A rectangle has an area of 897.6 square inches and a base of 34 inches. what is the height? anwser.com

What’s the answers? How do i know it’s correct.

snow_tiger [21]

The unit rate would be the slope, which is given in the equation as 18.25

To find if the given point would be on the graph, replace x with the x value of the point, solve the equation and see if it equals the Y value of the point:

Equation: y = 18.25x

Point: (8,146)

Solve: y = 18.25(8) = 146

This equals the y value of the given point, so this would be on the graph.

8 0

3 years ago

Plz help me I will be giving a brainliest to the first person to answer

pishuonlain [190]

15 is a positive value, -3 is a negative value.

A positive value is always greater than a negative value.

The answer is: 15 > -3

8 0

3 years ago

Read 2 more answers

Help please I will mark brainless

BARSIC [14]

Answer:

your answer ill be d hope ur help nad mark me brainlist

7 0

3 years ago

What is the justification for each step in the solution of the equation?

jeka57 [31]

13x−4=−36−27x+12: Distributive property

13x−4=−24−27x: Combine like terms

40x=−20: Addition Property of equality

6 0

2 years ago

Read 2 more answers

A consulting firm has received 2 Super Bowl playoff tickets from one of its clients. To be fair, the firm is randomly selecting

Soloha48 [4]

Answer:

Therefore the only statement that is not true is b.)

Step-by-step explanation:

There employees are 6 secretaries, 5 consultants and 4 partners in the firm.

a.) The probability that a secretary wins in the first draw

= $\frac{number \hspace{0.1cm} of \hspace{0.1cm} secreataries}{total \hspace{0.1cm} number \hspace{0.1cm} of \hspace{0.1cm} employees} = \frac{6}{15}$

b.) The probability that a secretary wins a ticket on second draw. It has been given that a ticket was won on the first draw by a consultant.

p(secretary wins on second draw | consultant wins on first draw)

= $\frac{p((consultant \hspace{0.1cm} wins \hspace{0.1cm} on \hspace{0.1cm} first \hspace{0.1cm}draw)\cap( secretary\hspace{0.1cm} wins\hspace{0.1cm} on second \hspace{0.1cm}draw))}{p(consultant \hspace{0.1cm} wins \hspace{0.1cm} on \hspace{0.1cm} first \hspace{0.1cm}draw)}$

= $\frac{\frac{5}{15} \times \frac{6}{14}}{\frac{5}{15} } = \frac{6}{14}$ .

The probability that a ticket was won on the first draw by a consultant a secretary wins a ticket on second draw = $\frac{6}{15}$ is not true.

The probability that a secretary wins on the second draw = $\frac{number \hspace{0.1cm} of \hspace{0.1cm} secretaries \hspace{0.1cm} remaining } { number \hspace{0.1cm} of \hspace{0.1cm} employees \hspace{0.1cm} remaining} = \frac{6 - 1}{15 - 1} = \frac{5}{14}$

c.) The probability that a consultant wins on the first draw =

$\frac{number \hspace{0.1cm} of \hspace{0.1cm} consultants \hspace{0.1cm} } { number \hspace{0.1cm} of \hspace{0.1cm} employees \hspace{0.1cm} } = \frac{5 }{15} = \frac{1}{3}$

d.) The probability of two secretaries winning both tickets

= (probability of a secretary winning in the first draw) × (The probability that a secretary wins on the second draw)

= $\frac{6}{15} \times \frac{5}{14} = \frac{1}{7}$

Therefore the only statement that is not true is b.)

5 0

3 years ago