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3 years ago

A group of people bought movies tickets at the AMC Century City. They bought a total of 7 tickets, some

2 answers:

7 0

Hello!,

First you would divide 72 by 12 which would be....6 okay 6 adults.

Second you would divide 72 by 9 witch is...... 8 kids

6 adults 8 kids

DedPeter [7]3 years ago

7 0

You would just plug in numbers until you find two the work in the equation:

12x+9y=72

x=3 y=4

The correct answer is 3 adult tickets and 4 child tickets.

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At homecoming, 125 tickets where sold. Tickets purchased for homecoming cost $20.5 without an activity card, and $15.75 with an

ale4655 [162]

Answer:

Nearly 83 students did not had activity card.

Nearly 42 students had an activity card.

Step-by-step explanation:

Total number of tickets sold at homecoming = 125

Cost of each ticket without activity card = $20.5

Cost of each ticket with activity card = $15.75

Total amount earned = $2365

Let, the number of tickets sold without activity car d = k

So, the number of tickets sold without card = (Total sold - k)

= 125 - k

Now, according to the question:

$20.5(k) + $15.75(125 - k) = $2365

or, 20.5 k + 1968.5 - 15.75 k = 2365

⇒ 4.75 k = 2365 - 1968.5 = 396.5

⇒ $k = \frac{396.5}{4.75 } = 83.47$

Hence, nearly 83 students did not had activity card.

So,the number of students that had activity card = 125 - 83.47 = 41.53 ~ 42

Hence, nearly 42 students had an activity card.

4 0

3 years ago

Given that f(x)=2x/3 evaluate f^-1(6)

abruzzese [7]

Answer:

Step-by-step explanation:

we know by definition that

$f^{-1}(f(x))=x\\\\f^{-1}(\frac{2x}{3})=x\\\\f^{-1}(x)=ax\\\\f^{-1}(\frac{2x}{3})=a(\frac{2x}{3})=x\\\\f^{-1}(x)=\frac{3x}{2}$

so now we evaluate

$f^{-1}(6)=\frac{3*6}{2}\\\\=3*3\\\\f^{-1}(6)=9$

and if we want to do an extra step

$f^{-1}(f(6))=6\\\\f^{-1}(\frac{2*6}{3})=f^{-1}(4)=\frac{3*4}{2}=6$

which works.

4 0

3 years ago

I need help with #9 please

zepelin [54]

The median would decrease because the temperature given is lower than the median

5 0

3 years ago

According to the Mortgage Bankers Association, 8% of U.S. mortgages were delinquent in 2011. A delinquent mortgage is one that h

shepuryov [24]

Answer:

The probability that exactly one of these mortgages is delinquent is 0.357.

Step-by-step explanation:

We are given that according to the Mortgage Bankers Association, 8% of U.S. mortgages were delinquent in 2011. A delinquent mortgage is one that has missed at least one payment but has not yet gone to foreclosure.

A random sample of eight mortgages was selected.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 8 mortgages

r = number of success = exactly one

p = probability of success which in our question is % of U.S.

mortgages those were delinquent in 2011, i.e; 8%

LET X = Number of U.S. mortgages those were delinquent in 2011

So, it means X ~ $Binom(n=8, p=0.08)$

Now, Probability that exactly one of these mortgages is delinquent is given by = P(X = 1)

P(X = 1) = $\binom{8}{1}\times 0.08^{1} \times (1-0.08)^{8-1}$

= $8 \times 0.08 \times 0.92^{7}$

= 0.357

Hence, the probability that exactly one of these mortgages is delinquent is 0.357.

4 0

3 years ago

The lines shown below are parallel.if the green line has a slope of -1/2, what is the slope of the red line?

Delvig [45]

The anwser is D

because it is very easy

4 0

3 years ago

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