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Lunna [17]
2 years ago
8

Maya had $27. She spent all the money on buying 3 burgers for $x each and 2 sandwiches for $y each. If Maya had bought 2 burgers

and 1 sandwich, she would have been left with $11.
A student concluded that the price of each burger is $5 and the price of each sandwich is $6. Which statement best justifies whether the student's conclusion is correct or incorrect?

1) The student's conclusion is correct because the solution to the system of equations 3x + 2y = 11 and 2x + y = 16 is (5, 6).
2) The student's conclusion is incorrect because the solution to the system of equations 3x − 2y = 11 and 2x − y = 16 is (5, 6).
3) The student's conclusion is correct because the solution to the system of equations 3x + 2y = 27 and 2x + y = 16 is (5, 6).
4) The student's conclusion is incorrect because the solution to the system of equations 2x + 3y = 27 and x + 2y = 16 is (5, 6).
Mathematics
1 answer:
Ber [7]2 years ago
3 0
Let
x-------> the cost of the burger
y-------> the cost of the <span>sandwiches

we know that
3x+2y=27-----> equation 1
2x+y=27-11-----> 2x+y=16-----> multiply by -2-----> -4x-2y=-32--> equation 2

adds equation 1 and equation 2
</span> 3x+2y=27
-4x-2y=-32
<span>----------------
-x=27-32-----> x=5
3*5+2y=27----> 2y=27-15-----> y=6

therefore

the answer is the option
</span><span>3) The student's conclusion is correct because the solution to the system of equations 3x + 2y = 27 and 2x + y = 16 is (5, 6).</span>
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Suppose a certain airline uses passenger seats that are 16.2 inches wide. Assume that adult men have hip breadths that are norma
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Answer:

Each adult male has a 5.05% probability of having a hip width greater than 16.2 inches.

There is a 0.01% probability that the 110 adult men will have an average hip width greater than 16.2 inches.

Step-by-step explanation:

Problems of normally distributed samples can be 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}

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. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem

Assume that adult men have hip breadths that are normally distributed with a mean of 14.4 inches and a standard deviation of 1.1 inches. This means that \mu = 14.4, \sigma = 1.1.

What is the probability that any one of those adult male will have a hip width greater than 16.2 inches?

For each one of these adult males, the probability that they have a hip width greater than 16.2 inches is 1 subtracted by the pvalue of Z when X = 16.2. So:

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

Z = \frac{16.2 - 14.4}{1.1}

Z = 1.64

Z = 1.64 has a pvalue of 0.9495.

This means that each male has a 1-0.9495 = 0.0505 = 5.05% probability of having a hip width greater than 16.2 inches.

For the average of the sample

What is the probability that the 110 adult men will have an average hip width greater than 16.2 inches?

Now, we need to find the standard deviation of the sample before using the zscore formula. That is:

s = \frac{\sigma}{\sqrt{110}} = 0.1.

Now

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

Z = \frac{16.2 - 14.4}{0.1}

Z = 18

Z = 18 has a pvalue of 0.9999.

This means that there is a 1-0.9999 = 0.0001 = 0.01% probability that the 110 adult men will have an average hip width greater than 16.2 inches.

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2 years ago
In general, the probability that a blood donor has Type A blood is 0.40.Consider 8 randomly chosen blood donors, what is the pro
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Answer:

The probability that more than half of them have Type A blood in the sample of 8 randomly chosen donors is P(X>4)=0.1738.

Step-by-step explanation:

This can be modeled as a binomial random variable with n=8 and p=0.4.

The probability that k individuals in the sample have Type A blood can be calculated as:

P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}\\\\\\P(x=k) = \dbinom{8}{k} 0.4^{k} 0.6^{8-k}\\\\\\

Then, we can calculate the probability that more than 8/2=4 have Type A blood as:

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Answer:

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Intervals of increase and decrease: increasing on (-∞, 0) and (1.25, 3.25), decreasing on (0, 1.25) and (3.25, ∞)

Positive and negative intervals: positive on (2, 4), negative on (-∞, 0), (0, 2), and (4, ∞)

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Step-by-step explanation:

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Answer:

He is missing 1500, the final solution would be 2090

Step-by-step explanation:

5 times 8 equals 40

5 times 30 equals 150

50 times 8 equals 400

50 times 30 equals 1500

40+150+400+1500= 2090

6 0
2 years ago
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