What are the coordinates of the hole in the graph of the function f(x) ?
f(x)=x2−9x−3

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).

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

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.

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.

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

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)

Answer:

Step-by-step explanation:

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

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.