ITS Urgent. jjdjwjehvd

tino4ka555 [31]

Answer:

0,-3

Step-by-step explanation:

See the steps below:) you can use the app photo math. you just take a picture of the problem and it gives you the answer and explains the steps:)

7 0

2 years ago

Read 2 more answers

Anyone know the answer?

SIZIF [17.4K]

Answer:

No

Step-by-step explanation:

No function, because the same output can't have two different inputs.

7 0

3 years ago

A certain class has 25 students. If 8 students play football and, 10 plays baseball and 6 plays both, how many students play nei

NNADVOKAT [17]

Answer:1

Step-by-step explanation:

8+10=18 18+6=24 25-24=1

6 0

3 years ago

Read 2 more answers

Please help me on this.

ValentinkaMS [17]

Answer:

first one: function

second one: not a function

third one: not a function

fourth one: not a function

Step-by-step explanation:

it is never a function if two of the x inputs are the same or if one x input has two y outputs..

Hope this helps!! Pls mark me brainliest!!

8 0

3 years ago

The manager of a music store has kept records of

Marysya12 [62]

Answer:

(a) $P(x\le 3) = 0.75$

(b) $P(x\le 3) = 0.75$

(b) is the same as (a)

(c) $P(x \ge 5) = 0.10$

(d) $P(x = 1\ or\ 2) = 0.55$

(e) $P(x > 2) = 0.45$

Step-by-step explanation:

Given

$\begin{array}{ccccccc}{CDs} & {1} & {2} & {3} & {4} & {5} & {6\ or\ more}\ \\ {Prob} & {0.30} & {0.25} & {0.20} & {0.15} & {0.05} & {0.05}\ \ \end{array}$

Solving (a): Probability of 3 or fewer CDs

Here, we consider:

$\begin{array}{cccc}{CDs} & {1} & {2} & {3} \ \\ {Prob} & {0.30} & {0.25} & {0.20} \ \ \end{array}$

This probability is calculated as:

$P(x\le 3) = P(1) + P(2) + P(3)$

This gives:

$P(x\le 3) = 0.30 + 0.25 + 0.20$

$P(x\le 3) = 0.75$

Solving (b): Probability of at most 3 CDs

Here, we consider:

$\begin{array}{cccc}{CDs} & {1} & {2} & {3} \ \\ {Prob} & {0.30} & {0.25} & {0.20} \ \ \end{array}$

This probability is calculated as:

$P(x\le 3) = P(1) + P(2) + P(3)$

This gives:

$P(x\le 3) = 0.30 + 0.25 + 0.20$

$P(x\le 3) = 0.75$

(b) is the same as (a)

Solving (c): Probability of 5 or more CDs

Here, we consider:

$\begin{array}{ccc}{CDs} & {5} & {6\ or\ more}\ \\ {Prob} & {0.05} & {0.05}\ \ \end{array}$

This probability is calculated as:

$P(x \ge 5) = P(5) + P(6\ or\ more)$

This gives:

$P(x\ge 5) = 0.05 + 0.05$

$P(x \ge 5) = 0.10$

Solving (d): Probability of 1 or 2 CDs

Here, we consider:

$\begin{array}{ccc}{CDs} & {1} & {2} \ \\ {Prob} & {0.30} & {0.25} \ \ \end{array}$

This probability is calculated as:

$P(x = 1\ or\ 2) = P(1) + P(2)$

This gives:

$P(x = 1\ or\ 2) = 0.30 + 0.25$

$P(x = 1\ or\ 2) = 0.55$

Solving (e): Probability of more than 2 CDs

Here, we consider:

$\begin{array}{ccccc}{CDs} & {3} & {4} & {5} & {6\ or\ more}\ \\ {Prob} & {0.20} & {0.15} & {0.05} & {0.05}\ \ \end{array}$

This probability is calculated as:

$P(x > 2) = P(3) + P(4) + P(5) + P(6\ or\ more)$

This gives:

$P(x > 2) = 0.20+ 0.15 + 0.05 + 0.05$

$P(x > 2) = 0.45$

3 0

3 years ago

ITS Urgent.jjdjwjehvd​

ITS Urgent.jjdjwjehvd