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enyata [817]
2 years ago
8

It costs 135.25 for 15 child tickets ti the museum.Adult tickets cost 12.95 each. Find the cost for 40 children to attend along

with 5 adult chaperones
Mathematics
1 answer:
Masja [62]2 years ago
4 0

Answer:the cost for 40 children to attend along with 5 adult tickets = $ 417.55

Step-by-step explanation:

It costs 135.25 for 15 child tickets to the museum. Let us determine the cost per child ticket to the museum.

cost per child ticket to the museum.

= total cost of 15 child tickets / number of child tickets.

= 132.25 / 15 = 8.82

Adult tickets cost 12.95 each.

The cost of 40 child tickets is

40 × 8.82 = $352.8

The cost of 5 adult tickets = 5 × 12.95

=$ 64.75

the cost for 40 children to attend along with 5 adult tickets would be

352.8 + 64.75 = $ 417.55

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If h(x)=x^2-5x, g(x)=2x, f(x)=3x-2, what is h(f(g(x)))
Aleonysh [2.5K]

Step-by-step explanation:

start from f(g(x))

f(g(x))= 3(g(x)) -2

f(g(x))= 6x -2

h(f(g(x)))=(f(g(x)))² -5(f(g(x)))

h(f(g(x)))=(6x-2)² -5(6x-2)

h(f(g(x)))=36x²-24x+4-(30x-10)

h(f(g(x)))=36x²-24x+4-30x+10

h(f(g(x)))=36x²-54x+14

6 0
3 years ago
In a large population, 82% of the households have cable tv. A simple random sample of 225 households is to be contacted and the
anzhelika [568]

Answer:

The mean of the sampling distribution of the sample proportions is 0.82 and the standard deviation is 0.0256.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For proportions, the mean is \mu = p and the standard deviation is s = \sqrt{\frac{p(1-p)}{n}}

In this problem, we have that:

p = 0.82, n = 225.

So

\mu = 0.82

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.82*0.18}{225}} = 0.0256

The mean of the sampling distribution of the sample proportions is 0.82 and the standard deviation is 0.0256.

7 0
3 years ago
Read 2 more answers
Jules has $1,200 and is saving $40 per week. Kelsey has $400 and is saving $50 a week. In how many weeks will Jules and Kelsey h
enyata [817]
80 weeks 
1,200+40x=400+50x
800+40x=50x
800=10x
x=80
3 0
3 years ago
Let s represent the number of games a baseball fan attends. Enter an
jek_recluse [69]

Answer:3240<50s*81=4050

6 0
3 years ago
QUESTION
Lorico [155]
1. For this item we just refer to the prompt to know the conjectures of Ernest and Denise. According to Ernest, they should swim 1 kilometer on the first week then add 0.25km every week while Denise believes that they should swim 1 kilometer on the first week then add 0.5km every week.

2. Yes, these distances make an arithmetic sequence. It's because an arithmetic sequence is defined as a group of increasing or decreasing numbers where the difference between any two consecutive numbers is constant. This just means that every number has the same interval. In the case of their schedule, this is true.

3. For this item we just follow the descriptions of Ernest's and Denise's schedule in item number 1. For Ernest, we just keep adding 0.25 from 1 kilometer until we added it thrice. For Denise, we also keep adding a number thrice but this time it's 0.5 instead of 0.25.

Ernest's Schedule: 1, 1.25, 1.5, 1.75
Denise's Schedule: 1, 1.5, 2, 2.5

4. Here we are asked to determine a formula that will describe the schedules of Ernest and Denise. In the given formula a_{n}= a_{n-1}+d, a_{n} refers to the next term in the sequence, a_{n-1} refers to the previous term, while d refers to the common difference. In the recursive formula all we need is to insert the value of d to the equations.

Ernest: a_{n}= a_{n-1}+0.25
Denise: a_{n}= a_{n-1}+0.5

5. For this item we basically do the same thing but this time we are given another formula. Our formula is in the form a_{n}= a_{1}+(n-1)d where a_{n} is still the nth term of the sequence, a_{1} is the very first time, n is the number of terms, and d is the common difference. 

Ernest: a_{n}= 1.0+0.25(n-1)
Denise: a_{n}= 1.0+0.5(n-1)

6. In this item we will just basically substitute numbers to one of the equations that we've set up in item #5. For this we need Ernest's explicit formula first. To know how far they will be swimming on week 10, the number of elements (n) must be 10.

a_{10}= 1.0+0.25(10-1)
a_{10}= 1.0+0.25(9)
a_{10}= 1.0+2.25
a_{10}= 3.25

7. Here, we just do the same thing as item #6 but this time we will consider Denise's explicit formula. Since we are also asked how far the students will be swimming on week 10, the number of elements would also be 10 and this would also be our value for n.

a_{10}= 1.0+0.5(10-1)
a_{10}= 1.0+0.5(9)
a_{10}= 1.0+4.5
a_{10}= 5.5

8. The answer for this question is obvious. You would just need to look at the 10th element in Ernest's and Denise's sequences and tell whose schedule had more than or equal to 5 as an answer. Following Ernest's schedule, you will just get 3.5 kilometers on the 10th week so it's definitely a no. Denise's schedule, on the other hand, would get you to 5.5 kilometers on week 10 so her training schedule should be followed.
7 0
3 years ago
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