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Natali5045456 [20]
3 years ago
6

Raymond just got done jumping at Super Bounce Trampoline Center. The total cost of his session was \$43.25$43.25dollar sign, 43,

point, 25. He had to pay a \$7$7dollar sign, 7 entrance fee and \$1.25$1.25dollar sign, 1, point, 25 for every minute he was on the trampoline. Write an equation to determine the number of minutes (to)(, t, )that Raymond was on the trampoline. Find the number of minutes he was on the trampoline.
Mathematics
1 answer:
rodikova [14]3 years ago
5 0

Answer:

Raymond was for 29 minutes on the trampoline.

Step-by-step explanation:

We are given the following in the question:

Total cost of session = $43.25

Entrance fee = $7

Cost per minute = $1.25

Let t be the minutes that Raymond was on the trampoline.

Thus, we can write

Total cost =

Entrance fee + cost per mute(minutes that Raymond was on trampoline)

C(t) = 7 + 1.25t

Putting the values, we get,

43.25 = 7 + 1.25(t)\\1.25t = 43.25 - 7\\1.25t = 36.25\\\\t = \dfrac{36.25}{1.25}\\\\t = 29

Thus, Raymond was for 29 minutes on the trampoline.

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A food processor packages orange juice in small jars. The weights of the filled jars are approximately normally distributed with
Ratling [72]

Answer:

P(X>10.983)=P(\frac{X-\mu}{\sigma}>\frac{10.983-\mu}{\sigma})=P(Z>\frac{10.983-10.5}{0.3})=P(z>1.61)

And we can find this probability using the complement rule and with excel or the normal standard table:

P(z>1.61)=1-P(z

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

X \sim N(10.5,0.3)  

Where \mu=10.5 and \sigma=0.3

We are interested on this probability

P(X>10.983)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(X>10.983)=P(\frac{X-\mu}{\sigma}>\frac{10.983-\mu}{\sigma})=P(Z>\frac{10.983-10.5}{0.3})=P(z>1.61)

And we can find this probability using the complement rule and with excel or the normal standard table:

P(z>1.61)=1-P(z

8 0
3 years ago
First correct response gets brainliest
Simora [160]

Answer:

x = 13

Step-by-step explanation:

This question is based on Secant Secant theorem.

Secant Secant theorem gives us the following formula:

(AB + BD)AB = (AC + CE).AC

From the above question we have the following parameters

AB = 5

BD = x

AC = 7.5

CE = 4.5

Hence,

(AB + BD)AB = (AC + CE).AC

(5 + x)5 = (7.5 + 4.5)7.5

25 + 5x = 90

Collect like terms

5x = 90 - 25

5x = 65

x = 65/5

x = 13

5 0
2 years ago
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