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3 years ago

Which graph could represent a constant blalance in a bank account?

Mathematics

2 answers:

Crank3 years ago

8 0

Answer:

the third graph

Step-by-step explanation:

Natalka [10]3 years ago

4 0

<h2>Answer:</h2>

The third graph ( i.e. a horizontal line ) represents a constant balance in a bank account.

<h2>Step-by-step explanation:</h2>

Constant balance in a bank amount means that the amount of balance in the account is same over different time period.

i.e. it did not increase or decrease over a period of time.

Hence, the graph of such a situation is a horizontal line parallel to the axis labeled as Time period.

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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

A pair of fair dice is cast. Let Edenote the event that the number landing uppermost on the first die is a 1 and let Fdenote the

iris [78.8K]

Answer:

They are not independent

Step-by-step explanation:

Given

E = Occurrence of 1 on first die

F = Sum of the uppermost occurrence in both die is 5

Required

Are E and F independent

First, we need to list the sample space of a roll of a die

$Event\ 1 = \{1,2,3,4,5,6\}$

Next, we list out the sample space of F

$Event\ 2 = \{2,3,4,5,6,7,3,4,5,6,7,8,4,5,\$

$6,7,8,9,5,6,7,8,9,10,6,7,8,9,10,11,7,8,9,10,11,12\}$

In (1): the sample space of E is:

$E = \{1\}$

So:

$P(E) = \frac{n(E)}{n(Event\ 1)}$

$P(E) = \frac{1}{6}$

In (2): the sample space of F is:

$F = \{5,5,5,5\}$

So:

$P(F) = \frac{n(F)}{n(Event\ 2)}$

$P(F) =\frac{4}{36}$

$P(F) =\frac{1}{9}$

For E and F to be independent:

$P(E\ and\ F) = P(E) * P(F)$

Substitute values for P(E) and P(F)

This gives:

$P(E\ and\ F) = \frac{1}{6} * \frac{1}{9}$

$P(E\ and\ F) = \frac{1}{54}$

However, the actual value of P(E and F) is 0.

This is so because $E = \{1\}$ and $F = \{5,5,5,5\}$ have 0 common elements:

So:

$P(E\ and\ F) = 0$

Compare $P(E\ and\ F) = \frac{1}{54}$ and $P(E\ and\ F) = 0$ .

These values are not equal.

Hence: the two events are not independent

6 0

3 years ago

A 120-inch strip of metal 12 inches wide is to be made into a small open trough by bending up two sides on the long side, at rig

mezya [45]

Answer:

x= 3 inch should be turned up on each side

Step-by-step explanation:

Let the height of trough be x.

Width of trough be 12 - 2x.

and length of trough = 120 inch

Volume of trough, V = L×W×H = 120 × (12-2x) × x = 120x(12 - 2x)

For maximum volume, we find V' = 0

i.e 1440 -480x = 0

or x = $\frac{1440}{480}$

or x= 3

Hence x= 3 inch should be turned up on each side

3 0

3 years ago

-4004 + 2023 – 24x2 – 2 / – 2x2 + 2x + 1?

Kamila [148]

Answer:

388.5/x

Step-by-step explanation:

multiply:

(-4004+2023-48-2)/ -4 + 2x +1

Simplify:

-2031/ -3 +2x

divide by -3

677/2x

388.5 / x

as simple as you can go

hope this helps!

5 0

3 years ago

Hi! i don’t really understand this question. if someone could help me i’d appreciate it!

Shkiper50 [21]

Answer:

Step-by-step explanation:

cute it into separate parts then add the areas together

3 0

3 years ago