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alexandr402 [8]

3 years ago

10

One Monday, Oscar sold 15 cups of lemonade. On Tuesday, he sold 5 more cups than he sold on Monday. On Wednesday and Thursday, h

e sold 25 cups of lemonade. Which graph shows the number of lemonade cups that Oscar sold each day? A) A B) B C) C D) D

2 answers:

larisa86 [58]3 years ago

5 0

It’s c hopefully that helps

blsea [12.9K]3 years ago

4 0

The answer is C..... :)

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liubo4ka [24]

Answer:

The equation that represents the motion of the string is given by:

$y =Ae^{-kt}\cos(2\pi ft)$ .....[1] where t represents the time in second.

Given that: A = 0.6 cm (distance above its resting position) , k = 1.8(damping constant) and frequency(f) = 105 cycles per second.

Substitute the given values in [1] we get;

$y =0.6e^{-1.8t}\cos(2\pi 105t)$

or

$y =0.6e^{-1.8t}\cos(210\pi t)$

(a)

The trigonometric function that models the motion of the string is given by:

$y =0.6e^{-1.8t}\cos(210\pi t)$

(b)

Determine the amount of time t that it takes the string to be damped so that $-0.24\leq y \leq0.24$

Using graphing calculator for the equation

$y =0.6e^{-1.8x}\cos(210\pi x)$

let x = t (time in sec)

Graph as shown below in the attachment:

we get:

the amount of time t that it takes the string to be damped so that $-0.24\leq y \leq0.24$ is, 0.5 sec

4 0

3 years ago

Suppose that the weights of passengers on a flight to Greenland on Frigid Aire Lines are normal with mean 175 pounds and standar

Natali5045456 [20]

Answer:

The minimum weight for a passenger who outweighs at least 90% of the other passengers is 203.16 pounds

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:

$\mu = 175, \sigma = 22$

What is the minimum weight for a passenger who outweighs at least 90% of the other passengers?

90th percentile

The 90th percentile is X when Z has a pvalue of 0.9. So it is X when Z = 1.28. So

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

$1.28 = \frac{X - 175}{22}$

$X - 175 = 22*1.28$

$X = 203.16$

The minimum weight for a passenger who outweighs at least 90% of the other passengers is 203.16 pounds

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3 years ago

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Louis wrote the correct answer.

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