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3 years ago

Jake has just become the senior marketing rep for a new chain of upscale hotels. His first responsibility is to develop a

Mathematics

1 answer:

cluponka [151]3 years ago

5 0

Answer:

Startup.

Step-by-step explanation:

If Jake has just become the senior marketing rep for a new chain of upscale hotels, his first responsibility is to develop a plan, commonly referred to as the startup.

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3. Find the area of the shaded region. Leave your answer in terms of pi. (1 point)

anygoal [31]

ANSWER

$(4 - \pi) {ft}^{2}$

EXPLANATION

The area of the square is

${l}^{2}$

The length of the square is

$l = 2ft$

The area of the square is

$= {2}^{2} = 4 {ft}^{2}$

Area of the circle is

$= \pi \: {r}^{2}$

The radius of the circle is

$r = \frac{2}{2} = 1ft$

Area of circle is

$= \pi \: \times {1}^{2} = \pi {ft}^{2}$

The area of the shaded region is

$= (4 - \pi) {ft}^{2}$

5 0

3 years ago

Kelly earns 15% more than her sister, and together they earn $999.75. How much does Kelly earn per week?

valina [46]

Answer:

534. 75

Step-by-step explanation:

let the income of kellys sister be x

kellys income=x+15% of x

=23x over 20

x+23x/20=999. 75

or, x=465

kellys income=23×465/20

=534. 75

5 0

3 years ago

Suppose the number of children in a household has a binomial distribution with parameters n=12n=12 and p=50p=50%. Find the proba

nadya68 [22]

Answer:

a) 20.95% probability of a household having 2 or 5 children.

b) 7.29% probability of a household having 3 or fewer children.

c) 19.37% probability of a household having 8 or more children.

d) 19.37% probability of a household having fewer than 5 children.

e) 92.71% probability of a household having more than 3 children.

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem, we have that:

$n = 12, p = 0.5$

(a) 2 or 5 children

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 5) = C_{12,5}.(0.5)^{5}.(0.5)^{7} = 0.1934$

$p = P(X = 2) + P(X = 5) = 0.0161 + 0.1934 = 0.2095$

20.95% probability of a household having 2 or 5 children.

(b) 3 or fewer children

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{12,0}.(0.5)^{0}.(0.5)^{12} = 0.0002$

$P(X = 1) = C_{12,1}.(0.5)^{1}.(0.5)^{11} = 0.0029$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.0537$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0002 + 0.0029 + 0.0161 + 0.0537 = 0.0729$

7.29% probability of a household having 3 or fewer children.

(c) 8 or more children

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 8) = C_{12,8}.(0.5)^{8}.(0.5)^{4} = 0.1208$

$P(X = 9) = C_{12,9}.(0.5)^{9}.(0.5)^{3} = 0.0537$

$P(X = 10) = C_{12,10}.(0.5)^{10}.(0.5)^{2} = 0.0161$

$P(X = 11) = C_{12,11}.(0.5)^{11}.(0.5)^{1} = 0.0029$

$P(X = 12) = C_{12,12}.(0.5)^{12}.(0.5)^{0} = 0.0002$

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.1208 + 0.0537 + 0.0161 + 0.0029 + 0.0002 = 0.1937$

19.37% probability of a household having 8 or more children.

(d) fewer than 5 children

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{12,0}.(0.5)^{0}.(0.5)^{12} = 0.0002$

$P(X = 1) = C_{12,1}.(0.5)^{1}.(0.5)^{11} = 0.0029$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.0537$

$P(X = 4) = C_{12,4}.(0.5)^{4}.(0.5)^{8} = 0.1208$

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0002 + 0.0029 + 0.0161 + 0.0537 + 0.1208 = 0.1937$

19.37% probability of a household having fewer than 5 children.

(e) more than 3 children

Either a household has 3 or fewer children, or it has more than 3. The sum of these probabilities is 100%.

From b)

7.29% probability of a household having 3 or fewer children.

p + 7.29 = 100

p = 92.71

92.71% probability of a household having more than 3 children.

5 0

3 years ago

If y varies directly with x, the constant of variation is . The equation representing this relationship is . What is the value o

Julli [10]

Answer:

Y~x

Y = k × v

Y = 22

Step-by-step explanation:

Y~ x

Y = k × X

Since the question isn't complete,we are going to assign figures to complete this very question and move on to other things and you can as well solve questions that are similar to this ones.

Now let's assume that the question says that is 6 when x is 3,what is y when x is 11.

To solve this question above,we need to find the constant of variation which will then enable us to know what y is when x is 11

Y~ x

Y = k × V

When y is 6 and x is 3,k will be

6 = 3k

K= 2

Now when k is 2,x is 11,y Will then be

Y = kx

Y = 2 × 11

Y = 22.

With this procedure,one can find any value of x or y when they have the constant of variation.

If you have a figure assigned to x and y is to be found,the constant of variation will help to get y and vice versa

8 0

3 years ago

What what is the standard and expanded form of 3 billion 630 2900058

algol [13]

3,000,000,000,000+600,000,000+30,000,000+200,000+90,000+50+8

3 0

3 years ago