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1 year ago

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Sam and chad are ticket-sellers at their class play. sam is selling student tickets for $2.00 each, and chad selling adult ticke

ts for $5.50 each. if their total income for 24 tickets was $83.00, how many tickets did sam sell?

1 answer:

Nikolay [14]1 year ago

3 0

15 adult tickets, or 41 children tickets. but combined it could be, 13 adults and 5 children tickets.

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Janie ordered boxed lunches for a student advisory committee meeting. Each lunch costs $4.25. The total cost of the lunches is $

andriy [413]

Answer:

11 boxed lunches

Step-by-step explanation:

Full question

Janie ordered boxed lunches for a student advisory committee meeting. Each lunch cost 4.25. The total cost of the lunches is 53.75, including a 7$ delivery fee. Write and solve an equation to find x the number of boxed lunches Janie ordered

First of all subtract the delivery feesince it was inckuded in the total cost, this will now be the total cost of all the noxed lunches ordered by Janie, then divide the balance of the total cost by the cost of one boxed lunch to get thd total boxed kunches

X= 53.75-7/4.25

X= 53.75-7= 46.75/4.25

X=11

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1 year ago

Bria and Matt have a total of 400 stamps. Matt has three times as many stamps as Bria. Which system of equations can be used to

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M= 400÷ 4b
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Use the expression (a+b)(c-d)+9e(a+b)(c−d)+9eleft parenthesis, a, plus, b, right parenthesis, left parenthesis, c, minus, d, rig

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Step-by-step explanation:

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1 year ago

A simple random sample of 100 concert tickets was drawn from a normal population. The mean and standard deviation of the sample

alexandr402 [8]

Answer:

We accept the null hypothesis and the population mean is $120.

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 100

Sample mean, $\bar{x}$ = $120

Alpha, α = 0.01

Sample standard deviation, s = $25

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 125\\H_A: \mu \neq 125$

We use two-tailed t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = -2.0$

p-value one tail= 0.024

p-value two tail= 0.048

Conclusion:

Since the p-value for two tailed test is greater than the significance level, we fail to reject the null hypothesis and accept it.

Thus, the population mean is $120.

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1 year ago