Ron Co. sold sweatshirts ($20) and baseball caps ($10). If total sales were $2,860 and people bought 6 times as many sweatshirts as caps, how many of each were sold?

The percent of the decimal 2.39 can be expressed as 239%.

To express the decimal 2.39 as a percent, we need to multiply it by 100. This is because a percent is a fraction out of 100.

To do this, we can write 2.39 as 2.39/1 and multiply it by 100/1:

(2.39/1) * (100/1) = 239/1

Therefore, 2.39 as a percent is 239%.

To understand this concept further, let's break it down:

The number 2.39 represents 2 whole units and 0.39 of another unit. When we convert it to a percent, we are essentially comparing it to 100 units.

We can think of the decimal point as moving two places to the right when we multiply by 100. So, 2.39 becomes 239.

Finally, we add the percent symbol (%) to indicate that the value is a percentage.

Therefore, the decimal 2.39 can be expressed in percent as 239%.

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Answer:

Next two number would be –5, –10, –20,–40, –80, –160.

Step-by-step explanation:

Given : –5, –10, –20, –40, . . .

To find : Use inductive reasoning to describe the pattern. Then find the next two numbers in the pattern.

Solution : We have given that

–5, –10, –20, –40, . . .

We can see from given pattern

-5 * 2 = - 10.

- 10 * 2 = - 20 .

-20 * 2 = - 40

So, each number is multiplied by 2 to get next number.

-40 * 2 = - 80 .

- 80 * 2 = - 160 .

Therefore, Next two number would be –5, –10, –20,–40, –80, –160.

To find the total cost of three types of tickets priced at $8.75, $6.50, and $6.50, add the prices together, resulting in a total cost of $21.75.

To write an expression for the total cost of purchasing three types of tickets with prices $8.75, $6.50, and $6.50, you need to add these values together. Here is the step-by-step process of calculating the total cost:

The expression for the total cost of the three tickets is $21.75.

To find the total cost of the three types of tickets, you sum the individual costs to get a total of $21.75.

To write an expression for the total cost of the three types of tickets, we sum up the cost of each ticket. The three types of tickets cost $8.75, $6.50, and $6.50 each. Therefore, the expression for the total cost is:

Total Cost = $8.75 + $6.50 + $6.50

To calculate the total cost, we add these prices together:

$8.75 + $6.50 = $15.25

$15.25 + $6.50 = $21.75

So, the total cost of the three tickets is $21.75.

Keywords

linear equation, variables

Step 1

Define the variables

Let

x-----> number of erasers

y-----> total cost

we know that

[tex]y=3(0.28)+(0.38)x[/tex]

[tex]y=0.84+(0.38)x[/tex]  ------> this is the linear equation that represent the situation

therefore

the answer is

[tex]y=0.84+(0.38)x[/tex]

The total cost of roasting 110 lb of coffee is $765.

Based on the given function, the marginal cost of roasting x pounds of coffee is:

C'(x) = -0.018x + 4.75 for x ≤ 200.

To find the total cost of roasting 110 lb of coffee, we need to integrate the marginal cost function to get the total cost function C(x), and then evaluate C(110).

However, since we are disregarding any fixed costs, we can assume that the total cost function is simply the integral of the marginal cost function.

Integrating C'(x) = -0.018x + 4.75 with respect to x, we get:

C(x) = -0.009x² + 4.75x + C

where C is the constant of integration. Since we are disregarding any fixed costs, we can assume that C = 0.

Therefore, the total cost of roasting 110 lb of coffee is:

C(110) = -0.009(110)² + 4.75(110) = 765

The total cost of roasting 110 lb of coffee is $765.

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Final answer:

To find the total cost of roasting 110 lb of coffee, integrate the marginal cost function C'(x) = -0.018x + 4.75 over the range of x values. Substitute 110 for x in the total cost function to find the total cost.

Explanation:

To find the total cost of roasting 110 lb of coffee, we need to integrate the marginal cost function over the desired range of x values. The marginal cost function is given by C'(x) = -0.018x + 4.75. Integrating this function will give us the total cost function C(x).

Integrating C'(x) gives C(x) = -0.009x^2 + 4.75x + C, where C is the constant of integration. Since we are disregarding any fixed costs, we can ignore the constant term C.

Now we can substitute 110 for x in the total cost function to find the total cost of roasting 110 lb of coffee: C(110) = -0.009(110)^2 + 4.75(110) = -108.9 + 522.5 = $413.6.

The charge for using 50 therms is $16.803, while the charge for using 500 therms is $64.413. The model that relates the monthly charge for x therms of gas can be computed using specific conditions. Graphing the function will result in a piecewise linear graph with different slopes for different ranges of therms.

To find the charge for using 50 therms in a month, we can use the rate schedule provided. The charge for the first 50 therms is $0.33606 per therm. So, the charge for using 50 therms would be 50 x $0.33606 = $16.803.

For using 500 therms in a month, we need to consider both the charge for the first 50 therms and the charge for the additional therms. The charge for the first 50 therms is $0.33606 per therm, and for the additional therms (450 therms), the charge is $0.10580 per therm. So, the total charge would be (50 x $0.33606) + (450 x $0.10580) = $16.803 + $47.610 = $64.413.

The model that relates the monthly charge for x therms of gas can be represented as follows:

C(x) = $15.95 + ($0.33606 x therms) if x ≤ 50

C(x) = $15.95 + ($0.33606 x 50) + ($0.10580 x (therms-50)) if x > 50

Graphing this function will result in a piecewise linear graph with a slope of $0.33606 for the first 50 therms and a slope of $0.10580 for any additional therms beyond 50.

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The half-life of the element is approximately 542.78 million years.

To find out what percent of the radioactive element remains after 500 million years, we can use the exponential decay formula for radioactive decay:

[tex]\[ N(t) = N_0 \times e^{-kt} \][/tex]

Where:

- [tex]\( N(t) \)[/tex] is the amount of radioactive material remaining at time \( t \)

- [tex]\( N_0 \)[/tex] is the initial amount of radioactive material

- k is the decay constant

- t is the time elapsed

Given that 60% of the radioactive element remains after 400 million years, we know that [tex]\( N(t) = 0.60N_0 \)[/tex]  when [tex]\( t = 400 \)[/tex] million years. We also know that [tex]\( N_0 \)[/tex] represents the initial amount of radioactive material, which will cancel out when we're finding the ratio of remaining material, so we don't need its exact value.

Substituting these values into the exponential decay formula:

[tex]\[ 0.60N_0 = N_0 \times e^{-400k} \][/tex]

We can simplify this to find the decay constant k:

[tex]\[ 0.60 = e^{-400k} \][/tex]

Taking the natural logarithm (ln) of both sides to solve for k:

[tex]\[ \ln(0.60) = -400k \][/tex]

[tex]\[ k = \frac{\ln(0.60)}{-400} \][/tex]

Using this value of k, we can find out what percent of the radioactive element remains after 500 million years:

[tex]\[ N(500) = N_0 \times e^{-500k} \][/tex]

[tex]\[ N(500) = N_0 \times e^{-500 \times \frac{\ln(0.60)}{-400}} \][/tex]

Now, to find the half-life of the element, we know that the half-life [tex](\( T_{\frac{1}{2}} \))[/tex] is the time it takes for the amount of radioactive material to decrease by half. In other words, when [tex]\( N(t) = \frac{1}{2}N_0 \),[/tex] we have:

[tex]\[ \frac{1}{2}N_0 = N_0 \times e^{-kT_{\frac{1}{2}}} \][/tex]

[tex]\[ \frac{1}{2} = e^{-kT_{\frac{1}{2}}} \][/tex]

[tex]\[ \ln\left(\frac{1}{2}\right) = -kT_{\frac{1}{2}} \][/tex]

[tex]\[ T_{\frac{1}{2}} = \frac{\ln(2)}{k} \][/tex]

Now, we can calculate both the percentage remaining after 500 million years and the half-life of the element using the calculated value of k. Let's do that.

First, let's calculate the value of \( k \):

[tex]\[ k = \frac{\ln(0.60)}{-400} \][/tex]

[tex]\[ k ≈ \frac{-0.5108}{-400} \][/tex]

[tex]\[ k = 0.001277 \][/tex]

Now, let's use this value of k to find out what percent of the radioactive element remains after 500 million years:

[tex]\[ N(500) = N_0 \times e^{-500k} \][/tex]

[tex]\[ N(500) = N_0 \times e^{-500 \times 0.001277} \][/tex]

[tex]\[ N(500) ≈ N_0 \times e^{-0.6385} \][/tex]

Now, let's find out what percent this is of the original amount [tex](\( N_0 \))[/tex]:

[tex]\[ \frac{N(500)}{N_0} = e^{-0.6385} \][/tex]

[tex]\[ \frac{N(500)}{N_0} ≈ 0.5274 \][/tex]

So, approximately 52.74% of the radioactive element remains after 500 million years.

Now, let's calculate the half-life of the element:

[tex]\[ T_{\frac{1}{2}} = \frac{\ln(2)}{k} \][/tex]

[tex]\[ T_{\frac{1}{2}} = \frac{\ln(2)}{0.001277} \][/tex]

[tex]\[ T_{\frac{1}{2}} = \frac{0.6931}{0.001277} \][/tex]

[tex]\[ T_{\frac{1}{2}} = 542.78 \text{ million years} \][/tex]

So, the half-life of the element is approximately 542.78 million years.

Final answer:

There is a 40% chance that one of the five doors will have a car behind it as there are two doors with cars and five doors in total.

Explanation:

To find the percentage of doors that have cars behind them in the 21st century version of Let’s Make a Deal with five doors, we can use a simple ratio. Since two out of the five doors have cars behind them, we can set up the ratio as 2 cars to 5 total doors.

The calculation for the percentage would be: (Number of doors with cars / Total number of doors) × 100. Plugging in the numbers gives us:

(2 / 5) × 100 = 40%

Therefore, the percentage of doors with cars behind them is 40%.

Final answer:

The correct answer is option b. percentage, probability, and proportion, because these terms relate to ratios or fractions of a whole and can often be used interchangeably in various statistical contexts. So the correct option is b.

Explanation:

The question is asking which of the following groups contain terms that can be used interchangeably. The correct answer is b. percentage, probability, proportion. These terms can often be used interchangeably because a percentage is a way of expressing a proportion as a number out of 100, and probability is the measure of the likelihood that an event will occur, often expressed as a decimal or a proportion. None of the other options provided (a, c, and d) contain directly interchangeable terms.

Steps:

The probability that the second number rolled is greater than or equal to the first number when rolling a fair eight-sided die twice is 1/16.

To find the probability that the second number rolled is greater than or equal to the first number when rolling a fair eight-sided die twice, we need to determine the favorable outcomes and the total number of possible outcomes.

There are 8 possible outcomes for the first roll and 8 possible outcomes for the second roll. However, since the die is fair and has different numbers on each side, the second roll can only be greater than or equal to the first roll for 4 of the outcomes.

Therefore, the probability is 4 favorable outcomes out of 64 total outcomes, which simplifies to a probability of B

Answer:

[tex]90< \gamma< 180[/tex]

Step-by-step explanation:

An acute angle is an angle that measures more than 0º and less than 90º. And a right angle is an angle that measures exactly 90º. Thus:

Let:

[tex]\alpha = Right\hspace{3}angle=90^{\circ}\\\beta= Acute\hspace{3}angle\hspace{10}0^{\circ}<\beta<90^{\circ}\\\gamma= Sum\hspace{3} of\hspace{3} the\hspace{3} two\hspace{3} angles =\alpha +\beta[/tex]

So, in this sense, the sum of the two angles can´t be equal or greater than 180º, since [tex]\beta<90^{\circ}[/tex] , also it has to be at least greater than 90º since [tex]\beta>0^{\circ}[/tex]

Therefore the sum of the two angles is:

[tex]90^{\circ}< \gamma< 180^{\circ}[/tex]

In another words, the sum is greater than 90º and less than 180º

The ratio will remain constant. Then the number of cups of sugar used in 1 and 1/2 batches of these cookies will be 1.125.

A ratio is an ordered set of integers a and b expressed as a/b, with b never equaling 0. A percentage is a mathematical expression in which two things are equal.

A recipe for 1 batch of cookies uses 3/4 cup of sugar.

Then the number of cups of sugar used in 1 and 1/2 batches of these cookies will be

We know that the ratio will remain constant. Then we have

1 and 1/2 can be written as 3/2.

Let x be the number of cups of sugar.

Then the ratio will be

[tex]\rm \dfrac{x}{3/2} = \dfrac{3/4}{1}\\\\x = \dfrac{3 \times 3}{2 \times 4}[/tex]

Then we have

x = 9/8

x = 1.125 cups of sugar

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The mathematical symbol that makes the expression -3/4 -11/12 true is subtraction.

The student is likely asking which mathematical symbol (operation) would make the expression -3/4 -11/12 a true sentence. The symbols that could apply are addition (+), subtraction (-), multiplication (x), or division (). To determine which symbol makes the sentence true, one would typically compare the result of applying each operation to the numeric values given.

For example, to test addition:

-3/4 + (-11/12) = (-9/12) + (-11/12) = -20/12 = -5/3

To test subtraction:

-3/4 - (-11/12) = (-9/12) - (-11/12) = 2/12 = 1/6

Therefore, subtraction makes the original expression a true one, because subtracting a negative is the same as adding the positive equivalent, which would simplify to an expression that equals 1/6.