if Anne's age is represented by the variable (a) then.... Epress Jenny's age in terms of (a) if Jenny is 3 years less than 1/2 of Anne's age.




Express Fraz's age in terms of (a) if Fraz is twice the sum of Anne's age and 7

double x + 3= 2x + 6. double four is 8. 8 * 2x + 6 = 16x + 48

For this case we have that if 20 trees produce 1 gallon of syrup each, then we have 20 gallons of syrup.

We must find the number of containers for the 20 gallons, knowing that each container has a capacity of 3 gallons.

[tex]\frac {20} {3} = 6.67[/tex]

So, we need 7 containers.

Answer:

7 containers

Answer:

(2,3)

Step-by-step explanation:

Given : [tex]x + y = 5[/tex]

           [tex]y=\frac{x}{2}+2[/tex]

To find: what point do the lines representing the two equations intersect

Solution:

[tex]x + y = 5[/tex]

[tex]y=\frac{x}{2}+2[/tex]

Plot the lines on the graph

Green line denotes  [tex]x + y = 5[/tex]

Purple line denotes [tex]y=\frac{x}{2}+2[/tex]

So, Intersection point = (2,3)

Refer the attached figure

So, option C is true

Hence Intersection point is (2,3)

A prism with a volume of 270 m³ is dilated by a factor of 3. What is the volume of the dilated prism? Enter your answer in the box.

If we dilate each dimension by 3, then the volume of the figure will be dilated by [tex] 3^{3} [/tex] .

[tex] 3^{3} [/tex] =3*3*3 =27

So, the volume of the dilated prism is [tex] 3^{3} [/tex] *volume of original prism

Volume of dilated prism=27*270[tex] m^{3} [/tex]

Volume of dilated prism=7290[tex] m^{3} [/tex]

The volume of a prism when dilated by a factor of 3 increases to 7290 m³,

A prism with a volume of 270 m³ that is dilated by a factor of 3 will have its new volume calculated by cubing the scale factor and multiplying it by the original volume.

   Original Volume (V1) = 270 m³

   Scale Factor (k) = 3

   Dilated Volume (V2) = V1 × k³

  So, V2 = 270 m³ × 3³

  V2 = 270 m³ × 27

  V2 = 7290 m³
Hence, the volume of the dilated prism is 7290 m³.

Final answer:

The probability of selecting five defective batteries from a set of seven batteries, five of which are defective, is calculated by multiplying the individual probabilities of each selection. This results in an approximate probability of 0.048.

Explanation:

The probability of selecting all defective batteries from a group where 7 batteries are available and 5 are defective is obtained by calculating the probability of selecting a defective battery five times in a row without replacement.

The probability of the first battery being defective is 5/7, as there are 5 defective out of 7 total. If the first one is defective, there are now 4 defective batteries left out of 6. This makes the probability of the second being defective 4/6. We continue this pattern until we have:

Probability of 1st defective battery = 5/7

Probability of 2nd defective battery = 4/6

Probability of 3rd defective battery = 3/5

Probability of 4th defective battery = 2/4

Probability of 5th defective battery = 1/3

To find the overall probability of all being defective, we multiply these probabilities together:

P(all defective) = (5/7) × (4/6) × (3/5) × (2/4) × (1/3)

When we calculate this, we get:

P(all defective) = (5×4×3×2×1) / (7×6×5×4×3) = 120 / 2520 = 0.0476

Therefore, the probability is approximately 0.048 (rounded to three decimal places).

Graphing x \<= -8 involves placing a solid circle at -8 on a number line and shading everything to the left to represent all numbers less than or equal to -8.

To graph x \<= -8 on a number line, you need to include every number that is less than or equal to -8. Start with a number line and locate the point -8 on it. Then, draw a solid circle at -8 to indicate that -8 is included in the solution (because of the '<=' symbol). Next, shade or draw a thick line to the left of -8, going towards the negative direction, to cover all numbers that are less than -8. This shaded area represents all the possible values for x that satisfy the inequality x \<= -8.

Visual Representation on Number Line

A visual representation helps to understand the solutions to the inequality. When graphing, remember that the area to the left of the chosen point on a number line includes all numbers less than the indicated value, and if the circle is solid, that point is also part of the solution set.

The process described here provides a straightforward approach to graphing linear inequalities on a number line, which is a fundamental concept in algebra.

On first day the tank had 8000 liters of water.

Half is removed so half of 8000 is 4000.

Subtracting 8000-4000 =4000

So at the end of first day 4000 liters of water is there.

Water in the tank at the end of second day:

Again half of 4000 is removed.

half of 4000 is 2000.

subtracting 4000-2000 =2000.

At the end of second day 2000 l of water is present in tank.

At the end of third day:

Again half of 2000 is removed.

half of 2000 is 1000.

subtracting 2000-1000 =1000.

At the end of third day 1000 l of water is present in tank.

At the end of fourth day:

Again half of 1000 is removed.

half of 1000 is 500.

subtracting 1000-500 =500.

At the end of fourth day 500 l of water is present in tank.

At the end of fifth day:

Again half of 500 is removed.

half of 500 is 250.

subtracting 500-250=250.

At the end of fifth day 250 l of water is present in tank.

At the end of sixth day:

Again half of 250 is removed.

half of 250 is 125.

subtracting 250 - 125 =125.

At the end of sixth day 125 l of water is present in tank.

At the end of seventh day:

Again half of 125 is removed.

half of 125 is 62.5.

subtracting 125-62.5 =62.5.

At the end of seventh day 62.5 l of water is present in tank.

At the end of eighth day:

Again half of 62.5 is removed.

half of 62.5 is 31.25

subtracting 62.5 - 31.25 =31.25.

At the end of eighth day 31.25 l of water is present in tank.

We have been given that Tina's favorite shade of teal is made with 7 ounces of blue paint for every 5 ounces of green paint and Tyler's favorite shade of teal is made with 5 ounces of blue paint for every 7 ounces of green paint.

We can express the ratio of blue paint to green paint for both Tina as well as Tyler's favorite shades of teal as shown below.

For Tina: [tex]\frac{7}{5}[/tex]

For Tyler: [tex]\frac{5}{7}[/tex]

In order to compare the two ratios, we will have to make the denominators equal. Let us make denominators equal for both these ratios.

For Tina: [tex]\frac{7\cdot7}{5\cdot7}=\frac{49}{35}[/tex]

For Tyler: [tex]\frac{5\cdot5}{7\cdot5}=\frac{25}{35}[/tex]

Therefore, we can see that for equal amount of Green paint, Tina's favorite shade of teal contains [tex]\frac{49}{25}[/tex] times, that is almost twice the Blue paint that Tyler's favorite shade of teal contains.

Tina's favorite shade is more blue than Tyler's

Answer:

Step-by-step explanation:

The edge length of a cube with a volume of 64 cubic units can be expressed as 4 units, which is the cube root of 64, or algebraically as V⅛ or 64⅛.

To express the edge length of a cube with a volume of 64 cubic units, we can use the formula for the volume of a cube, which is V = L³, where V is volume and L is the length of an edge. Since we have the volume (V), we can solve for L (the edge length) by taking the cube root of the volume. Consequently, the cube root of 64 is 4, so one way to express the edge length is simply as 4 units.

Alternatively, we can express the edge length in terms of the formula for volume. If the volume V is a constant 64 cubic units, the edge length can be represented as V⅛, or 64⅛ in this specific case. This second expression emphasizes the relationship between volume and edge length in a cube and is particularly useful in algebraic and geometric applications.

Answer: A. I only

Step-by-step explanation:  

Most of the observations are on the lower end, so the distribution is skewed right. The interquartile range is indicated by the length of the box, about 14. The median is indicated by the vertical line running through the box, about 13