Roxanne wants to save up enough money so that she can buy a video game. The game costs $59. Roxanne has $22 saved from mowing the grass. In order to make more money, she plans to pressure wash her neighbors’ driveways. She plans to charge $5 for each driveway she pressure washes, and any extra money she makes beyond $59 she will save to buy another video game. How many driveways does she need to pressure wash to reach her goal?

The work done to lift the coal is 6.8*10^5 ft-lb

Data;

• weight of cable = 6lb/ft
• weight of coal = 500lb
• length of mine shaft = 400ft

Let the distance (ft) below top of the shaft be represented by x

The weight of the coal  to be lifted from mine = 500lb

Work Done

The work done to lift the coal is

Taking the summation limit

we can take the integration of both sides where x will the range from 0 to 400.

From the calculations above, the work done is 6.8*10^5 ft-lb

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Answer:

8.2 *10^5 ft lb

Step-by-step explanation:

Wcoal = 800*500  = 400000 = 4*10^5 ft lb

Using a right Riemann sum

The width of the entireregion to be estimated = 400 - 0 = 400

Considering 8 equal subdivisions, then the width of each rectangular division is 400/8 = 50

F(x) = 6(400-y)

Wrope = 50(2100) + 50(1800) + 50(1500) + 50(1200) + 50(900) + 50(600)

+ 50(300) + 50(0) = 420000 = 4.2 * 10^5 ft lb

Note: Riemann sum is an approximation, so may not give a accurate value

work done =  Wcoal + Wrope= 4*10^5+ 4.2 * 10^5 = 8.2 *10^5 ft lb

Answer:

a. Null hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is equal to the mean plasma aluminum level of the population of infants not receiving antacids.

Complementary alternative hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is different from the mean plasma aluminum level of the population of infants not receiving antacids.

b. (32.1, 42.3)

c. p-value < .00001

d. The null hypothesis is rejected at the α=0.05 significance level

e. Reformulated null hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is equal to the mean plasma aluminum level of the population of infants not receiving antacids.

Reformulated complementary alternative hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is higher than the mean plasma aluminum level of the population of infants not receiving antacids.

p-value equals < .00001. The null hypothesis is rejected at the α=0.05 significance level. This suggests that being given antacidsgreatly increases the plasma aluminum levels of children.

Step-by-step explanation:

a. Null hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is equal to the mean plasma aluminum level of the population of infants not receiving antacids.

Complementary alternative hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is different from the mean plasma aluminum level of the population of infants not receiving antacids. This may imply that being given antacids significantly changes the plasma aluminum level of infants.

b. Since the population standard deviation σ is unknown, we must use the t distribution to find 95% confidence limits for μ. For a t distribution with 10-1=9 degrees of freedom, 95% of the observations lie between -2.262 and 2.262. Therefore, replacing σ with s, a 95% confidence interval for the population mean μ is:

(X bar - 2.262\frac{s}{\sqrt{10} } , X bar + 2.262\frac{s}{\sqrt{10} })

Substituting in the values of X bar and s, the interval becomes:

(37.2 - 2.262\frac{7.13}{\sqrt{10} } , 37.2 + 2.262\frac{7.13}{\sqrt{10} })

or (32.1, 42.3)

c. To calculate p-value of the sample , we need to calculate the t-statistics which equals:

t=\frac{(Xbar-u)}{\frac{s}{\sqrt{10} } } = \frac{(37.2-4.13)}{\frac{7.13}{\sqrt{10} } } = 14.67.

Given two-sided test and degrees of freedom = 9, the p-value equals < .00001, which is less than 0.05.

d. The mean plasma aluminum level for the population of infants not receiving antacids is 4.13 ug/l - not a plausible value of mean plasma aluminum level for the population of infants receiving antacids. The 95% confidence interval for the population mean of infants receiving antacids is (32.1, 42.3) and does not cover the value 4.13. Therefore, the null hypothesis is rejected at the α=0.05 significance level. This suggests that being given antacids greatly changes the plasma aluminum levels of children.

e. Reformulated null hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is equal to the mean plasma aluminum level of the population of infants not receiving antacids.

Reformulated complementary alternative hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is higher than the mean plasma aluminum level of the population of infants not receiving antacids.

Given one-sided test and degree of freedom = 9, the p-value equals < .00001, which is less than 0.05. This result is similar to result in part (c). the null hypothesis is rejected at the α=0.05 significance level. This suggests that being given antacids greatly increases the plasma aluminum levels of children.

Answer:

710 hairbands

Step-by-step explanation:

Total hairbands = (5*dollars spent) + 90

Max hairbands = (5 * 124) + 90

The length of sides x in the simplest radical form is 2√2

Properties of a right angled triangle:

• One of the angle is equals to 90 degree.
• The sides can be found using Pythagoras theorem
• Trigonometric ratios can be used to find the angles.

Let's find x using trigonometric ratios

sin ∅ = opposite / hypotenuse

Therefore,

sin 30° = √2 / x

x sin 30° = √2

x = √2 / sin 30°

x =

x = × 2

x = 2√2

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Final answer:

Probability of not witnessing a major tornado hit over the next ten years would be derived from historical data based on frequency of such events. Probability is a measure of uncertainty and these estimates can change as more information becomes available. Actual probability calculations can involve complex concepts such as Poisson distributions and conditional probabilities.

Explanation:

The question asks for the probability of not witnessing a major tornado hit over the next ten years. Probability in this context would typically be derived from historical data on frequency of such events. The data provided seems unrelated, so I'll provide a hypothetical example instead. If historical data shows that the state gets hit by a major tornado every 20 years on average, you could argue there's a 50% chance (1 out of 2 decades) that the state will not be hit by a major tornado in the next ten years.

However, keep in mind that probability is a measure of uncertainty, not certainty. Probabilities are estimates based on what we currently know and can change as more information becomes available. For example, if climate change leads to more severe weather events, the chance of a tornado might increase.

This is a simplified model and actual probability calculations can be much more complex, involving things like Poisson distributions and conditional probabilities which are often used in analyses of rare events such as natural disasters.

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