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Naddik [55]
2 years ago
10

A taxi charges a flat rate of $3.00, plus an additional $0.50 per mile. Carl will only take the taxi home if the cost is under $

10, otherwise he will take a bus. Carl is 16 miles from home. Explain how to write and solve an inequality to determine if Carl will take the taxi or a bus.
Mathematics
2 answers:
marysya [2.9K]2 years ago
7 0

Answer:Let the variable x be the number of miles. Then the problem can be modeled with the inequality 0.5x + 3 < 10,="" in="" which="">x is the number of miles. To solve, first subtract 3 from each

Step-by-step explanation:

Ahat [919]2 years ago
6 0

Answer: Take the bus to save money

Step-by-step explanation:

you plug x for 15, the inequality would be ($0.50 * 15) + $3 <  $10. We need a less than sign (<) because Carl says the price has to be under and not equal to.

$7.5 + $3 < $10. After you do simple math and add, you will know that Carl is short of money by 50 cents.

Hope this helps :)

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Troyanec [42]

Answer:

b_2 =  -2.5

Step-by-step explanation:

Given

\frac{1}{2}(4.5 + b_2) * 6 = 21

Required

Determine the value of T

\frac{1}{2}(4.5 + b_2) * 6 = 21

Multiply both sides by 2

2 * \frac{1}{2}(4.5 + b_2) * 6 = 21 * 2

(4.5 + b_2) * 6 = 42

Divide through by 6

4.5 + b_2 = 42/6``

4.5 + b_2 = 7

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2 years ago
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it might 1700

Step-by-step explanation:

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2 years ago
Emily wants to hang a painting in a gallery. The painting and frame must have an area of 31 square feet. The painting is 5 feet
seropon [69]
Area=legnth times width=31
the frame must go around the frame with equal legnth, x
so
the total area (meaning the frame +picture) is 5+x by 6+x
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horrorfan [7]

Answer:

a. X\sim N(\mu = 6.1, \sigma = 1.9) b. 6.1 c. 0.6842 d. 0.4166 e. 0.1194 f. 8.5349

Step-by-step explanation:

a. The distribution of X is normal with mean 6.1 kg. and standard deviation 1.9 kg. this because X is the weight of a randomly selected seedless watermelon and we know that the set of weights of seedless watermelons is normally distributed.

b. Because for the normal distribution the mean and the median are the same, we have that the median seedless watermelong weight is 6.1 kg.

c. The z-score for a seedless watermelon weighing 7.4 kg is (7.4-6.1)/1.9 = 0.6842

d. The z-score for 6.5 kg is (6.5-6.1)/1.9 = 0.2105, and the probability we are seeking is P(Z > 0.2105) = 0.4166

e. The z-score related to 6.4 kg is z_{1} = (6.4-6.1)/1.9 = 0.1579 and the z-score related to 7 kg is z_{2} = (7-6.1)/1.9 = 0.4737, we are seeking P(0.1579 < Z < 0.4737) = P(Z < 0.4737) - P(Z < 0.1579) = 0.6821 - 0.5627 = 0.1194

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