How many whole numbers between −32 and 77?

Given the graph below, what is the zero(s)?

JulsSmile [24]

Answer:

Great job!

Step-by-step explanation:

6 0

3 years ago

Loretta’s income last year was $81,300. She made $56,800 at her salaried job and had additional passive income. If Loretta earne

aalyn [17]

Answer:

Option a. $2,040

Step-by-step explanation:

step 1

To find out the amount of the additional passive income last year, subtract the amount earned at her salaried job from Loretta’s income last year

so

$\$81,300-\$56,800=\$24,500$

step 2

Divide the additional passive income last year by 12 (number of months in a year)

$\$24,500/12=\$2,041.67$

therefore

approximately $2,400 per month

5 0

2 years ago

Read 2 more answers

. Laura and Alicia both exercise 5 days a week. Laura exercises for 30 minutes each day. Alicia exercises for 45 minutes each da

statuscvo [17]

Answer: 1 hour 15 minutes

Step-by-step explanation:

From the question, we are informed that Laura and Alicia both exercise 5 days a week and that Laura exercises for 30 minutes each day. For the 5 days, she'll exercise for:

= 5 × 30 minutes

= 150 minutes

= 2 hours 30 minutes

Alicia exercises for 45 minutes each day. Fir the 5 days, she'll exercise for:

= 5 × 45 minutes

= 225 minutes

= 3 hours 45 minutes

We then calculate the difference in their exercise per week which will be:

= 3 hours 45 minutes - 2 hours 30 minutes

= 1 hour 15 minutes

6 0

2 years ago

Let X be a set of size 20 and A CX be of size 10. (a) How many sets B are there that satisfy A Ç B Ç X? (b) How many sets B are

Svetlanka [38]

Answer:

(a) Number of sets B given that

A⊆B⊆C: 2¹⁰. (That is: A is a subset of B, B is a subset of C. B might be equal to C)
A⊂B⊂C: 2¹⁰ - 2. (That is: A is a proper subset of B, B is a proper subset of C. B≠C)

(b) Number of sets B given that set A and set B are disjoint, and that set B is a subset of set X: 2²⁰ - 2¹⁰.

Step-by-step explanation:

Let $x_1, x_2, \cdots, x_{20}$ denote the 20 elements of set X.

Let $x_1, x_2, \cdots, x_{10}$ denote elements of set X that are also part of set A.

For set A to be a subset of set B, each element in set A must also be present in set B. In other words, set B should also contain $x_1, x_2, \cdots, x_{10}$ .

For set B to be a subset of set C, all elements of set B also need to be in set C. In other words, all the elements of set B should come from $x_1, x_2, \cdots, x_{20}$ .

$\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}& \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{Maybe} & \cdots & \text{Maybe}\end{array}$ .

For each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus $2^{10} = 1024$ possibilities for set B.

In case the question connected set A and B, and set B and C using the symbol ⊂ (proper subset of) instead of ⊆, A ≠ B and B ≠ C. Two possibilities will need to be eliminated: B contains all ten "maybe" elements or B contains none of the ten "maybe" elements. That leaves $2^{10} -2 = 1024 - 2 = 1022$ possibilities.

Set A and set B are disjoint if none of the elements in set A are also in set B, and none of the elements in set B are in set A.

Start by considering the case when set A and set B are indeed disjoint.

$\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}& \text{No}&\text{No}&\cdots &\text{No}& \text{Maybe} & \cdots & \text{Maybe}\end{array}$ .

Set B might be an empty set. Once again, for each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus $2^{10} = 1024$ possibilities for a set B that is disjoint with set A.

There are 20 elements in X so that's $2^{20} = 1048576$ possibilities for B ⊆ X if there's no restriction on B. However, since B cannot be disjoint with set A, there's only $2^{20} - 2^{10}$ possibilities left.

5 0

2 years ago

Find the Solution of the system of equations.<br><br> -4x+3y=-21<br> -2x-3y=3

LUCKY_DIMON [66]

Answer:

(3, - 3)

Step-by-step explanation: that might be it

8 0

2 years ago