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2 years ago

It costs twelve dollars to get in to the fair. Tickets for rides cost extra and are d dollars each.

Mathematics

2 answers:

Anton [14]2 years ago

7 0

Answer:

Answer:

12 + 20d

Step-by-step explanation:

OG fare = 12

Ticket price =d

20 Tickets =20d

Paraphin [41]2 years ago

6 0

Answer:

$12 + 20 d

Step-by-step explanation:

Right of the bat, we know she‘ll pay $12 just to get in to the fair.

If Tyra wanted to buy 20 tickets, it’ll cost her d dollars each. Which is 20 d.

Add 12 and 20(d) together and it’ll determine how much Tyra will spend in total at the fair.

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Can someone please help me with my Imagine Math? My teacher assigned are class to do this even though we haven't learned about t

Daniel [21]

Answer:

first graph is not a function, second graph is a function, 3rd is not enough information, 4th is a function, 5th is not a function, 6th is a function

Step-by-step explanation:

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2 years ago

A manager at a local manufacturing company has been monitoring the output of one of the machines used to manufacture chromium sh

denpristay [2]

Answer:

0.682 = 68.2% probability that the average length of these 16 shells will be between 116 and 120 centimeters when the machine is operating "properly".

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Normally distributed with a mean of 118 centimeters and a standard deviation of 8 centimeters.

This means that $\mu = 118, \sigma = 8$

Sample of 16 shells

This means that $n = 16, s = \frac{8}{\sqrt{16}} = 2$

What is the probability that the average length of these 16 shells will be between 116 and 120 centimeters when the machine is operating "properly?"

This is the pvalue of Z when X = 120 subtracted by the pvalue of Z when X = 116.

X = 120

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

By the Central Limit Theorem

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

$Z = \frac{120 - 118}{2}$

$Z = 1$

$Z = 1$ has a pvalue of 0.841

X = 116

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

$Z = \frac{116 - 118}{2}$

$Z = -1$

$Z = -1$ has a pvalue of 0.159

0.841 - 0.159 = 0.682

0.682 = 68.2% probability that the average length of these 16 shells will be between 116 and 120 centimeters when the machine is operating "properly".

3 0

2 years ago

Weight of a rock: In a geology course, students are learning to use a balance scale to accurately weigh rocks. One student plans

statuscvo [17]

Answer:

The student who weighted the rock 5 times has a 95% confidence interval of (25.2, 29.1) which is guaranteed to be more wider (less precise) than the other student who weighted the rock 20 times.

Step-by-step explanation:

What is Confidence Interval?

The confidence interval represents an interval that we can guarantee that the target variable will be within this interval for a given confidence level.

The confidence interval is given by

$CI = \bar{x} + t_{\alpha/2}(\frac{\sigma}{\sqrt{n} } ) \\$

Where $\bar{x}$ is the mean weight $\sigma$ is the standard deviation $t_{\alpha/2}$ is the critical value from t-table and n is the sample size.

The term $t_{\alpha/2}(\frac{\sigma}{\sqrt{n} } )$ is known as margin of error.

As the sample size is decreased the corresponding margin of error increases which results in wider confidence interval which means smaller precision.

The student who weighted the rock 5 times has a 95% confidence interval of (25.2, 29.1) which is guaranteed to be more wider (less precise) than the other student who weighted the rock 20 times.

We can say with 95% confidence that the true mean weight of the rock is within the interval of (25.2, 29.1).

7 0

2 years ago

In a lecture demonstration, an object is suspended from a spring scale which reads 8N when the object is in air. The object is t

katovenus [111]

Answer:

a) density of the object is 3995.01, b) the weight scale reads 22N c) the sum individually will be the same with when added together.

Step-by-step explanation:

The weight of the object in air is 8N,

and weight = Mass * acceleration due to gravity = m * 9.81

8/9.81 = 0.815,

upthrust( force acting on the body from the liquid impeding the immersion) on the body when fully submerged = weight in air - weight in water = 8N - 6N =2N

Upthrust = weight of water displaced = 2N = mass * acceleration

2/9.81 = 0.204kg

density of water(1000kg/m^3) = mass of water / volume of water

volume of water displaced = 0.204/1000 = 0.000204m^3 (204cm^3)

volume of water displaced = volume of the solid

density of solid = mass/ volume = 0.815/0.000204 = 3995.01kg/m^3

b) when fully submerge in water the the scale experience according to newton third law of motion ( equal and opposite reaction of forces) additional 2N push so that total weight with the fully submerge solid = 20N + upthrust = 20N + 2N =22N

c) the of two scale reading is before (8N + 20N = 28N) and after (6N + 22N = 28) since there is no loss of matter; the demonstration was in equilibrium.

6 0

2 years ago

NEED HELP!

TEA [102]

Answer:

30 minutes

Step-by-step explanation:

20 * 96 / 64 = 30

8 0

2 years ago