Jesse bought 24 gallons of gas for $45.58. What was the price per gallon?

Two possibilities  --- see below

Step-by-step explanation:

So it could be GREATER than 46

  .85 x = 46    x = 54.11 cm^3

Or it could be  smaller than 46

  46= (1.15)  x      x = 40 cm^3

1a) 256/9

2a) 0.0048

3a) 0.64

4a) 2/5

5a) 4.6875

1b) 125/343

2b) 32/3125

3b) 0.003125

4b) 48.875

5b) 1.39

Whew! Hope this helps :)

Answer:

27

Step-by-step explanation:

(3^−9)(3^6)(3^6)

=1/19683(3^6)(3^6)

=1/19683(3^6)(3^6)

=(1/19683(729))(3^6)

=1/27(3^6)

=1/27(729)

=27

Answer:

27^3

Step-by-step explanation:

Answer:

c 1:4

Step-by-step explanation:

Gradient = rise/ run

rise is d

run is the difference in our x coordinates of c & e

(or c - e)

So, it's d/C - e

(3rd option)

Hope this helps!

Answer:

3rd option

Step-by-step explanation:

calculate the slope m using the slope formula

m = [tex]\frac{y_{2}-y_{1} }{x_{2}-x_{1} }[/tex]

with (x₁, y₁ ) = E (e, 0 ) and (x₂, y₂ ) = (c, d )

m = [tex]\frac{d-0}{c-e}[/tex] = [tex]\frac{d}{c-e}[/tex]

Answer: 17.3 cm

Step-by-step explanation:

[tex]\tan 68^{\circ}=\frac{AC}{7}\\\\AC=7\tan 68^{\circ}\\\\AC \approx 17.3[/tex]

Answer:

Greetings !

check the attachment above☝️ but i haven't done the second question wait a moment. thx

Answer: (x + 11) cm²

Step-by-step explanation:

The formula for the area of a trapezoid is [tex]\frac{1}{2}(a+b)h[/tex], where

For this figure, [tex]a = 5[/tex] and [tex]b= x+6[/tex] (the top and bottom sides), while [tex]h=2[/tex] (the height).

If we suppose [tex]A[/tex] is the area, plugging these values in, we get

[tex]A = \frac{1}{2}(5 + x + 6)*2[/tex]

[tex]A = \frac{1}{2} * 2 * (x + 11)[/tex]

[tex]A = x + 11[/tex]

All the lengths are in cm, so the area will be (x + 11) cm².

No this does not indicate that the population proportion of women athletes who graduate from the university is now less than 67%.

Given long term graduation rate of 67%, sample size of 36  and the women athletes graduated is 23.

We have to find whether  the given information shows  that the population proportion is less than 67%.

First we have to create hypothesis for this :

[tex]H_{0}[/tex]:P=0.67

[tex]H_{1}[/tex]:P<0.67

Under null hypothesis the test statistic is

z=p bar-p/[tex]\sqrt{p(1-p)/n}[/tex]

where p bar=23/36

=0.638

z=(0.638-0.67)/[tex]\sqrt{0.67(1-0.67)/36[/tex]

=-0.032/[tex]\sqrt{0.67*0.33/36}[/tex]

=-0.032/[tex]\sqrt{0.0064}[/tex]

=-0.032/0.078

=-0.41

Now we have to find the left tailed critical at 0.01 significance level using z table.

z=-2.33

Since the z value does not fall in the critical region,therefore we fail to reject the null hypothesis. So we can conclude that there is not sufficient evidence to say that the population proportion of women athletes who graduate from the university is now less than 67%.

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Question is incomplete as it should specify the signficance level of 0.01 to be used.

Answer:

- 4/7 , -1/9, 7/4

Step-by-step explanation:

You know that 7/4 is the largest number because it is the only positive number.  To compare the two negative numbers you need common denominators.  The common denominator would be 133 (19 x 7).  Now make equivalent fractions.-4/7 = x/133  133 divided by 7 is 19.  Multiple the -4 by 19 to get -76/133.  Next, make an equivalent fraction for -1/19.  The new denominator will be 133.  If I divide 133 by 19, I will get 7, so I multiple -1x7 and get -7 and the new fraction -7/133.

I have two fractions to compare -7/133 and -76/133.  -76/133 if farthest from 0 so it has the least value.

Answer:

L (1, - 7 )

Step-by-step explanation:

give endpoints (x₁, y₁ ( and (x₂, y₂ )  then the midpoint is

( [tex]\frac{x_{1}+x_{2} }{2}[/tex] , [tex]\frac{y_{1}+y_{2} }{2}[/tex] )

here (x₁, y₁ ) = K (9, - 7 ) and (x₂, y₂ ) = L (x, y )

use the midpoint formula and equate to corresponding coordinates of M

[tex]\frac{9+x}{2}[/tex] = 5 ( multiply both sides by 2 to clear the fraction )

9 + x = 10 ( subtract 9 from both sides )

x = 1

and

[tex]\frac{-7+y}{2}[/tex] = - 7 ( multiply both sides by 2 )

- 7 + y = - 14 ( add 7 to both sides )

y = - 7

Then L  = (1, - 7 )

Using the laws of indices , 4⁵ ÷ 4² can rewritten as 4³.

Given the expression; 4⁵ ÷ 4²

To perform the division operation, we use one of the laws of indices;

[tex]n^a/n^b = n^{a-b}[/tex]

Given that;

Now, we apply the laws of indices ;

4⁵ ÷ 4² = 4⁵⁻² = 4³

Therefore, using the laws of indices, 4⁵ ÷ 4² can rewritten as 4³.

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Answer:

X=4 easy blah blah blah it’s 4. jdjjdjfnfvbdnnfn

Rolle's Theorem does not apply to the function because there are points on the interval (a,b) where f is not differentiable.

Given the function is [tex]f(x)=\sqrt{(2-x^{\frac{2}{3}})^{3}}[/tex]  and the Rolle's Theorem does not apply to the function.

Rolle's theorem is used to determine if a function is continuous and also differentiable.

The condition for Rolle's theorem to be true as:

To apply the Rolle’s Theorem we need to have function that is differentiable on the given open interval.

If we look closely at the given function we can see that the first derivative of the given function is:

[tex]\begin{aligned}f(x)&=\sqrt{(2-x^{\frac{2}{3}})^3}\\ f(x)&=(2-x^{\frac{2}{3}})^{\frac{3}{2}}\\ f'(x)&=\frac{3}{2}(2-x^{\frac{2}{3}})^{\frac{1}{2}}\cdot \frac{2}{3}\cdot (-x)^{\frac{1}{3}}\\ f'(x)&=\frac{-\sqrt{2-x^{\frac{2}{3}}}}{\sqrt[3]{x}}\end[/tex]

From this point of view we can see that the given function is not defined for x=0.

Hence, all the assumptions are not satisfied we can reach a conclusion that we cannot apply the Rolle's Theorem.

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The weighted mean grade for the semester is 8.22

It is calculated by dividing the total number of values in a set of data, such as measurements or numbers, by the total number of values.

No. of credits for Spanish, History and English = 6

No. of credits for Statistics = 8

Grade points for A = 4

Grade points for B = 3

Grade points for C = 2

Weighted mean = (6 * 2 + 6 * 3 + 8 * 4 + 6 * 2) / 9

Weighted mean = (12  + 18 + 32 + 12) / 9

Weighted mean = 74 / 9 = 8.22

Hence, the weighted mean grade for the semester is 8.22.

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Answer:

Translation of 3 units to the left.

Vertical stretch by a factor of 2.

Translation of 5 units down.

Step-by-step explanation:

Transformations

For a > 0

[tex]f(x+a) \implies f(x) \: \textsf{translated}\:a\:\textsf{units left}[/tex]

[tex]y=a\:f(x) \implies f(x) \: \textsf{stretched parallel to the y-axis (vertically) by a factor of}\:a[/tex]

[tex]f(x)-a \implies f(x) \: \textsf{translated}\:a\:\textsf{units down}[/tex]

Parent function:

[tex]y=x^2[/tex]

Translate 3 units left

Add 3 to the variable of the function

[tex]\implies y=(x+3)^2[/tex]

Stretch vertically by a factor of 2

Multiply the whole function by 2:

[tex]\implies y=2(x+3)^2[/tex]

Translate 5 units down

Subtract 5 from the whole function:

[tex]\implies y=2(x+3)^2-5[/tex]

Please see the attached graphs for the final transformed function (as well as the graphed steps).

Answer:

  a) vertical expansion by a factor of 2; translation 3 units left and 5 units down

  b) see attached

Step-by-step explanation:

Describing transformations is all about matching patterns. The elements of the transformed function are matched with the elements of a transformation.

A function is scaled vertically by multiplying each function value by some scale factor. In generic terms, the function f(x) is scaled vertically by the factor 'c' in this way:

If we want the function f(x) = x² scaled vertically by a factor of 2, then we have

  f(x) = x² . . . . . . . original function

  2·f(x) = 2x² . . . . scaled vertically by a factor of 2

On a graph, each point is vertically twice as far vertically from some reference point (the vertex, for example) as it is in the original function graph.

A function is translated to the right by 'h' units when x is replaced by (x -h).

If we want the function f(x) = x² translated right by 3 units, we will have ...

  f(x) = x² . . . . . . . . . . . original function

  f(x -3) = (x -3)² . . . . . .translated right 3 units

Note that translation left by 3 units would give ...

  f(x -(-3)) = f(x +3) = (x +3)² . . .  . translated left 3 units

On a graph, each point of the left-translated function is 3 units left of where it was on the original function graph.

A function is translated upward by 'k' units when k is added to the function value.

The value of k will be negative for a translation downward.

If we want the function f(x) = x² translated down by 5 units, we will have ...

  f(x) = x² . . . . . . . . . . . original function

  f(x) = x² -5 . . . . . . . . .translated down 5 units

Using all of these transformations at once, we have ...

  f(x) = x² . . . . . . . . . . . . . . . . . original function

  c·f(x -h) +k = c·(x -h)² +k . . . scaled by 'c', translated h right and k up

Compare this to the given function:

 y = 2(x +3)² -5

and we can see that ...

This is the pattern matching that is described at the beginning.

__

When graphing a transformed function, it is often useful to start with a distinctive feature and work from there. The vertex of a parabola is one such distinctive feature.

The transformations move the vertex 3 units left and 5 units down from its original position at (0, 0). The location of the vertex on the transformed function graph will be at (x, y) = (-3, -5).

The graph of the parent function parabola (y= x²) goes up from the vertex by the square of the number of units right or left. That is, 1 unit right or left of the vertex, the graph is 1 unit above the vertex. 2 units right or left, the graph is 2² = 4 units above the vertex.

The scaled graph will have these vertical distances multiplied by 2:

The graph of the transformed function is shown in blue in the attachment.

__

Additional comment

The vertical scale factor 'c' may have any non-zero value, positive or negative, greater than 1 or less than 1. When the magnitude is less than 1, the scaling is a compression, rather than an expansion. When the sign is negative, the graph is also reflected across the x-axis, before everything else.

Using the t-distribution, it is found that there is not enough evidence that the diet is helping people lose weight.

At the null hypothesis, it is tested if the mean hasn't decreased, that is:

[tex]H_0: \mu_B \leq \mu_A[/tex]

[tex]H_0: \mu_B - \mu_A \leq 0[qtex]

At the alternative hypothesis, it is tested if the mean has decreased, that is:

[tex]H_1: \mu_B - \mu_A > 0[qtex]

For each sample, they are given as follows:

Hence, for the distribution of differences, they are:

The test statistic is:

[tex]t = \frac{\overline{x} - \mu}{s}[/tex]

In which [tex]\mu = 0[/tex] is the value tested at the null hypothesis.

Hence:

[tex]t = \frac{\overline{x} - \mu}{s}[/tex]

[tex]t = \frac{9.88 - 0}{11.66}[/tex]

t = 0.85.

Considering a right-tailed test with 9 + 9 - 2 = 16 df, with a standard significance level of 0.05, the critical value is t = 1.7459. Since t = 0.85 < 0.85, there is not enough evidence that the diet is helping people lose weight.

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Answer:

4ab2(7a2b2 5a - 4b)

Step-by-step explanation:

28a3b4 20a2b2 − 16ab3

The GCF of 28, 20 and 16 is 4.

GCF of the variables is ab2

So the answer is

4ab2(7a2b2 5a - 4b)

Answer:

4ab2(7a2b2 5a - 4b)

Step-by-step explanation:

Answer:

20

Steps-by-step explanation:

sorry mistake

Answer:

12

Explanation: I just counted them! :) and there was 6 in the middle and on the outside so 6 + 6 = 12! :)

By critically observing and comparing the functions for the given graphs, we can logically deduce that the widest graph is: "Function 1: y = 0.25x²."

In Geometry, a graph which has the smallest positive coefficient of x² in absolute value is generally considered as the widest graph while the graph with the largest positive coefficient of x² in absolute value is the narrowest graph.

By critically observing and comparing the functions for the given graphs, we can logically deduce that "Function 1: y = 0.25x²" is the widest.

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Complete Question:

Compare the graphs of the functions listed below.

Function 1: y = 0.25x²

Function 2: y = 4x²

Function 3: y = -½x²

Function 4: y = -16x²

The graph of ______ is the widest?

Answer:

A:(1/3x)+(x-2/4x^2+9)

Step-by-step explanation:

Edg2022

The option which makes more sense for Jake to choose is option (1) He can accept the tie, and the game is over.

The complete question is added as an attachment

From the complete question, the probability in part D is

Probability = 33.5%

The options to choose one from are:

The calculated probability is less than 0.5.

This means that Jake is more likely to lose than win

So, the best option for Jake is option (A) walk away

Hence, the option which makes more sense for Jake to choose is option (1) He can accept the tie, and the game is over.

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Given that (p - 1/p) = 4, the value of p² + 1/p² is 18. Detail below

(p - 1/p) = 4

Square both sides

(p - 1/p)² = (4)²

(p - 1/p)² = 16 ....(1)

Recall

(a - b)² = a² + b² - 2ab

Thus,

(p - 1/p)² = p² + 1/p² - (2 × p × 1/p)

(p - 1/p)² = p² + 1/p² - 2

From equation (1) above,

(p - 1/p)² = 16

Therefore,

p² + 1/p² - 2 = 16

Rearrange

p² + 1/p² = 16 + 2

p² + 1/p² = 18

Thus, the value of p² + 1/p² is 18

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Answer: 6x

Step-by-step explanation:

The  expression represents the radius of the cone's base is √111/π x = r

a three dimensional shape can be defined as a solid figure or an object or shape that has three dimensions—length, width, and height.

The formula for the volume of a cone is V = (1/3)πr²h, where V is the volume,

r is the radius of the base,

h is the height, and π is a constant approximately equal to 3.14.

We are given that the volume of the cone is 37x³ cubic units and the height is x units.

Using the formula for the volume of a cone, we can write:

37x³ = (1/3)πr²x

37x² = πr²/3

37×3 x²/π = r²

111x²/π  = r²

Take square root on both sides

√111/π x = r

Hence, the  expression represents the radius of the cone's base is √111/π x = r

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The transformation for circle 1 exists (x+9), (y+5) and the scale factor of circle 1 exists 2.

Scale exists described as the ratio of the length of any object on a model to the actual length of the exact object in the entire world.

The function transformation rules: f(x)+b changes the function b units upward. f(x)−b shifts the function b units downward. f(x + b) shifts the function b units to the left.

Circle 1

center: (−4, −2)

Radius: 3 centimeters

Circle 2

center: (5, 3)

Radius: 6 centimeters

Transformations can be used to circle 1 to verify that the circles exist similar.

As circle 2 and circle 1 do not contain the exact coordinates

So, circle 1 has to utilize transformation rules.

The difference between the coordinates of circle 2 and circle 1 exists

[tex]$x_2 - x_1 = 5 - (-4 ) = 9[/tex]

[tex]y_2 - y_1 = 3 - (-2) = 5[/tex]

Therefore, transformation for circle 1: (x+9), (y + 5)

The scale factor between circle 2 and circle 1 exists

The radius of Circle 2 = 2 [tex]*[/tex] radius of circle 1

Therefore, the scale factor of circle 1 = 2

Therefore, the transformation for circle 1 exists (x+9), (y+5), and the scale factor of circle 1 exists 2.

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The equation of the parallel line is  2y-x=-7

A parallel line will have the same slope, but different intercept. Let's assume C in the intercept of this parallel line. Therefore, equation of the parallel would be,

2y-x=C, coefficients of x and y remain the same, since parallel lines.

This line passes through (3, -2). Therefore, substituting those values, we will get value of C.

2.(-2) -3=C

C=(-7)

The equation of the parallel line is therefore 2y-x=-7

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The expression which is equivalent to [tex]5 log_{10}x+log_{10}20-log_{10}10[/tex] is

[tex]$5 log_{10}x+log_{10}20-log_{10}10 = log_{10}(20x^5)-1[/tex].

A logarithmic equation exists as an equation that applies the logarithm of an expression having a variable.

Product Rule Law: [tex]log_{a} (MN) = log_{a} M + log_{a} N[/tex]

Quotient Rule Law: [tex]log_{a} (M/N) = log_{a} M - log_{a} N[/tex]

Power Rule Law: [tex]log_{a}M^{n} = n log_{a} M[/tex]

Given: [tex]5 log_{10}x+log_{10}20-log_{10}10[/tex]

[tex]5 log_{10}x+log_{10}20-log_{10}10 = log_{10}(x^5) +log_{10}20 -log_{10}10[/tex]

apply the law of logarithm

[tex]5 log_{10}x+log_{10}20-log_{10}10 = log_{10}(x^5*20/10)[/tex]

[tex]5 log_{10}x+log_{10}20-log_{10}10 = log_{10}(x^5*2)[/tex]

[tex]5 log_{10}x+log_{10}20-log_{10}10 = log_{10}(2x^5)[/tex]

Another possible equivalent expression is:

[tex]$5 log_{10}x+log_{10}20-log_{10}10 = log_{10}(x^5*20)-log_{10}10[/tex]

[tex]$5 log_{10}x+log_{10}20-log_{10}10 = log_{10}(20x^5)-log_{10}10[/tex]

Substitute the value of [tex]log_{10}10 = 1[/tex]

[tex]$5 log_{10}x+log_{10}20-log_{10}10 = log_{10}(20x^5)-1[/tex]

Therefore, the correct answer

[tex]$5 log_{10}x+log_{10}20-log_{10}10 = log_{10}(20x^5)-1[/tex].

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The equation will be like this,

6.89 + t = 8.00

t = 8 - 6.89

t = 1.11

tips = $1.11

6.89 + t = 8.00

To sum up the total amount of cash and coins you have, first, sort each invoice and coin by denomination. Create a separate stack for each denomination and count how many denominations or coins you have.

For each denomination of banknotes and coins, multiply the number you have by the denomination. For example, if you have 4 out of $ 10, multiply by 4x10 to get $ 40. If you have three $ 5 bills, multiply by 3x5 to get $ 15.

Sum the totals to calculate the total amount.

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