Solve the following expression when
k= 12 and y = 6
+ 5 – 4v + 12

Answer:

[tex]36\:\mathrm{in^3}[/tex]

Step-by-step explanation:

The area of a prism is given by the product of the base and the height. In this triangular prism, the area of the either base is:

[tex]A=\frac{1}{2}\cdot 3\cdot 4=6\:\mathrm{in^2}[/tex]

The height is marked as [tex]6[/tex] inches. Thus, the volume of the prism is:

[tex]\frac{1}{2}\cdot 3\cdot 4\cdot 6=\boxed{36\:\mathrm{in^3}}[/tex]

Answer:

x = 3

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Equality Properties

Algebra I

Step-by-step explanation:

Step 1: Define

Identify

12x = 36

Step 2: Solve for x

Answer:

18.84

Step-by-step explanation:

6 times 3.14

= 18.84

Answer:

56.52

Step-by-step explanation:

1/2(π*radius^2)

Answer:

D. 4 root 113

Step-by-step explanation:

we have to use Pythagorean Theorem which is

a^2 + b^2 = c^2

where a and b are legs of the triangle that are 45° apart and c is the hypotenuse.

now we just plug in our values.

a^2 + (9)^2 = (47)^2

a^2 + 81 = 2209

now we're gonna solve for A (the missing side)

a^2 = 2128

now we square root both sides to undo the square on A

root(a^2) = root(2128)

because 2128 isn't a pretty root, we have to simplify it

a = 4 root(133)

Answer:

The answer would be A.) 4 inches

Step-by-step explanation:

It is A. because if the base of the triangle is 10 inches and one side is 4 then the two sides of the trianlge would be the same they are congruent so if one side of the triangle is 4 then the other side of the triangle is 4.

Hope this helps

[tex] \huge \mathfrak{ Answer : - }[/tex]

60 students walk to school, which is 20 % of the total students studying there,

so, let the the total number of students be " x "

now, we know

=》

[tex]20 \% \: \: of \: \: x = 60[/tex]

=》

[tex] \dfrac{20}{100} \times x = 60[/tex]

=》

[tex] \dfrac{1}{5} \times x = 60[/tex]

=》

[tex]x = 60 \times 5[/tex]

=》

[tex]x = 300[/tex]

hence, the total number of students :

x = 300

Answer:

simple answer 300

Step-by-step explanation:

20%=60 students. 100%/20%=5. 60*5=300 students

Answer:

3kg:0.8kg

Step-by-step explanation:

I'm not sure if it's right

For the given condition the age of the son will be 69 years.

A mathematical equation asserts the equality of two expressions, which may or may not contain variables or integers. Equations are fundamental concerns, and efforts to rationally address them have served as the primary driving forces behind the creation of mathematics.

Suppose the two-digit number is x,

The father's and son's ages are 10x+y and 10y+x respectively.According to the given condition,

(10x+y) -(10y+x)=27

9x-9y=27

x-y=27/9

x-y=3

The value of x and y will be 4 and 1.

At x=9 and y=6, the age of the son is obtained as,

=10×6+9

=69 years

Thus, the age of the son will be 69 years.

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The median of cumulative frequency is 67.86 and the interquartile range is 32.86.

The mean is "the middle number is sorter, ascending or descending, list of numbers and can be more descriptive of that data set than the average".

The interquartile range is "measure of where the 'middle fifty' is in data set".

According to the question,

Score               Mid value    Frequency       Cumulative frequency

10 ≤ x < 20            15                       1                                 1

20≤x < 30             25                      3                                 4

30≤x < 40             35                      6                                 10

40≤x < 50            45                       15                               25

50≤x < 60            55                       14                               39

60≤x < 70            65                        28                             67

70≤x < 80            75                         18                              85

80≤x < 90            85                         11                               96

90≤x < 100           95                         4                               100

Uniform interval is 10.

a) The median of examination mark

Frequency = 1+3+6+15+14+28+18+11+4 = 100

Median = [tex]l+\frac{(\frac{N}{2} - m )}{f}[/tex] × c       where l = lower limit of the median class, N = total frequency, m = cumulative frequency of the pre median class,

f = frequency of the median class, c = uniform class interval.  

            = 50 + [tex]\frac{\frac{100}{2} - 25 }{14}[/tex]  × 10

            = 50 + [tex]\frac{50 - 25}{14}[/tex]  × 10

            = 50 +[tex]\frac{25}{14}[/tex] × 10

           = 50 + 1.786 × 10

           = 50 + 17.86  

           = 67.86.

Thus, the median of examination mark is 67.86.

b) The number of students score less than 65 marks.

In order to find the number of students score less than 65 marks sum of all frequency below 65.

= 1 + 3 + 6 + 15 + 14

= 39

Thus, the number of students score less than 65 marks is 39.

c) The number of students scored between 50 and 80 marks

In order to find the number of students scored between 50 and 80 marks only if add frequency between 50 and 80 marks.

= 14 + 28 + 18 + 11

= 71

Thus, the number of students scored between 50 and 80 marks is 71.

d) The number of students failed, given that the pass mark is 50.

In order to find the number of students failed, given that the pass mark is 50 only if add frequency below 50.

= 1 + 3 + 6 + 15

= 25.

Thus,  the number of students failed, given that the pass mark is 50 is 25.

f) The interquartile range

Interquartile range =  Q₃ - Q₁    [In words, difference between upper quartile range and lower quartile range]

Q₁ =  l₁ + [tex]\frac{\frac{N}{4} - m }{f}[/tex] × c  where l₁ is the lower limit of the Q₁ class, m₁ is the cumulative frequency of the pre median class, f₁ is the Frequency of the Q₁ class, c is the class interval, N  is the total frequency.

[tex]\frac{N}{4}[/tex] = [tex]\frac{100}{4} = 25[/tex]; [tex]\frac{3N}{4}[/tex] = 75

     = 40 + [tex]\frac{25 - 25}{6}[/tex] * 10 = 40

Q₁ = 40

Q₃ = l₃ + [tex]\frac{3\frac{N}{4} - m3 }{f3}[/tex]  × c

   = 60 + [tex]\frac{75 - 39}{28}[/tex] × 10

   = 60 + [tex]\frac{36}{28}[/tex] × 10

   = 60 + 1.2857 × 10

   =   60 + 12.857

   = 72.857

Interquartile range = upper quartile range - lower quartile range

                                 = 72.857 - 40

                                 = 32.8571  ≈ 32.86.

Thus, interquartile range is 32.86.

Hence, a) The median of examination mark is 67.86.

b) The number of students score less than 65 marks is 39.

c) The number of students scored between 50 and 80 marks is 71.

d) The number of students failed, given that the pass mark is 50 below 50 is 25.

f)  The interquartile range is 32.86.

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Answer:

The p-value of the test is of 0.0045 < 0.05, which means that there is enough evidence to reject the null hypothesis and accept the alternate hypothesis that the mean net weight is less than advertised.

Step-by-step explanation:

Bags of a certain brand of tortilla chips claim to have a net weight of 14 ounces. A representative of a consumer advocacy group wishes to see if there is any evidence that the mean net weight is less than advertised.

At the null hypothesis, we test that the mean is of 14, that is:

[tex]H_0: \mu = 14[/tex]

At the alternate hypothesis, we test tha tthe mean is less than advertised, that is, less than 14. So:

[tex]H_a: \mu < 14[/tex]

The test statistic is:

[tex]t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}[/tex]

In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, s is the standard deviation of the sample and n is the size of the sample.

14 is tested at the null hypothesis:

This means that [tex]\mu = 14[/tex]

For this, the representative randomly selects 16 bags of this brand and determines the net weight of each. He finds the sample mean to be 13.82 and the sample standard deviation to be 0.24.

This means that [tex]n = 16, X = 13.82, s = 0.24[/tex]

Test statistic:

[tex]t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}[/tex]

[tex]t = \frac{13.82 - 14}{\frac{0.24}{\sqrt{16}}}[/tex]

[tex]t = -3[/tex]

P-value of the test and decision:

The p-value of the test is the probability of finding a sample mean below 13.82, which is the p-value of t = -3, using a left-tailed test with 16 - 1 = 15 degrees of freedom.

With the help of a calculator, the p-value is of 0.0045.

The p-value of the test is of 0.0045 < 0.05, which means that there is enough evidence to reject the null hypothesis and accept the alternate hypothesis that the mean net weight is less than advertised.

Answer:

d.

Step-by-step explanation:

The gradient is -5 meaning that it's going downwards as x increases

The y-intercept is 3

Feel free to mark as brainliest :D

Answer:

53.8%

Step-by-step explanation:

Answer:

u already got ur awnser so hi

Step-by-step explanation:

ui

Answer:

$9

Step-by-step explanation:

The linear system is going to be:

16m + 13s + 29c = $419

26m + 19s + 45c = $649

6m + 11s + 0c = $96

I don't really know why it is bringing up the determinant, all you need to do is put this into reduced row echelon form.  I personally used a calculator for this, but it could be done painstakingly by hand.

RREF gives me:

m=5

s=6

c=9

So a chocolate milkshake is $9

Step-by-step explanation:

its the same ratio 4/7

cos(theta) = adjacent /hypotenuse

Answer:

[tex]C.[/tex] 5/7 × 2/3

Step-by-step explanation:

Answer:

117.5 cm that the answer on your question

Answer: 4

Step-by-step explanation: 16/4 =4 brainlyest plz plz I need it for my rank

-0.2

'0

-0.2 = - .2 = - 2/10 = - 1/5

so

answer is -1/5

Answer:

-0.2 = -1/5

Step-by-step explanation:

you have to turn the decimal to a fraction.

0.2/1

multiply 0.2/1 by 10/10 which is 2/10

Since the GCF is 2...

You want to divide 2/10 by 2/2

that is 1/5

Now add the negative number sign

BOOM

Now you got yourself -1/5

I hope I helped! :)

Answer:

We all have the same I ready's lol

Anyway, i did it and, The CORRECT answer is:

$4.95

Answer:

Solution given:

radius of small circle[r]=16-4.5ft=11.5ft

radius of large circle[R]=16ft

nowcircumference of small circle=2πr=2*3.14*11.5=72.22ft

circumference of large circle=2πR=2*3.14*16=100.48ft

distance that the outside tires must travel to the distance the inside tires must travel in order to make a complete circle

=[tex]\frac{100.48}{72.22}=\frac{32}{23}[/tex]

ratio is 32:23.

The surface area of the given composite figure will be 384 square cm.

The space occupied by any two-dimensional figure in a plane is called the area. The area of the outer surface of any body is called the surface area.

In the given image we have a composite figure one is a rectangular cuboid and the other is a triangular prism. The surface area will be equal to the sum of all the outer sides of the figure.

The surface area of the triangular shape will be calculated as:-

SA = 2( Area of triangle ) + Area of rectangular surfaces

SA = 2 ( (1/2) B x H )  + (5 x 6 ) + ( 6 x 3  )

SA = ( 3 x 4 ) + (  30 ) + ( 18 )

SA = 12 + 30 + 18

SA = 60 square cm

The surface area of the rectangular cuboid will be calculated as:-

SA = ( 4 x 6 ) + 2 ( 158 x 4 ) + 2 ( 15 X 6 )

SA = 24 + 120 + 180

SA  = 324 square cm

Total surface area will be = 324 + 60 = 384 square cm.

Therefore the surface area of the given composite figure will be 384 square cm.

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Answer:

A

Step-by-step explanation:

The pattern is blue red blue red

+

The pattern is vertical horizontal vertical horizontal

Answer:

A.

The second figure adds another red horizontal line to the first. The third adds a vertical blue line to the second. The fourth must add a red horizontal line to the third.

Answer:

a triangle

Step-by-step explanation:

the definition of a triangle is a shape with 3 sides and the shape that is shown has 3 sides.

Answer:

Step-by-step explanation:

Answer:

(-[tex]\frac{\sqrt{3}}{2}[/tex],-[tex]\frac{1}{2}[/tex])

Step-by-step explanation:

The corresponding point on the unit circle is ( -√3/2 , -1/2).

A unit circle is just a circle that has a radius of 1 unit. A unit circle can be used to define right triangle relationships known as sine, cosine and tangent. These relationships describe how angles and sides of a right triangle relate to one another.

For the given situation,

The radian measure is θ = 7pi/6

In unit circle, the radian θ = [tex]\frac{7\pi }{6}[/tex] lies in the third quadrant and it's equal degree measure is [tex]210[/tex]°.

The corresponding point for this radian measure is

( [tex]\frac{-\sqrt{3} }{2} ,\frac{-1}{2}[/tex] )

Hence we can conclude that the corresponding point on the unit circle is ( -√3/2 , -1/2).

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Answer:

I believe it is 9

Step-by-step explanation:

If you turn that into a triangle then find the area it is 9.

Hope this Helps!

Answer:

Think of it as a right triangle. You can use pythagorean theorem.

a^2 + b^2 = c^2

6^2 + 3^2 = c^2

36 + 9 = c^2

45 = c^2

sqrt45 = c

Step-by-step explanation: