Jill has $8.75. Jack has three times as much as Jill, but he spent $5.00 to buy a book. How much money does Jack have?

A shorter style of cup is stacked tall. The graph displays the height of the stack in centimeters for different numbers of cups.

Strike441 [17]

Answer:

0.5

Step-by-step explanation:

We are looking for how much height is gained per cup added.

Height per cup added can be calculated by finding the slope of the line that runs through the two given points on the graph, (3, 5.5) and (8, 8).

Formula for slope =

Let,

3 0

3 years ago

Read 2 more answers

Find the value of this expression if x = 5 and y = -1

geniusboy [140]

Answer:

5/6

Step-by-step explanation:

8 0

2 years ago

I need help with premetier and area for 7 8 9

SCORPION-xisa [38]

The answers to the three shapes are:

Figure 7; perimeter = 16.2mm, area = 12.64 square mm

Figure 8; perimeter = 15.2inches, area = 9.61 square inches

Figure 9; perimeter = 21.47yards, area = 16.81 square yards

<h3>What is the perimeter of a shape?</h3>

Perimeter is the outside boundary of a plane shape.

Analysis:

for figure 7, perimeter = s + s + s = 3s = 3(5.4) = 16.2mm

Height of triangle = $\sqrt{((5.4)^{2} - (2.7)^{2} }$ = 4.68

Area of triangle = 1/2 base x height = 1/2 x 5.4 x 4.68 = 12.64 $mm^{2}$

For figure 8, perimeter = b + b + a = 5.9 + 5.9 + 3.4 = 15.2 inches

Height of triangle = $\sqrt{(5.9)^{2} - (1.7)^{2} }$ = 5.65inches

Area of triangle = 1/2 base x height = 1/2 x 3.4 x 5.65 = 9.61 $inches^{2}$

For figure 9, perimeter = b + a + c = 8.2 + 4.1 + 9.17 = 21.47yards

Area of triangle = 1/2 x base x height = 1/2 x 8.2 x 4.1 = 16.81 $yards^{2}$

In conclusion, the perimeter and area of the given shapes are:

Figure 7; perimeter = 16.2mm, area = 12.64 square mm

Figure 8; perimeter = 15.2inches, area = 9.61 square inches

Figure 9; perimeter = 21.47yards, area = 16.81 square yards

Learn more about perimeter of plane shapes: brainly.com/question/2569205

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5 0

2 years ago

The weight of an organ in adult males has a bell-shaped distribution with a mean of 310 grams and a standard deviation of 25 gra

Shkiper50 [21]

Answer:

About 68% of organs will be between 300 grams and 320 grams, about 95% of organs will be About 68% of organs will be between 300 grams and 320 grams, About 68% of organs will be between 300 grams and 320 grams, about 95% of organs will be between 280 grams and 360 grams, the percentage of organs weighs less than 280 grams or more than 360 grams is 5%, and the percentage of organs weighs between 300 grams and 360 grams is 81.5%.

Given :

The weight of an organ in adult males has a bell-shaped distribution with a mean of 320 grams and a standard deviation of 20 grams.

A) According to the empirical rule, using the values of mean and standard deviation:

\rm \mu-\sigma = 320-20=300 \; gramsμ+σ=320+20=320grams

Therefore, about 68% of organs will be between 300 grams and 320 grams.

B) Again according to the empirical rule, using the values of mean and standard deviation:

\rm \mu-2\times \sigma = 320-40=280 \; gramsμ−2×σ=320−40=280grams

\rm \mu+2\times \sigma = 320+40=360 \; gramsμ+2×σ=320+40=360grams

Therefore, according to the empirical rule, about 95% of organs will be between 280 grams and 360 grams.

C)

The percentage of organs weighs less than 280 grams or more than 360 grams = 100 - (The percentage of organs weighs between 280 grams and 360 grams)

The percentage of organs weighs less than 280 grams or more than 360 grams = 100 - 95 = 5%

D)

The percentage of organs weighs between 300 grams and 360 grams = 0.5 \times× ( percentage of organs weighs between 280 grams and 360 grams + percentage of organs weighs between 300 grams and 320 grams)

The percentage of organs weighs between 300 grams and 360 grams = 0.5 \times× (95 + 68)

So, the percentage of organs weighing between 300 grams and 360 grams is 81.5%.

For more information, refer to the link given below:

brainly.com/question/23017717 95% of organs will be between 280 grams and 360 grams, the percentage of organs weighs less than 280 grams or more than 360 grams is 5%, and the percentage of organs weighs between 300 grams and 360 grams is 81.5%.

Given :

The weight of an organ in adult males has a bell-shaped distribution with a mean of 320 grams and a standard deviation of 20 grams.

A) According to the empirical rule, using the values of mean and standard deviation:

\rm \mu-\sigma = 320-20=300 \; gramsμ−σ=320−20=300grams

\rm \mu+\sigma = 320+20=320 \; gramsμ+σ=320+20=320grams

Therefore, about 68% of organs will be between 300 grams and 320 grams.

B) Again according to the empirical rule, using the values of mean and standard deviation:

\rm \mu-2\times \sigma = 320-40=280 \; gramsμ−2×σ=320−40=280grams

\rm \mu+2\times \sigma = 320+40=360 \; gramsμ+2×σ=320+40=360grams

Therefore, according to the empirical rule, about 95% of organs will be between 280 grams and 360 grams.

C)

The percentage of organs weighs less than 280 grams or more than 360 grams = 100 - (The percentage of organs weighs between 280 grams and 360 grams)

The percentage of organs weighs less than 280 grams or more than 360 grams = 100 - 95 = 5%

D)

The percentage of organs weighs between 300 grams and 360 grams = 0.5 \times× ( percentage of organs weighs between 280 grams and 360 grams + percentage of organs weighs between 300 grams and 320 grams)

The percentage of organs weighs between 300 grams and 360 grams = 0.5 \times× (95 + 68)

So, the percentage of organs weighing between 300 grams and 360 grams is 81.5%.

For more information, refer to the link given below:

brainly.com/question/23017717

5 0

2 years ago

The plane x+y+2z=8 intersects the paraboloid z=x2+y2 in an ellipse. Find the points on this ellipse that are nearest to and fart

DiKsa [7]

Answer:

The minimum distance of √((195-19√33)/8) occurs at ((-1+√33)/4; (-1+√33)/4; (17-√33)/4) and the maximum distance of √((195+19√33)/8) occurs at (-(1+√33)/4; - (1+√33)/4; (17+√33)/4)

Step-by-step explanation:

Here, the two constraints are

g (x, y, z) = x + y + 2z − 8

and

h (x, y, z) = x ² + y² − z.

Any critical point that we find during the Lagrange multiplier process will satisfy both of these constraints, so we actually don’t need to find an explicit equation for the ellipse that is their intersection.

Suppose that (x, y, z) is any point that satisfies both of the constraints (and hence is on the ellipse.)

Then the distance from (x, y, z) to the origin is given by

√((x − 0)² + (y − 0)² + (z − 0)² ).

This expression (and its partial derivatives) would be cumbersome to work with, so we will find the the extrema of the square of the distance. Thus, our objective function is

f(x, y, z) = x ² + y ² + z ²

and

∇f = (2x, 2y, 2z )

λ∇g = (λ, λ, 2λ)

µ∇h = (2µx, 2µy, −µ)

Thus the system we need to solve for (x, y, z) is

2x = λ + 2µx (1)

2y = λ + 2µy (2)

2z = 2λ − µ (3)

x + y + 2z = 8 (4)

x ² + y ² − z = 0 (5)

Subtracting (2) from (1) and factoring gives

2 (x − y) = 2µ (x − y)

so µ = 1 whenever x ≠ y. Substituting µ = 1 into (1) gives us λ = 0 and substituting µ = 1 and λ = 0 into (3) gives us 2z = −1 and thus z = − 1 /2 . Subtituting z = − 1 /2 into (4) and (5) gives us

x + y − 9 = 0

x ² + y ² + 1 /2 = 0

however, x ² + y ² + 1 /2 = 0 has no solution. Thus we must have x = y.

Since we now know x = y, (4) and (5) become

2x + 2z = 8

2x ² − z = 0

so

z = 4 − x

z = 2x²

Combining these together gives us 2x² = 4 − x , so

2x² + x − 4 = 0 which has solutions

x = (-1+√33)/4

and

x = -(1+√33)/4.

Further substitution yeilds the critical points

((-1+√33)/4; (-1+√33)/4; (17-√33)/4) and

(-(1+√33)/4; - (1+√33)/4; (17+√33)/4).

Substituting these into our objective function gives us

f((-1+√33)/4; (-1+√33)/4; (17-√33)/4) = (195-19√33)/8

f(-(1+√33)/4; - (1+√33)/4; (17+√33)/4) = (195+19√33)/8

Thus minimum distance of √((195-19√33)/8) occurs at ((-1+√33)/4; (-1+√33)/4; (17-√33)/4) and the maximum distance of √((195+19√33)/8) occurs at (-(1+√33)/4; - (1+√33)/4; (17+√33)/4)

4 0

3 years ago