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2 years ago

To raise money for a local charity, students organized a wake-a-thon

Mathematics

1 answer:

arsen [322]2 years ago

6 0

Answer:

a) T= 3.125h

b) $75

Step-by-step explanation:

Given: We know Tania raised $50 for staying awake 16 hours. We also know the amount of money raised varies directly with the time.

We can divide $50 into 16 hours.

50/16 = 25/8

25/8 = 3.125

3.125 = 3 1/8 (You can use a fraction or decimal, whichever is preferred)

That's how much money she earns per hour. It may not make sense because you have 1/8 of a dollar, so it's okay if you're a little confused on that.

Now let's answer the first question: Write an equation relating the money Tania raised and the amount of time, in hours, she stayed awake.

We know Tania earns $3.125 per hour, so we can make it this equation

T= 3.125h

T can represent Tania's earnings, and 3.125 is how much she earns h, hourly.

Now onto the second question: How much would she have raised by staying awake for 24 h?

We need to multiply 3.125, or 3 1/8, 24 times. Your answer should be $75.

(I apologize in advance if I misread something or made a mistake)

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The mean gross annual incomes of certified welders are normally distributed with the mean of $20,000 and a standard deviation of

grandymaker [24]

Answer:

At the 0.10 level of significance the z table gives critical values of -1.645 and 1.645 for two-tailed test.

Step-by-step explanation:

We are given that the mean gross annual incomes of certified welders are normally distributed with the mean of $20,000 and a standard deviation of $2,000.

The ship building association wishes to find out whether their welders earn more or less than $20,000 annually.

Let $\mu$ = mean gross annual incomes of certified welders

So, Null Hypothesis, $H_0$ : $\mu$ = $20,000

Alternate hypothesis, $H_A$ : $\mu \neq$ $20,000

Here, null hypothesis states that the mean income of welders is equal to $20,000.

On the other hand, alternate hypothesis states that the mean income of welders is not $20,000.

Also, the test statistics that would be used here is One-sample z test statistics as we know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean income

$\sigma$ = population standard deviation = $2,000

n = sample size

Now, at the 0.10 level of significance the z table gives critical values of -1.645 and 1.645 for two-tailed test.

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2 years ago

The data represents the body mass index (BMI) values for 20 females. Construct a frequency distribution beginning with a lower

Papessa [141]

Answer:

Given:

Body mass index values:

17.7

29.4

19.2

27.5

33.5

25.6

22.1

44.9

26.5

18.3

22.4

32.4

24.9

28.6

37.7

26.1

21.8

21.2

30.7

21.4

Constructing a frequency distribution beginning with a lower class limit of 15.0 and use a class width of 6.0.

we have:

Body Mass Index____ Frequency

15.0 - 20.9__________3( values of 17.7, 18.3, & 19.2 are within this range)

21.0 to 26.9__________8 values are within this range)

27.0 - 32.9____________ 5 values

33.0 - 38.9____________ 2 values

39.0 - 44.9 _____________2 values

The frequency distribution is not a normal distribution. Here, although the frequencies start from the lowest, increases afterwards and then a decrease is recorded again, it is not normally distributed because it is not symmetric.

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2 years ago

What are the attributes of fractions that are equivalent to 100%

castortr0y [4]

The denominator and numerator have to be the same. I'm sosorry if this is wrong but i really tried <3

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2 years ago

The pucks used by the National Hockey League for ice hockey must weigh between and ounces. Suppose the weights of pucks produced

Dahasolnce [82]

Answer:

$P(5.5$

And we can find this probability using the normal standard distribution or excel and we got:

$P(-2.769$

Step-by-step explanation:

For this case we assume the following complete question: "The pucks used by the National Hockey League for ice hockey must weigh between 5.5 and 6 ounces. Suppose the weights of pucks produced at a factory are normally distributed with a mean of 5.86 ounces and a standard deviation of 0.13ounces. What percentage of the pucks produced at this factory cannot be used by the National Hockey League? Round your answer to two decimal places. "

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:

$X \sim N(5.86,0.13)$

Where $\mu=5.86$ and $\sigma=0.13$

We are interested on this probability

$P(5.5$

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:

$P(5.5$

And we can find this probability using the normal standard distribution or excel and we got:

$P(-2.769$

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2 years ago

9. Your school offers two English classes, three math classes and three

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There are 12 different ways to organize your schedule

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