Larry received 1/3 pound of candy from his grandmother and 2/5 pound of candy from his best friend. Larry's sister ate 1/2 pound of candy. How many pounds of candy does Larry have?

The athlete spends 20 minutes on the treadmill and 10 minutes on the elliptical machine. This is determined by setting up and solving a system of linear equations based on the calorie burn rate of each machine and the total calories burned during the workout.

To solve the problem of how many minutes an athlete spends on each machine during their workout, we must set up a system of linear equations. The athlete burns 12 calories per minute on a treadmill and 8.5 calories per minute on an elliptical machine. In a 30-minute workout, the athlete burns a total of 325 calories. Let's denote the time spent on the treadmill as t minutes and the time spent on the elliptical machine as e minutes.

We now have two equations and two unknowns, which can be solved simultaneously.

Using these steps, we find that:

Therefore, the athlete spends 20 minutes on the treadmill and 10 minutes on the elliptical machine during their 30-minute workout.

The store sold a total of 948 balls on Saturday and Friday.

Given that,
A sporting good store sells golf balls and boxes each box has 4 compartments of golf balls each compartment has three golf balls the store sold 37 boxes on Friday and 42 boxes on Saturday how many golf balls did the store sell is to be detrmined.

In mathematics, it deals with numbers of operations according to the statements. There are four major arithmetic operators, addition, subtraction, multiplication and division,

Here,
A sporting goods store sells golf balls and boxes each box has 4 compartments of golf balls each compartment has three golf balls
Implies
each box has = 3 x 4 = 12 balls

Total box sold = 37 + 42 = 79
Total balls = 79 * 12 = 948

Thus, the store sold a total of 948 balls on Saturday and Friday.

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Final answer:

The t-distribution is employed when testing the difference between two population means if the data come from a simple random sample and the population is approximately normally distributed or the sample size is large. It is also used when the population standard deviation is unknown, and a one-tailed or two-tailed test is conducted based on expectations about the direction of the mean difference.

Explanation:

Conditions for Using the t-Distribution to Test Differences Between Two Means

The t-distribution is appropriately used to test the difference between two population means under specific conditions. Firstly, your data must be obtained from a simple random sample. Secondly, the sample should come from a population that is approximately normally distributed, or the sample size should be large enough to compensate for non-normality. This is because the t-test relies on the assumption of normality, especially with smaller sample sizes (<30). If the sample size is small and the population distribution is unknown or not normal, the use of a t-test may not be appropriate without transformation or other techniques to meet the assumptions.

It is also essential to use the t-distribution when the population standard deviation is unknown, and as a result, the sample standard deviation is used as an estimate. Moreover, the t-test can adapt to various research scenarios, considering that it is designed to handle different levels of variability and data types. This flexibility makes the t-distribution a key tool for many scientific studies where the variance is not known ahead of time.

In hypothesis testing, the direction of the mean difference is considered, such as when conducting a one-tailed test or a two-tailed test. A one-tailed test is conducted when an expectation about the direction of the difference exists, while a two-tailed test is used to determine if there is any statistically significant difference between group means without a predefined direction. Lastly, when employing the t-test for paired data, the assumption is that the population from which the differences are drawn is normally distributed.

Answer:

The LCM of 4, 8, and 14 = 56

Step-by-step explanation:

Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.

Multiples of 4:

4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, (56), 60, 64

Multiples of 8:

8, 16, 24, 32, 40, 48, (56), 64, 72

Multiples of 14:

14, 28, 42, (56), 70, 84

Hope this helps!

Answer:56

Step-by-step explanation:

Answer:

72=x3+y: nonlinear

y+1=5(x−9): linear

7y + 2x = 12: linear

4y = 24: linear

I know because I took the k12 test.

Answer:

There is 676,000 combinations

Step-by-step explanation:

VOTE ME FOR BRAINILEST

divide the two:

52.50 / 7 = 7.50 per hour

E) None is correct.

To solve this problem, we need to divide the total volume of 67 gallons 2 quarts by 3.

Convert the total volume to only quarts since both gallons and quarts are involved in the problem. There are 4 quarts in a gallon.

67 gallons = 67 * 4 quarts = 268 quarts

Add the additional 2 quarts: 268 quarts + 2 quarts = 270 quarts

Divide the total quarts by 3 to find the result:

270 quarts ÷ 3 = 90 quarts

Convert the result back to gallons and quarts. Since there are 4 quarts in a gallon:

90 quarts ÷ 4 = 22 gallons with a remainder of 2 quarts

sqrt(86) = 9.273

which is between 9 & 10

To find the initial size of the grey squirrel population, we used the exponential growth formula. After calculating, we determined that the initial population size was 1,875 squirrels.

The student asked how to find the initial size of a grey squirrel population that doubles every 7 years and now has a population of 60,000 after 35 years.

To calculate the initial population size, we'll use the formula for exponential growth: P(t) = P0 × (2t/T), where P(t) is the population at time t, P0 is the initial population size, 2 is the base because the population doubles, t is the number of years, and T is the time it takes for the population to double.

The problem gives us 35 years (t) and a doubling time (T) of 7 years. Plugging these into the equation, we have 60,000 = P0 × (235/7). Simplifying, we get 60,000 = P0 × 25 or 60,000 = P0 × 32. Dividing both sides by 32, we find the initial population size is 1,875 squirrels.

Thus, the initial size of the grey squirrel population was 1,875.

The correct expression is equivalent to 18a+12b+9a+24b is,

⇒ 27a + 36b

Mathematical expression is defined as the collection of the numbers variables and functions by using operations like addition, subtraction, multiplication, and division.

Given that;

The expression is,

⇒ 18a + 12b + 9a + 24b

Now, We can simplify as;

⇒ 18a + 12b + 9a + 24b

⇒ 18a + 9a + 12b + 24b

⇒ 27a + 36b

Thus, The correct expression is equivalent to 18a+12b+9a+24b is,

⇒ 27a + 36b

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"The correct counterexample to disprove the inequality [tex]\( n^3 \leq 3n^2 \)[/tex]for real numbers [tex]\( n \)[/tex]is any real number[tex]\( n \)[/tex]greater than 3.

To find a counterexample, we need to show that there exists at least one real number[tex]\( n \)[/tex] for which the inequality [tex]\( n^3 \leq 3n^2 \)[/tex] does not hold. Let's solve the inequality to find the values of[tex]\( n \)[/tex] that satisfy it:

 [tex]\[ n^3 \leq 3n^2 \][/tex]

[tex]\[ n^3 - 3n^2 \leq 0 \][/tex]

[tex]\[ n^2(n - 3) \leq 0 \][/tex]

The critical points of this inequality are [tex]\( n = 0 \) and \( n = 3 \)[/tex]. We can divide the real number line into three intervals:[tex]\( n < 0 \), \( 0 \leq n \leq 3 \)[/tex], and[tex]\( n > 3 \).[/tex]Let's test each interval:

 1. For[tex]\( n < 0 \), \( n^2 \)[/tex] is positive, but [tex]\( n - 3 \)[/tex] is negative, so the product [tex]\( n^2(n - 3) \)[/tex]is negative, and the inequality holds.

 2. For [tex]\( 0 \leq n \leq 3 \)[/tex], both [tex]\( n^2 \) and \( n - 3 \)[/tex] are non-negative, so the product [tex]\( n^2(n - 3) \)[/tex]is non-positive, and the inequality holds.

3. For [tex]\( n > 3 \)[/tex], both [tex]\( n^2 \)[/tex] and [tex]\( n - 3 \)[/tex] are positive, so the product[tex]\( n^2(n - 3) \)[/tex] is positive, and the inequality does not hold.

Therefore, any real number [tex]\( n \)[/tex] greater than 3 will serve as a counterexample to the inequality[tex]\( n^3 \leq 3n^2 \)[/tex]. For instance, if we take [tex]\( n = 4 \),[/tex] we get:

[tex]\[ 4^3 \leq 3 \cdot 4^2 \][/tex]

[tex]\[ 64 \leq 48 \][/tex]

 This statement is false, as 64 is not less than or equal to 48. Thus, \( n = 4 \)[tex]\( n = 4 \)[/tex] is a valid counterexample."