If you roll 2 dice, what is the probability the sum is a 4, 7 or 10?

Answer:

VERY NICE RACK U HAVE MAM

Step-by-step explanation:

Answer:

Its a

Step-by-step explanation:

found on another thing and im taking test

Answer:

arc length   =   circumference • [central angle (degrees) ÷ 360]

central angle (degrees)  = 360 * arc length / circumference

circumference = PI * 2 * 4 = 25.1327412287

central angle (degrees)  = 360 * 10 / 25.1327412287

central angle (degrees)  = 143.2394487828

answer is A--------

Answer:

the number before the first variable (first term)

Step-by-step explanation:

this appears to be an incomplete question. The numerical coefficient of a term is the number before the variable.

the constant is the number without a variable.

Answer:

The numerical limits for a C grade are 60.6 and 69.1.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Scores on the test are normally distributed with a mean of 67 and a standard deviation of 7.3.

This means that [tex]\mu = 67, \sigma = 7.3[/tex]

Find the numerical limits for a C grade.

Below the 100 - 38 = 62th percentile and above the 18th percentile.

18th percentile:

X when Z has a p-value of 0.18, so X when Z = -0.915.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-0.915 = \frac{X - 67}{7.3}[/tex]

[tex]X - 67 = -0.915*7[/tex]

[tex]X = 60.6[/tex]

62th percentile:

X when Z has a p-value of 0.62, so X when Z = 0.305.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]0.305 = \frac{X - 67}{7.3}[/tex]

[tex]X - 67 = 0.305*7[/tex]

[tex]X = 69.1[/tex]

The numerical limits for a C grade are 60.6 and 69.1.

Answer:

There is significant evidence to conclude that carina e in New section is greater than 42.3

Step-by-step explanation:

Given :

Sample variance, s² = 48.3

Population variance, σ² = 42.3

Sample size, n = 28

α = 0.05

The hypothesis :

H0 : σ² = 42.3

H0 : σ² > 42.3

The test statistic, χ² : (n-1)*s²/σ²

χ² = [(28 - 1) * 48.3] / 42.3

χ² = (27 * 48.3) / 42.3

χ² = 1304.1 / 42.3

χ² = 30.829787

χ² = 30.830

The Critical value at α = 0.05 ; df = (28-1) = 27

Critical value(0.05, 27) = 27.587

Reject H0 if χ² statistic > Critical value

Since, 30.830 > 27.587 ; Reject H0 ; and conclude that variance in the new section is greater than 42.3

Answer:

C. 11x² - 3x

Step-by-step explanation:

(12x² - 5x) - (x² - 2x)

12x² - 5x - x² + 2x

12x - x² - 5x + 2x

11x² - 3x

Step-by-step explanation:

a)

For each value of x, we need to solve for y.

Our equation is x² -2. This means that we multiply x by itself ( x * x = x²), and then subtract that value by 2.

Here is a list of the unfilled answers:

x = -2:

-2²-2 = -2 * -2 -2 = 4-2 = 2. -2 * -2 = 4 because two negatives multiplied together makes a positive.

x=-1:

-1²-2 = 1 -2 = -1

x=1

1²-2 = 1-2= -1

x=3:

3²-2  = 9-2 = 7

b)

In this graph, the y represents the vertical part and x represents the horizontal. For each value of y, there is a horizontal line that can be drawn. For example, when y=0, the bolded horizontal line near the bottom of the graph represents that. We then need to find the points on the line representing x²-2 where it touches the horizontal line of y=0. This happens when x=-1.5 and x=1.5, as it is between the positive/negative 1 and 2 values of x

Answer:

x = 2 - sqrt(19/6)

x = 2 + sqrt(19/6)

Step-by-step explanation:

Answer:

add 4

subtract 24 from 5

2

Step-by-step explanation:

Answer:

y = 2x + 10

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Here m = 2 , then

y = 2x + c ← is the partial equation

To find c substitute (- 4, 2 ) into the partial equation

2 = - 8 + c ⇒ c = 2 + 8 = 10

y = 2x + 10 ← equation of line

Answer:

5 cm

Step-by-step explanation:

V = l*w*h

360 = l*8*9

360= 72l

Divide each side by 72

360/72 = 72l/72

5=l

Answer:

[tex]length = 5cm[/tex]

Step-by-step explanation:

[tex]volume \: = l \times b \times h \\ l = \frac{v}{b \times h} \\ l = \frac{360 {cm}^{3} }{9cm \times 8cm} \\ l = \frac{360}{72} \\ l = 5cm[/tex]

Hope this helps you.

have a nice day!

Answer:

please someone help me with my latest question

Answer:

41 units

Step-by-step explanation:

Its the reflection the length are same

Answer:

3/11

Step-by-step explanation:

Solution :

Given :

2y''' + 15y'' + 24y' + 11y= 0

Let x = independent variable

[tex](a_0D^n + a_1D^{n-1}+a_2D^{n-2} + ....+ a_n) y) = Q(x)[/tex]  is a differential equation.

If [tex]Q(x) \neq 0[/tex]

It is non homogeneous then,

The general solution  = complementary solution + particular integral

If Q(x) = 0

It is called the homogeneous then the general solution =  complementary solution.

2y''' + 15y'' + 24y' + 11y= 0

[tex]$(2D^3+15D^2+24D+11)y=0$[/tex]

Auxiliary equation,

[tex]$2m^3+15m^2+24m +11 = 0$[/tex]

-1  | 2    15    24     11

    | 0   -2    - 13    -11  

      2    13    11       0

∴ [tex]2m^2+13m+11=0[/tex]

The roots are

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

[tex]$=\frac{-13\pm \sqrt{13^2-4(11)(2)}}{2(2)}$[/tex]

[tex]$=\frac{-13\pm9}{4}$[/tex]

[tex]$=-5.5, -1$[/tex]

So, [tex]m_1, m_2, m_3 = -1, -1, -5.5[/tex]

Then the general solution is :

[tex]$= (c_1+c_2 x)e^{-x} + c_3 \ e^{-5.5x}$[/tex]

Answer:

Step-by-step explanation:

5t² + 4t = 5t² + 4 if and only if t=1.

A zero coefficient makes the value of the term equal to zero.

Answer: 5t to the second power and 5t to the second power. I don't know the answer to b.

Step-by-step explanation:

Answer:

sorry I don't know the answer

Answer:

For the equation y=log(x-k), the domain depends on the value of K. Sliding K moves the left bound of the domain interval. The range and the right end behavior stay the same. For the equation y=log x+k, the domain is fixed, starting at an x-value of 0. The vertical asymptote is also fixed. The range of the equation depends on K.

Step-by-step explanation:

Answer:The value of x is 4.

Step-by-step explanation:

Given : Expression  .

To find : Solve for x ?

Solution :

Applying exponential property,  

Comparing the base,

Step-by-step explanation:

Step-by-step explanation:

You didn't put the graph, but you can compare between your graphs and the picture.

Brainliest please

The graph that represents this inequality y ≥ 4x − 3 is attached below.

If the equation or inequality contains variable terms, then there might be some values of those variables for which that equation or inequality might be true. Such values are called solution to that equation or inequality. Set of such values is called solution set to the considered equation or inequality.

We are given that the inequality is;

y ≥ 4x − 3

The slope of the inequality is 4.

The equation of the red line is y = 4x − 3  

The shading is above the line and the line is solid, that means  y is greater than or equal 4x − 3

The graph of this inequality y ≥ 4x − 3 is attached below.

Learn more about inequalities here:

https://brainly.com/question/27425770

#SPJ2

Answer:

The solutions are w=0 ,7

Step-by-step explanation:

3w^2 – 21w = 0

Factor out 3w

3w(w-7) =0

Using the zero product property

3w=0  w-7=0

w =0    w=7

The solutions are w=0 ,7

In an arithmetic sequence, every pair of consecutive terms differs by a fixed number c, so that the n-th term [tex]a_n[/tex] is given recursively by

[tex]a_n=a_{n-1}+c[/tex]

Then for n ≥ 2, we have

[tex]a_2=a_1+c[/tex]

[tex]a_3=a_2+c = (a_1+c)+c = a_1 + 2c[/tex]

[tex]a_4=a_3+c = (a_1 + 2c) + c = a_1 + 3c[/tex]

and so on, up to

[tex]a_n=a_1+(n-1)c[/tex]

Given that [tex]a_3=126[/tex] and [tex]a_{64}=3725[/tex], we can solve for [tex]a_1[/tex]:

[tex]\begin{cases}a_1+2c=126\\a_1+63c=3725\end{cases}[/tex]

[tex]\implies(a_1+63c)-(a_1+2c)=3725-126[/tex]

[tex]\implies 61c = 3599[/tex]

[tex]\implies c=59[/tex]

[tex]\implies a_1+2\times59=126[/tex]

[tex]\implies a_1+118 = 126[/tex]

[tex]\implies \boxed{a_1=8}[/tex]

Answer:

Complementary angle

<y+61° =90°

<y = 90°- 61°

<y =29°

I Hope this helps.

Answer:

72 in. sq.

Step-by-step explanation:

6 x 6 = 36

6 x 6 = 36 / 2 = 18

6 x 6 = 36 / 2 = 18

18 + 18 + 36 = 72.

Hope this helps!

If there is something wrong just let me know.

Answer:

201.088 cubic units

Step-by-step explanation:

Given data

height= 12 units

radius= 4 units

The expression for the volume of a cone is given as

Volume = 1/3*πr^2h

Substitute the given data to find the volume of the cone

Volunme= 1/3* 3.142* 4^2* 12

Volume = 1/3* 3.142* 16*12

Volume= 1/3* 603.264

Volume= 201.088 cubic units

Answer:

A) quadratic

Step-by-step explanation:

ax2 + bx+c=0

Since the highest power of the equation is 2

A) quadratic -2

B) quartic- 4

C) linear- 1

D) cubic-3

Answer:

ik only 1 of them and i dont even know if this is correct...

a) A

Given:

A man owns 3/4 of the share of a business and sells 1/3 of his  shares for Bir 10,000.

To find:

The value of the business in Bir.

Solution:

Let x be the value of the business.

It is given that a man owns 3/4 of the share of a business and sells 1/3 of his  shares for Bir 10,000.

[tex]x\times \dfrac{3}{4}\times \dfrac{1}{3}=10000[/tex]

[tex]x\times \dfrac{1}{4}=10000[/tex]

Multiply both sides by 4.

[tex]x=4\times 10000[/tex]

[tex]x=40000[/tex]

Therefore, the value of the business is 40,000 Bir.

Answer:

Assuming the base number is 1, if it becomes 5 then it is stretched. If the number become 1/5(less than 1) it is called compressed.

Step-by-step explanation:

We know

[tex]\boxed{\sf P(x,y)=\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)}[/tex]

[tex]\\ \sf\longmapsto p(x,y)=\left(\dfrac{3(7)+5(-1)}{3+5},\dfrac{3(7)+5(1)}{3+5}\right)[/tex]

[tex]\\ \sf\longmapsto p(x,y)=\left(\dfrac{21-5}{8},\dfrac{21+5}{8}\right)[/tex]

[tex]\\ \sf\longmapsto p(x,y)=\left(\dfrac{16}{8},\dfrac{26}{8}\right)[/tex]

[tex]\\ \sf\longmapsto p(x,y)=\left(2,\dfrac{13}{4}\right)[/tex]

m:n=3:5

We know

[tex]\boxed{\sf M(x,y)=\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)}[/tex]

[tex]\\ \sf\longmapsto M(x,y)=\left(\dfrac{3(7)+5(-1)}{3+5},\dfrac{3(7)+5(1)}{3+5}\right)[/tex]

[tex]\\ \sf\longmapsto M(x,y)=\left(\dfrac{21-5}{8},\dfrac{21+5}{8}\right)[/tex]

[tex]\\ \sf\longmapsto M(x,y)=\left(\dfrac{16}{8},\dfrac{26}{8}\right)[/tex]

[tex]\\ \sf\longmapsto M(x,y)=\left(2,\dfrac{13}{4}\right)[/tex]