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PolarNik [594]
2 years ago
8

Arnob, Bella, Colin, Dante, and Erin are going to a baseball game. They have a total budget of $100.00 to spend. Each game ticke

t costs $17.50 and each drink costs $2.00. The inequality below relates x, the number of drinks the 5 friends could buy in all, with their ticket costs and budget.
Mathematics
2 answers:
Sliva [168]2 years ago
8 0

They can buy only from 0 to 6 drinks, but no more.


yuradex [85]2 years ago
4 0

Answer:

6

Step-by-step explanation:

Given :

Arnob, Bella, Colin, Dante, and Erin are going to a baseball game.

They have a total budget of $100.00 to spend.

Each game ticket costs $17.50 and each drink costs $2.00

To Find: the number of drinks the 5 friends could buy in all, with their ticket costs and budget.

Solution :

Cost of each ticket is $17.50

cost of 5 tickets = $17.50*5 = $87.50

Let they buy x drinks .

Cost of each drink is $2

so, cost of x drinks = $2x

Total money they spent =  $87.50+$2x

Since they can spend only $100

⇒100=87.50+2x

Thus the inequality is 100=87.50+2x

Now, 100=87.50+2x

100-87.50=2x

12.5=2x

x= 6.25

Thus the number of bottles friends could buy to maintain the budget is 6

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lina2011 [118]

Answer:

a) 6.68th percentile

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Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 550, \sigma = 100

a) A student who scored 400 on the Math SAT was at the ______ th percentile of the score distribution.

Z = \frac{X - \mu}{\sigma}

Z = \frac{400 - 550}{100}

Z = -1.5

Z = -1.5 has a pvalue of 0.0668

So this student is in the 6.68th percentile.

b) To be at the 75th percentile of the distribution, a student needed a score of about ______ points on the Math SAT.

He needs a score of X when Z has a pvalue of 0.75. So X when Z = 0.675.

Z = \frac{X - \mu}{\sigma}

0.675 = \frac{X - 550}{100}

X - 550 = 0.675*100

X = 617.5

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You are selling raffle tickets to raise money. Each ticket costs $5. Which equation solves for the number of tickets you must se
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D. 5x = 45 because you are trying to find x, the number of tickets that must be sold to earn $45.

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3 years ago
a test consists of 10 true false questions to pass a test a student must answer at least six questions correctly if a student gu
Delicious77 [7]

Answer:

P(X \geq 6) = P(X=6) +P(X=7) +P(X=8) +P(X=9) +P(X=10)

And using the probability mass function we got:

P(X=6)=(10C6)(0.5)^6 (1-0.5)^{10-6}=0.205

P(X=7)=(10C7)(0.5)^7 (1-0.5)^{10-7}=0.117

P(X=8)=(10C8)(0.5)^8 (1-0.5)^{10-8}=0.0439

P(X=9)=(10C9)(0.5)^9 (1-0.5)^{10-9}=0.0098

P(X=10)=(10C10)(0.5)^{10} (1-0.5)^{10-10}=0.000977

And adding the values we got:

P(X\geq 6) = 0.377

The best answer would be:

D. 0.377

Step-by-step explanation:

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=10, p=0.5)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

For this case in order to pass he needs to answer at leat 6 questions and we can rewrite this:

P(X \geq 6) = P(X=6) +P(X=7) +P(X=8) +P(X=9) +P(X=10)

And using the probability mass function we got:

P(X=6)=(10C6)(0.5)^6 (1-0.5)^{10-6}=0.205

P(X=7)=(10C7)(0.5)^7 (1-0.5)^{10-7}=0.117

P(X=8)=(10C8)(0.5)^8 (1-0.5)^{10-8}=0.0439

P(X=9)=(10C9)(0.5)^9 (1-0.5)^{10-9}=0.0098

P(X=10)=(10C10)(0.5)^{10} (1-0.5)^{10-10}=0.000977

And adding the values we got:

P(X\geq 6) = 0.377

The best answer would be:

D. 0.377

8 0
3 years ago
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