Need help ASAP math 2

In a movie theater, there are children and adults in the ratio 4:7. If there are 21 adults, how many people total are there in t

lozanna [386]

Answer:

33

Step-by-step explanation:

7x3=21

4x3=12

21+12=33

8 0

3 years ago

What is the average rate of change of the line whose equation is:

dedylja [7]

Answer:

Rate of change = -12

Step-by-step explanation:

y − 3 = −12(x + 4)

Expanding, we have;

y - 3 = -12x - 48

y = -12x - 48 + 3

y = -12x - 45

Now, formula for rate of change is;

Rate of change = (y(x + h) - y)/h

y(x + h) is; y = -12(x + h) - 45

This gives;

Rate of change = (-12(x + h) - 45 + 12x + 45)/h

Rate of change = (-12x - 12h - 45 + 12x + 45)/h

Rate of change = -12h/h

Rate of change = -12

3 0

3 years ago

Please help me with this homework

devlian [24]

Answer:

Well rewriting literal equations can help you understand better and you'll have more room to write the answer and figure it out.

Step-by-step explanation:

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

Ten tickets at $75 each cost EXPLAINNN and answer please !

nataly862011 [7]

Answer:

75*10=750

Step-by-step explanation:

10 tickets, each 75 dollars, so u multiply 75 by 10 to get 750, just write 75 and rop the 0. $750

5 0

2 years ago

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