What’s an example for y=1/2x-6 real world problems

12z - 21 = 92

12z - 21 + 21 = 92 + 21

12z = 113

z = 113/12

Answer:

1/8 Of the animals are rabbits.

Step-by-step explanation:

3/8 Cats + 4/8 Dogs = 7/8

1/8 Is all you have room left for so 1/8 would be rabbits.

Answer:

1/8.

Step-by-step explanation:

Fraction of rabbits = 1 - (3/7 + 4/8)

= 1 - 7/8

= 8/8 - 7/8

= 1/8.

Answer:

2.45.250- 94.750

= 92. 2975. this is the learners who got the admission

Answer:

1,50,500 students

Step-by-step explanation:

Hope this helps... vote as brainliest

Answer:

4)8 goats, 3 hens.

Step-by-step explanation:

8*4=32

3*2=6

32+6=38

Answer:

1.tenisha

2.1\5+1\3=1\×

3.n-12=1\3(n+2)

4.500×=120

Answer and Step-by-step explanation:

The answer is the second answer choice. y = 2x + 1

By looking at the graph, we see that there is a y-intercept at (0, 1), and it has a positive slope of 2.

#teamtrees #PAW (Plant and Water)

C.(f-g)(x) = 4x^3 +5x²-7x-1

Step-by-step explanation:

Given information :

[tex]f(x) = 4 {x}^{3} + 5 {x}^{2} - 3x - 6 \\ g(x) = 4x - 5[/tex]

Find :

[tex](f - g)(x) = \\ (4 {x}^{3} + 5 {x}^{2} - 3x - 6) \\ - 4x -5[/tex]

Open bracket and Simplify

[tex]4 {x}^{3} + 5 {x}^{2} - 3x - 6 - 4x + 5 \\ 4 {x}^{3} + 5 {x}^{2} - 7x - 1[/tex]

Answer

19.013.153,42629466 giây

Step-by-step explanati

van toc ánh sán  =299.792.458 m/s

s= v*t

t=s/v

t= 5.699.999.999.999.999/299.792.458= 19.013.153,42629466 giây

Answer:

The lorry travels by 6 mph more than the expected speed.

Step-by-step explanation:

Velocity formula:

Velocity is distance divided by time, that is:

[tex]v = \frac{d}{t}[/tex]

Shortest route:

400 miles in 10 hours, which means that [tex]d = 400, v = 10[/tex]. So

[tex]v = \frac{d}{t} = \frac{400}{10} = 40[/tex]

In mph.

The lorry actually travels by a different route which increases the distance by 15%, but it still arrives in 10 hours.

Distance is multiplied by 100% + 15% = 115% = 1.15, so:

[tex]d = 1.15*400 = 460[/tex]

Then

[tex]v = \frac{d}{t} = \frac{460}{10} = 46[/tex]

46 mph

By how many more mph than the expected speed does the lorry travel?

46 - 40 = 6 mph

The lorry travels by 6 mph more than the expected speed.

Answer:

[tex] {x}^{2} + 26x = 11 \\ x = 0.4 \: and \: - 26.4[/tex]

Answer:

Step-by-step explanation:

Use the formula P=MX+B

Answer:

3:7

Step-by-step explanation:

We know that there are 4 girls and 3 boys and 4+3=7.

Answer: 5,508 m3

Step-by-step explanation:  V= 18 x 17 x 18 = 5,508 m3

Answer:5, 508

Step-by-step explanation:

V 18×18×17=5,508

Answer:

C. she is not beautiful and not clever.

Step-by-step explanation:

A. she is beautiful but not clever. B. she is beautiful and clever

C. she is not beautiful and not clever.

D. she is beautiful or not clever.

p: she is beautiful

q :she is clever

~p^ (~q) in verbal form

~p = she is not beautiful

~q = she is not clever

~p^ (~q) = she is not beautiful and not clever.

C. she is not beautiful and not clever.

Answer:

Balance will be £3.92

Step-by-step explanation:

Cost of  1 loaf = £1.52

Therefore, cost of 4 loaves = 4 x 1 . 52 = £6.08

So if you pay with £10 your balance will be = 10.00 - 6.08 =  £3.92

The solution is  £ 3.92

The amount of money after the purchase of 4 loaves of bread is given by the equation A = £ 3.92

What is an Equation?

Equations are mathematical statements with two algebraic expressions flanking the equals (=) sign on either side.

It demonstrates the equality of the relationship between the expressions printed on the left and right sides.

Coefficients, variables, operators, constants, terms, expressions, and the equal to sign are some of the components of an equation. The "=" sign and terms on both sides must always be present when writing an equation.

Given data ,

Let the equation be represented as A

Now , the value of A is

The total amount of money = £ 10

The cost of one loaf of bread = £ 1.52

Now , the number of loaves of bread = 4 loaves

So , the cost of 4 loaves of bread = 4 x cost of one loaf of bread

Substituting the values in the equation , we get

The cost of 4 loaves of bread = 4 x 1.52

The cost of 4 loaves of bread = £ 6.08

The amount of money after the purchase A = total amount of money - cost of 4 loaves of bread

Substituting the values in the equation , we get

The amount of money after the purchase A = £ 10 - £ 6.08

The amount of money after the purchase A = £ 3.92

Therefore , the value of A is £ 3.92

Hence , the amount is £ 3.92

To learn more about equations click :

https://brainly.com/question/19297665

#SPJ2

Answer:

I guess it is the Answer C

Answer:

C

The distance around the edge of the circle

Explanation:

The sample space is simply the list of possible outcomes. We list all the hair colors possible in this class.

Answer:

[tex]\frac{N}{N_o} = 0.726 = 72.6\%[/tex]

Step-by-step explanation:

The following formula can be utilized for this question:

[tex]N = N_o (\frac{1}{2})^{\frac{t}{t_{1/2}} } \\\\\frac{N}{N_o} = (\frac{1}{2})^{\frac{t}{t_{1/2}} } \\\\[/tex]

where,

[tex]\frac{N}{N_o}[/tex] = ratio of the remaining amount to the original amount = ?

t = tme passed = 6 days

[tex]t_{1/2}[/tex] = half-life = 13 days

Therefore,

[tex]\frac{N}{N_o} = (\frac{1}{2} )^{\frac{6\ days}{13\ days} }\\\\[/tex]

[tex]\frac{N}{N_o} = 0.726 = 72.6\%[/tex]

Answer:

48

Step-by-step explanation:

the answer is 48 I got this answer by multiplying 5 by 10 and subtracting is from 2 which gives me 50 - 2 which is 48

Answer:

48

Step-by-step explanation:

5×10-2=50-2

=48

hope it helps!!

Answer:

B.

Step-by-step explanation:

The equation of a circle with center at (h, k) and radius r is

[tex] (x - h)^2 + (y - k)^2 = r^2 [/tex]

We have center at (-5, 0). That makes h = -5, and k = 0.

The radius is 3, so r = 3.

[tex] (x - (-5))^2 + (y - 0)^2 = 3^2 [/tex]

[tex] (x + 5)^2 + y^2 = 9 [/tex]

Answer: B.

Answer:

B

Step-by-step explanation:

The equation of a circle has the form:

[tex](x-h)^2+(y-k)^2=r^2[/tex]

Where (h, k) is the center of the circle and r is the radius.

From the graph, we can see that the center of the circle is at (-5, 0). So, (h, k) is (-5, 0), where h = -5 and k = 0.

And by counting, we can determine that the radius of the circle is three units. Hence, r = 3.

Substitute the information into the equation:

[tex](x-(-5))^2+(y-(0))^2=(3)^2[/tex]

Simplify. Therefore, our equation is:

[tex](x+5)^2+y^2=9[/tex]

Our answer is B.

Answer:

(a):

Dannette                   Alphonso

[tex]\bar x_D = 4.33[/tex]                    [tex]\bar x_A = 5.17[/tex]

[tex]M_D = 2.5[/tex]                    [tex]M_A = 5[/tex]

[tex]\sigma_D = 3.350[/tex]                  [tex]\sigma_A = 1.951[/tex]

[tex]IQR_D = 7[/tex]                  [tex]IQR_A = 1.5[/tex]

(b):

Measure of center: Median

Measure of spread: Interquartile range

(c):

There are no outliers in Dannette's dataset

There are outliers in Alphonso's dataset

Step-by-step explanation:

Given

See attachment for the appropriate data presentation

Solving (a): Mean, Median, Standard deviation and IQR of each

From the attached plots, we have:

IQR_A = 1.5 ---- Dannette

[tex]A = \{3,4,4,4,4,5,5,5,5,6,6,11\}[/tex] ---- Alphonso

n = 12 --- number of dataset

Mean

The mean is calculated

[tex]\bar x = \frac{\sum x}{n}[/tex]

So, we have:

[tex]\bar x_D = \frac{1+1+1+1+2+2+3+7+8+8+9+9}{12}[/tex]

[tex]\bar x_D = \frac{52}{12}[/tex]

[tex]\bar x_D = 4.33[/tex] --- Dannette

[tex]\bar x_A = \frac{3+4+4+4+4+5+5+5+5+6+6+11}{12}[/tex]

[tex]\bar x_A = \frac{62}{12}[/tex]

[tex]\bar x_A = 5.17[/tex]  --- Alphonso

Median

The median is calculated as:

[tex]M = \frac{n + 1}{2}th[/tex]

[tex]M = \frac{12 + 1}{2}th[/tex]

[tex]M = \frac{13}{2}th[/tex]

[tex]M = 6.5th[/tex]

This implies that the median is the mean of the 6th and the 7th item.

So, we have:

[tex]M_D = \frac{2+3}{2}[/tex]

[tex]M_D = \frac{5}{2}[/tex]

[tex]M_D = 2.5[/tex] ---- Dannette

[tex]M_A = \frac{5+5}{2}[/tex]

[tex]M_A = \frac{10}{2}[/tex]

[tex]M_A = 5[/tex]  ---- Alphonso

Standard Deviation

This is calculated as:

[tex]\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n}}[/tex]

So, we have:

[tex]\sigma_D = \sqrt{\frac{(1 - 4.33)^2 +.............+(9- 4.33)^2}{12}}[/tex]

[tex]\sigma_D = \sqrt{\frac{134.6668}{12}}[/tex]

[tex]\sigma_D = 3.350[/tex] ---- Dannette

[tex]\sigma_A = \sqrt{\frac{(3-5.17)^2+............+(11-5.17)^2}{12}}[/tex]

[tex]\sigma_A = \sqrt{\frac{45.6668}{12}}[/tex]

[tex]\sigma_A = 1.951[/tex] --- Alphonso

The Interquartile Range (IQR)

This is calculated as:

[tex]IQR =Q_3 - Q_1[/tex]

Where

[tex]Q_3 \to[/tex] Upper Quartile       and        [tex]Q_1 \to[/tex] Lower Quartile

[tex]Q_3[/tex] is calculated as:

[tex]Q_3 = \frac{3}{4}*({n + 1})th[/tex]

[tex]Q_3 = \frac{3}{4}*(12 + 1})th[/tex]

[tex]Q_3 = \frac{3}{4}*13th[/tex]

[tex]Q_3 = 9.75th[/tex]

This means that [tex]Q_3[/tex] is the mean of the 9th and 7th item. So, we have:

[tex]Q_3 = \frac{1}{2} * (8+8) = \frac{1}{2} * 16[/tex]           [tex]Q_3 = \frac{1}{2} * (5+6) = \frac{1}{2} * 11[/tex]

[tex]Q_3 = 8[/tex] ---- Dannette                 [tex]Q_3 = 5.5[/tex] --- Alphonso

[tex]Q_1[/tex] is calculated as:

[tex]Q_1 = \frac{1}{4}*({n + 1})th[/tex]

[tex]Q_1 = \frac{1}{4}*({12 + 1})th[/tex]

[tex]Q_1 = \frac{1}{4}*13th[/tex]

[tex]Q_1 = 3.25th[/tex]

This means that [tex]Q_1[/tex] is the mean of the 3rd and 4th item. So, we have:

[tex]Q_1 = \frac{1}{2}(1+1) = \frac{1}{2} * 2[/tex]                  [tex]Q_1 = \frac{1}{2}(4+4) = \frac{1}{2} * 8[/tex]

[tex]Q_1 = 1[/tex] --- Dannette                   [tex]Q_1 = 4[/tex] ---- Alphonso

So, the IQR is:

[tex]IQR = Q_3 - Q_1[/tex]

[tex]IQR_D = 8 - 1[/tex]                                     [tex]IQR_A = 5.5 - 4[/tex]

[tex]IQR_D = 7[/tex] --- Dannette                      [tex]IQR_A = 1.5[/tex] --- Alphonso

Solving (b): The measures to compare

Measure of  center

By observation, we can see that there are outliers is the plot of Alphonso (because 11 is far from the other dataset) while there are no outliers in Dannette plot (as all data are close).

Since, the above is the case; we simply compare the median of both because it is not affected by outliers

Measure of  spread

Compare the interquartile range of both, as it is arguably the best measure of spread, because it is also not affected by outliers.

Solving (c): Check for outlier

To check for outlier, we make use of the following formulas:

[tex]Lower =Q_1 - 1.5 * IQR[/tex]

[tex]Upper =Q_3 + 1.5 * IQR[/tex]

For Dannette:

[tex]Lower = 1 - 1.5 * 7 = -9.5[/tex]

[tex]Upper = 8 + 1.5 * 7 = 18.5[/tex]

Since, the dataset are all positive, we change the lower outlier to 0.

So, the valid data range are:

[tex]Valid = 0 \to 18.5[/tex]

From the question, the range of Dannette's dataset is: 1 to 9. Hence, there are no outliers in Dannette's dataset

For Alphonso:

[tex]Lower = 4 - 1.5 * 1.5 =1.75[/tex]

[tex]Upper = 5.5 + 1.5 * 1.5 =7.75[/tex]  

So, the valid data range are:

[tex]Valid = 1.75\to 7.75[/tex]

From the question, the range of Alphonso's dataset is: 3 to 11. Hence, there are outliers in Alphonso's dataset

Answer:

x=-1

Step-by-step explanation:

the middlepoint is where its symetrical, and so you take the x part of the point. the point is (-1,4), and all we need is x, so you have x=-1

Answer:

0.73 = 73% probability that a randomly selected student is either male or would admit to lying to a teacher, during the past year.

Step-by-step explanation:

I am going to treat these events as Venn probabilities, considering that:

Event A: Lying to the teacher.

Event B: Male

48% of high school students would admit to lying at least once to a teacher during the past year and that 25% of students are male and would admit to lying at least once to a teacher during the past year

This means that [tex]P(A) = 0.48, P(A \cap B) = 0.25[/tex]

Assume that 50% of the students are male.

This means that [tex]P(B) = 0.5[/tex]

What is the probability that a randomly selected student is either male or would admit to lying to a teacher, during the past year?

This is:

[tex]P(A \cup B) = P(A) + P(B) - P(A \cap B)[/tex]

Considering the values we were given:

[tex]P(A \cup B) = 0.48 + 0.5 - 0.25 = 0.73[/tex]

0.73 = 73% probability that a randomly selected student is either male or would admit to lying to a teacher, during the past year.

Answer:

Eighty thousand

Step-by-step explanation:

Look at the place value chart.

Answer:

5⁴a²

Step-by-step explanation:

(5³a³)÷5a-¹×5-²a²

5³a³÷5a-¹×5-²a²

5³a³÷5¹-²×a-¹+²

5³a³÷5-¹a

5³a³/5-¹a

5³-(-¹)a³-¹

5⁴a²

Answer:

Ignore the A before the ±, it wouldn't let me type it correctly.

[tex]x=\frac{2±\sqrt{7} }{3}[/tex]

Step-by-step explanation:

3x + 4 = 1 ÷ x

3x + 4 - 4 = 1 ÷ x - 4

3x = 1 ÷ x - 4

[tex]3x=\frac{1}{x} +\frac{x(-4)}{x}[/tex]

[tex]3x=\frac{1+x(-4)}{x}[/tex]

[tex]3x=\frac{1-4x}{x}[/tex]

[tex]x(3x)=x(\frac{1-4x}{x})[/tex]

x · 3x = - 4x + 1

3x² = - 4x + 1

3x² - (- 4x + 1) = 0

3x² + 4x - 1 = 0

Ignore the A before the ±, it wouldn't let me type it correctly.

[tex]x=\frac{-b±\sqrt{b^{2}-4ac } }{2a}[/tex]

a = 3

b = 4

c = - 1

[tex]x=\frac{-4±\sqrt{4^{2}-4((3)(-1)) } }{2(3)}[/tex]

[tex]x=\frac{-4±\sqrt{16-4((3)(-1)) } }{2(3)}[/tex]

[tex]x=\frac{-4±\sqrt{16+12 } }{2(3)}[/tex]

[tex]x=\frac{-4±\sqrt{28 } }{2(3)}[/tex]

[tex]x=\frac{-4±\sqrt{(2)(14) } }{2(3)}[/tex]

[tex]x=\frac{-4±\sqrt{(2)(2)(7) } }{2(3)}[/tex]

[tex]x=\frac{-4±\sqrt{2 } \sqrt{2}\sqrt{7} }{2(3)}[/tex]

[tex]x=\frac{-4±2\sqrt{7} }{2(3)}[/tex]

[tex]x=\frac{-4±2\sqrt{7} }{6}[/tex]

Two separate equations

[tex]x=\frac{-4+2\sqrt{7} }{6}[/tex]

[tex]x=\frac{2+\sqrt{7} }{3}[/tex]

[tex]x=\frac{-4-2\sqrt{7} }{6}[/tex]

[tex]x=\frac{2-\sqrt{7} }{3}[/tex]

Answer:

[tex] x=\frac{[2] \sqrt {[21] }}{[3] }[/tex]

Step-by-step explanation:

Since, given is a 30°-60°-90° triangle.

[tex] \therefore \sqrt 7 = \frac{\sqrt3}{2} \times x[/tex]

[tex] \therefore 2\sqrt 7 = \sqrt3 \times x[/tex]

[tex] \therefore x=\frac{2\sqrt 7}{\sqrt 3}[/tex]

[tex] \therefore x=\frac{2\sqrt 7(\sqrt 3)}{\sqrt 3(\sqrt 3)}[/tex]

[tex] \huge \therefore x=\frac{[2] \sqrt {[21] }}{[3] }[/tex]