+1=5

a=5, b=(5^2-1)/2 =12, c = 13

a=7, b=24, c=25

a=9, b=40, c=41

a=11, b=60, c=61

a=13, b=84, c=85

It's good to be able to recognize Pythagorean Triples when we see them.

Otherwise we'd have to work the calculator:

√(9² + 40²) = √1681 = 41

8 0

2 years ago

Determine the t critical value(s) that will capture the desired t-curve area in each of the following cases. (Assume that centra

Triss [41]

Answer:

Step-by-step explanation:

a) this involves 2 tails

The critical value is determined from the t distribution table.

α = 1 - 0.95 = 0.05

1 - α/2 = 1 - 0.05/2 = 1 - 0.025 = 0.975

Looking at 0.975 with df 10

The critical value is 2.228

b) α = 1 - 0.95 = 0.05

1 - α/2 = 1 - 0.05/2 = 1 - 0.025 = 0.975

Looking at 0.975 with df 20

The critical value is 2.086

c) α = 1 - 0.99 = 0.01

1 - α/2 = 1 - 0.01/2 = 1 - 0.005 = 0.995

Looking at 0.995 with df 20

The critical value is 2.845

d) α = 1 - 0.99 = 0.01

1 - α/2 = 1 - 0.01/2 = 1 - 0.005 = 0.995

Looking at 0.995 with df 60

The critical value is 2.660

e) 1 - α = 1 - 0.01 = 0.99

Looking at 0.99 with df 10

The critical value is 2.764

f) 1 - α = 1 - 0.025 = 0.975

Looking at 0.975 with df 5

The critical value is 2.571

6 0

2 years ago