5. Two unbiased dice are rolled. Calculate the probability that the sum of the two dice is:
ws
(a) 4
(b) 7
(c) Less than 7

Answer:

the line of best fit can be approximated to:

y = -1.560 x + 22.105

Step-by-step explanation:

You are most likely expected to use a graphing tool are statistical program to calculate this. So enter the list of x-values separate from the list of y values and run the tool in linear regression mode.

Look at the attached image with the actual results including the line of best fit.

The equation can be written (rounding slope and y-intercept to 3 decimals) as:

y = -1.560 x + 22.105

Let numbers be x and y

ATQ

Adding both

[tex]\\ \sf\longmapsto 2x=58[/tex]

[tex]\\ \sf\longmapsto x=\dfrac{58}{2}[/tex]

[tex]\\ \sf\longmapsto x=29[/tex]

Now putting value in eq(2)

[tex]\\ \sf\longmapsto x-y=9[/tex]

[tex]\\ \sf\longmapsto 29-9=y[/tex]

[tex]\\ \sf\longmapsto y=20[/tex]

Answer:

Option B, x ≈ -2.25

Step-by-step explanation:

3^x-2=(x-1)/(x^2+x-1)

or x ≈ -2.21166

so it's closest to the answer of the 2nd option

Answer:

1/(x^3 + 6x^2 + 12x + 8)

Step-by-step explanation:

The first thing we do is rationalize this expression. (2+x)^-3 is written as

1/(2+x)^3

Then from there we can foil out the denominator. It is easiest to foil (2+x)(2+x) first and then multiply that product by (2+x).

(2+x)(2+x) = 4 + 4x + x^2

(4+4x+x^2)(2+x) = 8+8x+2x^2+4x+4x^2+x^3.

Then we combine like terms and put them in order to get:

x^3 + 6x^2 + 12x + 8

And of course we can't forget that this was raised to the negative third power, so our answer is 1/(x^3 + 6x^2 + 12x + 8)

Answer:

Hello,

Step-by-step explanation:

[tex](a+x)^n=a^n+\left(\begin{array}{c}n\\ 1\end{array}\right)*a^{n-1}*x+\left(\begin{array}{c}n\\ 2\end{array}\right)*a^{n-2}*x^2+\left(\begin{array}{c}n\\ 3\end{array}\right)*a^{n-3}*x^3+\left(\begin{array}{c}n\\ 4\end{array}\right)*a^{n-4}*x^4+...+\left(\begin{array}{c}n\\ n\end{array}\right)*a^{n-n}*x^n[/tex]

[tex]with \\\\\left(\begin{array}{c}n\\ 1\end{array}\right)=n\\\\\left(\begin{array}{c}n\\ 2\end{array}\right)=\dfrac{n(n-1)}{2!} \\\\\left(\begin{array}{c}n\\3 \end{array}\right)=\dfrac{n(n-1)(n-2)}{3!} \\\\...\\[/tex]

[tex]\dfrac{1}{(2+x)^3} =\dfrac{1}{8} +3*\dfrac{x}{4}+3\dfrac{x^2}{2}+x^3\\\\[/tex]

Answer:

(2) If 300 lunches were sold, then 120 chose tacos.

Step-by-step explanation:

We can evaluate each option and see if it makes it true.

For 1: If 200 lunches were served, 10 more students chose pizza over hotdogs.

We can find how many pizzas/hotdogs were given if 200 lunches were served by relating it to 100.

20% chose hotdog, which is [tex]\frac{20}{100}[/tex]. Multiply both the numerator and denominator by two: [tex]\frac{40}{200}[/tex] - so 40 students chose hotdogs.

Same logic for pizza: 30% chose pizza - [tex]\frac{30}{100} = \frac{60}{200}[/tex] so 60.

60 - 40 = 20, not 10, so 1 doesn't work.

2: If 300 lunches were sold, then 120 chose tacos.

Let's set up a proportion again. 40% of 100 is 40.

[tex]\frac{40}{100} = \frac{40\cdot3}{300} = \frac{120}{300}[/tex]

So 120 tacos were chosen - yes this works!

Hope this helped!

Answer:

 At the 5% level, BAOI can infer that the average time to complete does not exceeds 60 hours.

Step-by-step explanation:

From the question we are told that

   The  population mean is [tex]\mu = 60 \ hr[/tex]

    The sample size is  [tex]n = 16[/tex]

    The  sample mean is  [tex]\= x = 68 \ hr[/tex]

     The  standard deviation is  [tex]\sigma = 20 \ hr[/tex]

The  null hypothesis is  [tex]H_o : \mu = 60[/tex]

The  alternative [tex]H_a : \mu > 60[/tex]

Here we would assume the level of significance of this test to be  

         [tex]\alpha = 5\% = 0.05[/tex]

Next we will obtain the critical value of the level of significance from the normal distribution table, the value is    [tex]Z_{0.05} = 1.645[/tex]

  Generally the test statistics  is mathematically represented as

           [tex]t = \frac{ \= x - \mu}{ \frac{ \sigma }{\sqrt{n} } }[/tex]

substituting values

           [tex]t = \frac{ 68 - 60 }{ \frac{ 20 }{\sqrt{16} } }[/tex]

          [tex]t = 1.6[/tex]

Looking at the value of t and  [tex]Z_{\alpha }[/tex] we see that [tex]t< Z_{\alpha }[/tex] hence we fail to reject the null hypothesis

   This means that there no sufficient evidence to conclude that it takes more than 60 hours to complete the course

So

   At the 5% level, BAOI can infer that the average time to complete does not exceeds 60 hours.

Answer:

Length = Width = 7 cm

Step-by-step explanation:

Volume of a triangular prism is represented by the formula,

Volume = (Area of the triangular base) × height

588 = 49 × h

h = [tex]\frac{588}{49}[/tex]

h = 12 cm

We have to find the side length of a rectangular prism having same volume.

Volume = Area of the rectangular base × height

588 = (l × b) × h [l = length and b = width ]

588 = (l × b) × 12

l × b = 49 = 7 × 7

Therefore, length = width = 7 cm may be the side lengths of the rectangular prism to have the same volume.              

Answer:

the percentage of  bearings   that will  not be acceptable = 7.3%

Step-by-step explanation:

Given that:

Mean = 0.499

standard deviation = 0.002

if the true average diameter of the bearings it produces is 0.500 in and bearing is acceptable if its diameter is within 0.004 in.

Then the ball bearing acceptable range = (0.500 - 0.004, 0.500 + 0.004 )

= ( 0.496 , 0.504)

If x represents the diameter of the bearing , then the probability for the  z value for the random variable x with a mean and standard deviation can be computed as follows:

[tex]P(0.496\leq X \leq 0.504) = (\dfrac{0.496 - \mu}{\sigma} \leq \dfrac{X -\mu}{\sigma} \leq \dfrac{0.504 - \mu}{\sigma})[/tex]

[tex]P(0.496\leq X \leq 0.504) = (\dfrac{0.496 - 0.499}{0.002} \leq \dfrac{X -0.499}{0.002} \leq \dfrac{0.504 - 0.499}{0.002})[/tex]

[tex]P(0.496\leq X \leq 0.504) = (\dfrac{-0.003}{0.002} \leq Z \leq \dfrac{0.005}{0.002})[/tex]

[tex]P(0.496\leq X \leq 0.504) = (-1.5 \leq Z \leq 2.5)[/tex]

[tex]P(0.496\leq X \leq 0.504) = P (-1.5 \leq Z \leq 2.5)[/tex]

[tex]P(0.496\leq X \leq 0.504) = P(Z \leq 2.5) - P(Z \leq -1.5)[/tex]

From the standard normal tables

[tex]P(0.496\leq X \leq 0.504) = 0.9938-0.0668[/tex]

[tex]P(0.496\leq X \leq 0.504) = 0.927[/tex]

By applying the concept of probability of a  complement , the percentage of bearings will now not be acceptable

P(not be acceptable)  = 1 - P(acceptable)

P(not be acceptable)  = 1 - 0.927

P(not be acceptable)  = 0.073

Thus, the percentage of  bearings   that will  not be acceptable = 7.3%

Answer:

C. 1059.50

Step-by-step explanation:

Sales price x sales tax rate = sales tax

16299.99 x .065 (6.5%) = 1059.50

Answer:

3x - 4

Step-by-step explanation:

As Tina's age is 3 into x ( 3 x x= 3x)but 4years less (-4)

Therefore Tina's age is 3x - 4

Answer:

3x - 4

Step-by-step explanation:

Use these representations:  niece's age: x

We triple x and then subract 4 years from the result, obtaining:

Tina's age:  3x - 4

Split up the boundary of C (which I denote ∂C throughout) into the parabolic segment from (1, 1) to (0, 0) (the part corresponding to y = x ²), and the line segment from (1, 1) to (0, 0) (the part of ∂C on the line y = x).

Parameterize these pieces respectively by

r(t) = x(t) i + y(t) j = t i + t ² j

and

s(t) = x(t) i + y(t) j = (1 - t ) i + (1 - t ) j

both with 0 ≤ t ≤ 1.

The circulation of F around ∂C is given by the line integral with respect to arc length,

[tex]\displaystyle \int_{\partial C}\mathbf F\cdot\mathbf T \,\mathrm ds[/tex]

where T denotes the tangent vector to ∂C. Split up the integral over each piece of ∂C :

• on the parabolic segment, we have

T = dr/dt = i + 2t j

• on the line segment,

T = ds/dt = -i - j

Then the circulation is

[tex]\displaystyle \int_{\partial C}\mathbf F\cdot\mathbf T\,\mathrm ds = \int_0^1 (7t^3\,\mathbf i+5t^4\,\mathbf j)\cdot(\mathbf i+2t\,\mathbf j)\,\mathrm dt + \int_0^1 (7(1-t)^2\,\mathbf i+5(1-t)^2\,\mathbf j)\cdot(-\mathbf i-\mathbf j)\,\mathrm dt \\\\ = \int_0^1 (7t^3+10t^5)\,\mathrm dt - 12 \int_0^1 (1-t)^2\,\mathrm dt =\boxed{-\frac7{12}}[/tex]

Alternatively, we can use Green's theorem to compute the circulation, as

[tex]\displaystyle\int_{\partial C}\mathbf F\cdot\mathbf T\,\mathrm ds = \iint_C\frac{\partial(5y^2)}{\partial x} - \frac{\partial(7xy)}{\partial y}\,\mathrm dx\,\mathrm dy \\\\ = -7\int_0^1\int_{x^2}^x x\,\mathrm dx \\\\ = -7\int_0^1 xy\bigg|_{y=x^2}^{y=x}\,\mathrm dx \\\\ =-7\int_0^1(x^2-x^3)\,\mathrm dx = -\frac7{12}[/tex]

The flux of F across ∂C is

[tex]\displaystyle \int_{\partial C}\mathbf F\cdot\mathbf N \,\mathrm ds[/tex]

where N is the normal vector to ∂C. While T = x'(t) i + y'(t) j, the normal vector is N = y'(t) i - x'(t) j.

• on the parabolic segment,

N = 2t i - j

• on the line segment,

N = - i + j

So the flux is

[tex]\displaystyle \int_{\partial C}\mathbf F\cdot\mathbf N\,\mathrm ds = \int_0^1 (7t^3\,\mathbf i+5t^4\,\mathbf j)\cdot(2t\,\mathbf i-\mathbf j)\,\mathrm dt + \int_0^1 (7(1-t)^2\,\mathbf i+5(1-t)^2\,\mathbf j)\cdot(-\mathbf i+\mathbf j)\,\mathrm dt \\\\ = \int_0^1 (14t^4-5t^4)\,\mathrm dt - 2 \int_0^1 (1-t)^2\,\mathrm dt =\boxed{\frac{17}{15}}[/tex]

Answer:

Step-by-step explanation:

a. table

x = 100,y = 20*100+3000 = 2000+3000 = 5000

x = 200,y = 20*200+3000 = 4000+3000 = 7000

x = 300,y = 20*300+3000 = 6000+3000 = 9000

b：

y = 4300

4300 = 20x+3000

20x = 4300-3000

20x = 1300

x = 1300/20

x  = 65

so 65 computer desks can be produced.

Answer:

x= i√30

Step-by-step explanation:

I'm going to go into this under the assumption that you've covered imaginary numbers based on the question. If I'm wrong then sorry about that.

Okay, so first you want to subtract 30 from both sides

x^2=-30

Then you take the square root of each side.

√(x^2)=√-30

x=√-30

Since it's impossible to square a number to get a negative number, you'll end up with an imaginary number. You have to rewrite x=√-30 to get rid of the negative sign under the radical. Rewriting this will also indicate that it's an imaginary number.

Final answer: x = i√30

Answer: Option "a" is the only expression that represents a function.

Step-by-step explanation:

A function f(x) = y is a "operator" that takes an input element, x, and assigns it to only one output element, y.

So, if we have that for a given value of x.

f(x) = y and f(x) = h

where y and h are different values, then this is not a function, because is assigning the input value x to two different output values.

Let's see the different options:

a) {(1, 2), (4, −2), (8, 3), (9, −3)}

This points are of the form (x, y)

We can see that each value of x is assigned to only one value of y, so this can represent a function.

b)  y^2 = 16 − x^2

Ok, suppose that x = 0, then:

y^2 = 16 - 0 = 16

then we have that y*y = 16.

So y can take two different values:

y = 4 ---> 4*4 = 16

y = -4 ---> -4*-4 = 16.

So this is not a function.

c) 2x^2 + y^2 = 5

First, we want to isolate y in one side:

y^2 = 5 - 2*x^2

Here we have a similar case to the option b, and we can use a similar argument to prove that this is not a function, so we can discard this.

d) x = 7.

Ok, this is not a relation between two variables, so this is not a function, as if x is the input value, we have only one value of x that solves the equation.

Answer:

Yes, Rolle's theorem can be applied

There is only one value of c such that f'(c) = 0, and this is c = 1.5 (or 3/2 in fraction form)

Step-by-step explanation:

Yes, Rolle's theorem can be applied on this function because the function is continuous in the closed interval (it is a polynomial function) and differentiable  in the open interval, and f(a) = f(b) given that:

[tex]f(0)=-0^2+3\,(0)=0\\f(3)=-3^2+3\,(3)=-9+9=0[/tex]

Then there must be a c in the open interval for which f'(c) =0

In order to find "c", we derive the function and evaluate it at "c", making the derivative equal zero, to solve for c:

[tex]f(x)=-x^2+3\,x\\f'(x)=-2\,x+3\\f'(c)=-2\,c+3\\0=-2\,c+3\\2\,c=3\\c=\frac{3}{2} =1.5[/tex]

There is a unique answer for c, and that is c = 1.5

Rolle's theorem is applicable if [tex]f(a)=f(b)[/tex] and $f$ is differentiable in $(a,b)$

since it's polynomial function, it's always continuous and differentiable..

and you can easily check that $f(0)=f(-3)=0$

so it is applicable.

now, $f'(x)=-2x+3=0 \implies x=\frac32$

there is only once value (as you can imagine, the graph will be downward parabola)

Answer:

1/13

Step-by-step explanation:

There are 52 cards in a deck of cards and 13 of them are hears

P(heart) = hearts / total

              = 13/52 = 1/13

Using Pythagorean theorem

[tex]\\ \sf\longmapsto H^2=P^2+B^2[/tex]

[tex]\\ \sf\longmapsto H^2=5^3+3^2[/tex]

[tex]\\ \sf\longmapsto H^2=25+9[/tex]

[tex]\\ \sf\longmapsto H^2=34[/tex]

[tex]\\ \sf\longmapsto H=\sqrt{34}[/tex]

[tex]\\ \sf\longmapsto H=5.22[/tex]

Answer: 33% is moisture content

Step-by-step explanation:

120kg - 80kg = 40kg

40 of 120 is %

Work:

40/120 = 0.33

0.33x100

= 33%

Complete Question

The complete question is shown on the first uploaded image

Answer:

The  lower bound is  [tex]0.0234[/tex]

The  upper bound is  [tex]0.100[/tex]

So from the value obtained the solution to the question are

  1  Does not include

  2 sufficient

 3  not different  

Step-by-step explanation:

From the question we are told that

The  sample size of  individuals who earned in excess of​ $100,000 per​ year is   [tex]n_ 1 = 1205[/tex]

The  number of  individuals who earned in excess of​ $100,000 per​ year  that said yes is

    [tex]w = 712[/tex]

The  sample size  individuals who earned less than​ $100,000 per​ year is [tex]n_2 = 1310[/tex]

The  number of  individuals who earned less than​ $100,000 per​ year that said yes is

       [tex]v= 693[/tex]

The sample proportion of  individuals who earned in excess of​ $100,000 per​ year  that said yes is

           [tex]\r p _ 1 = \frac{w}{n_1 }[/tex]

substituting values

          [tex]\r p _ 1 = \frac{712}{1205}[/tex]

          [tex]\r p _ 1 =0.5909[/tex]

The sample proportion of  individuals who earned less than​ $100,000 per​ year that said yes is

          [tex]\r p _ 1 = \frac{v}{n_2 }[/tex]

substituting values

         [tex]\r p _ 1 = \frac{693 }{1310}[/tex]

        [tex]\r p _ 1 = 0.529[/tex]

Given that the confidence level is  95% then the level of significance is mathematically represented as

           [tex]\alpha = 1 -0.95[/tex]

           [tex]\alpha = 0.05[/tex]

 Next we obtain the critical value of  [tex]\frac{\alpha }{2}[/tex] from the normal distribution table the value is  [tex]Z_{\frac{\alpha }{2} } = 1.96[/tex]

    Generally the margin of error is  

   [tex]E = Z_{\frac{\alpha }{2} } * \sqrt{ \frac{ \r p _1 (1- \r p_1 )}{n_1} + \frac{ \r p _2 (1- \r p_2 )}{n_2} } }[/tex]

substituting values

   [tex]E = 1.96 * \sqrt{ \frac{ 0.5909 (1- 0.5909 )}{1205} + \frac{ 0.592 (1- 0.6592 )}{1310} } }[/tex]

    [tex]E =0.03846[/tex]

Generally the 95% confidence interval is  

        [tex](\r p_1 - \r p_2) - E < p_1 - p_2 <( \r p_1 - \r p_2 ) + E[/tex]

substituting values

        [tex](0.5909 - 0.529 ) - 0.03846 < p_1 - p_2 < (0.5909 - 0.529 ) + 0.03846[/tex]

         [tex]0.02344 < p_1 - p_2 < 0.10036[/tex]

The  lower bound is  [tex]0.0234[/tex]

The  upper bound is  [tex]0.100[/tex]

So from the value obtained the solution to the question are

  1  Does not include

  2 sufficient

 3  not different  

The lower bound is 0.0234 and the upper bound is 0.100. Then the 95% confidence interval is (0.0234, 0.100)

The probability or the chances of error while choosing or calculating a sample in a survey is called the margin of error.

The research group asked the following question of individuals who earned in excess of​ $100,000 per year and those who earned less than​ $100,000 per​ year.

The sample size of individuals who earned in excess of $100,000 per year will be

[tex]\rm n_1 =1205[/tex]

The sample size of individuals who earned less than $100,000 per year will be

[tex]\rm n_1 =1205[/tex]

The number of individuals who earn an excess of $100,000 per year that said yes will be

[tex]\rm w = 712[/tex]

The number of individuals who earn less than $100,000 per year that said yes will be

[tex]\rm v= 693[/tex]

Then the sample proportion of individuals who earned in excess of $100,000 per year that said yes will be

[tex]\rm \hat{p}_1=\dfrac{w}{n_1}\\\\\hat{p}_1=\dfrac{712}{1205}\\\\\hat{p}_1= 0.5909[/tex]

Then the sample proportion of individuals who earned less than $100,000 per year that said yes will be

[tex]\rm \hat{p}_2=\dfrac{v}{n_2}\\\\\hat{p}_2=\dfrac{693}{1310}\\\\\hat{p}_2= 0.529[/tex]

The confidence level is 95% then the level of significance is mathematically represented as

[tex]\alpha =1-0.95\\\\\alpha =0.05[/tex]

Then the critical value of α/2 from the normal distribution table. Then the value of z is 1.96, then the error of margin will be

[tex]E = z_{\alpha /2} \times \sqrt{\dfrac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \dfrac{\hat{p}_2(1-\hat{p}_2)}{n_2}}\\\\E = 1.96 \times \sqrt{\dfrac{05909(1-0.5909)}{1205} + \dfrac{0.529(1-0529)}{1310}}\\\\E = 0.03846[/tex]

The 95% confidence interval will be

[tex]\begin{aligned} (\hat{p}_1-\hat{p}_2)-E & < p_1-p_2 < (\hat{p}_1-\hat{p}_2) + E\\\\(0.5909 - 0.529) - 0.03846 & < p_1-p_2 < (0.5909 - 0.529) + 0.03846\\\\0.02344 & < p_1-p_2 < 0.10036 \end{aligned}[/tex]

More about the margin of error link is given below.

https://brainly.com/question/6979326

Answer:

Hey there!

We can use the midpoint formula to find that the midpoint is (-1, -9).

Let me know if this helps :)

The midpoint of the segment between the points (8,−10) and (−10,−8) will be (−1, −9). Then the correct option is A.

What is the midpoint of line segment AB?

Let C be the mid-point of the line segment AB.

A = (x₁, y₁)

B = (x₂, y₂)

C = (x, y)

Then the midpoint will be

x = (x₁ + x₂) / 2

y = (y₁ + y₂) / 2

The midpoint of the segment between the points (8,−10) and (−10,−8)

x = (8 – 10) / 2

x = –1

y = (– 10 – 8) / 2

y = –9

Then the correct option is A.

More about the midpoint of line segment AB link is given below.

https://brainly.com/question/17410964

#SPJ5

Answer:

∠NMC  = 50°

Step-by-step explanation:

The interpretation of the information given in the question can be seen in the attached images below.

In ΔABC;

∠ A + ∠ B + ∠ C = 180°    (sum of angles in a triangle)

∠ A + 70°  + 50°  = 180°

∠ A = 180° - 70° - 50°

∠ A =  180° - 120°

∠ A =  60°

In ΔAMN ; the base angle are equal , let the base angles be x and y

So; x = y   (base angle of an equilateral  triangle)

Then;

x + x + 60° = 180°

2x +  60° = 180°

2x = 180° - 60°

2x = 120°

x = 120°/2

x = 60°

∴ x = 60° , y = 60°

In ΔBQC

∠a + ∠e + ∠b = 180°

50° + ∠e + 40° = 180°

∠e = 180° - 50° - 40°

∠e = 180° - 90°

∠e = 90°

At point Q , ∠e = ∠f = ∠g = ∠h = 90°  (angles at a point)

∠i  = 50° - 40° = 10°

In ΔNQC

∠f + ∠i   + ∠j = 180°

90° + 10° + ∠j = 180°

∠j  = 180° - 90°-10°

∠j  = 180° - 100°

∠j  = 80°

From  line AC , at point N , ∠y + ∠c + ∠j = 180°   (sum of angles on a straight line)

60° + ∠c + ∠80° = 180°

∠c  = 180° - 60°-80°

∠c  = 180° - 140°

∠c  = 40°

Recall that :

At point Q , ∠e = ∠f = ∠g = ∠h = 90°  (angles at a point)

Then In Δ NMC ;

∠d + ∠h + ∠c = 180°   (sum of angles in a triangle)

∠d + 90° + 40° = 180°

∠d  = 180° - 90° -40°

∠d  = 180° - 130°

∠d  = 50°

Therefore, ∠NMC = ∠d  = 50°

Answer:

12). LM = 37.1 units

13). c = 4.6 mi

Step-by-step explanation:

12). LM² = 23² + 20² - 2(23)(20)cos(119)°

    LM² = 529 + 400 - 920cos(119)°

    LM² = 929 - 920cos(119)°

    LM = [tex]\sqrt{929+446.03}[/tex]

          = [tex]\sqrt{1375.03}[/tex]

          = 37.08

          ≈ 37.1 units

13). c² = 5.4² + 3.6² - 2(5.4)(3.6)cos(58)°

    c² = 29.16 + 12.96 - 38.88cos(58)°

    c² = 42.12 - 38.88cos(58)°

    c = [tex]\sqrt{42.12-20.603}[/tex]

    c = [tex]\sqrt{21.517}[/tex]

    c = 4.6386

    c ≈ 4.6 mi

Answer:

  x ≈ {0.653059729092, 3.75570086464}

Step-by-step explanation:

A graphing calculator can tell you the roots of ...

  f(x) = ln(x) -1/(x -3)

are near 0.653 and 3.756. These values are sufficiently close that Newton's method iteration can find solutions to full calculator precision in a few iterations.

In the attachment, we use g(x) as the iteration function. Since its value is shown even as its argument is being typed, we can start typing with the graphical solution value, then simply copy the digits of the iterated value as they appear. After about 6 or 8 input digits, the output stops changing, so that is our solution.

Rounded to 6 decimal places, the solutions are {0.653060, 3.755701}.

_____

A similar method can be used on a calculator such as the TI-84. One function can be defined a.s f(x) is above. Another can be defined as g(x) is in the attachment, by making use of the calculator's derivative function. After the first g(0.653) value is found, for example, remaining iterations can be g(Ans) until the result stops changing,

The value of y will be equal to 4. The correct option is D.

Expression in maths is defined as the collection of the numbers variables and functions by using signs like addition, subtraction, multiplication, and division.

Numbers (constants), variables, operations, functions, brackets, punctuation, and grouping can all be represented by mathematical symbols, which can also be used to indicate the logical syntax's order of operations and other features.

The given expression 8(y + 2) = 48 will be solved for y as below:-

8(y + 2) = 48

Divide both sides by 8 and solve.

[ 8 (y + 2) ] / 8 = 48 / 8

y + 2 = 6

Substract 2 from both the sides to get the value of y.

y + 2 - 2 = 6 -2

y = 4

Therefore, the value of y will be equal to 4. The correct option is D.

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Answer:

$6284.4≤μ≤$6313.6

Step-by-step explanation:

Using the formula for calculating confidence interval as shown:

CI = xbar ± Z×S/√n

xbar is the average premium

Z is the z-score at 95% confidence

S is the standard deviation

n is the sample size

Given parameters

xbar = $6600

Z score at 95% CI = 1.96

S = $800

n = 25

Substituting this parameters in the formula we have;

CI = 6600±1.96×800/√25

CI = 6600±(1.96×800/5)

CI = 6600±(1.96×160)

CI = 6600±313.6

CI = (6600-313.6, 6600+313.6)

CI = (6284.4, 6913.6)

Hence the 95% confidence interval for the mean premium amount paid by all workers who have employer provided health insurance is $6284.4≤μ≤$6313.6

It's given that the Bracelet uses 3 times the number of beads that's used in making a single earring.

It's also given that one single earing has 13 beads. So a single bracelet would have (3×13) beads .... and that's equal to 39.

Making a single set of jewellery needs a pair of earrings and a Bracelet.

So total number of required beads will be =

Answer:

345/5=69

Step-by-step explanation:

345/5=69

355/5=71

Answer:

Hello,

What is tan 30°?

[tex]tan(30^o)=\dfrac {\sqrt{3} }{3}[/tex]

Step-by-step explanation:

[tex]sin(30^o)=\dfrac{1}{2} \\\\cos(30^o)=\dfrac{\sqrt{3} }{2} \\\\\\tan(30^o)=\dfrac{sin(30^o)}{cos(30^o)} \\tan(30^o)=\dfrac{\dfrac{1}{2} } { \dfrac{\sqrt{3} }{2} }\\\\ =\dfrac {1*2}{2*\sqrt{3} }\\\\ =\dfrac {\sqrt{3} }{3}[/tex]

The value of tan 30° is 1/√3

The tangent of an angle is the ratio of the length of the opposite side to the length of the adjacent side.

In other words, it is the ratio of sine and cosine function of an acute angle such that the value of cosine function should not equal to zero.

Tan 30° = sin 30° / cos 30°

We know that, sin 30° = 1/2

cos 30° = √3/2

Therefore,

Tan 30° = 1/2 ÷ √3/2

Tan 30° = 1/2 x 2/√3

Tan 30° = 1/√3

Hence, the value of tan 30° is 1/√3

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Question

In a study of 100 new cars, 29 are white. Find p and q , where p is the proportion of new cars that are white.

Answer:

p = 0.29  and q = 0.71

Step-by-step explanation:

Given

Total new cars =  100

White new cars = 29

Required

Determine p and q

From the question;

p represents white new cars

Hence;

[tex]p = 29[/tex]

Note that;

[tex]p + q = 100[/tex]

Substitute 29 for p

[tex]29 + q = 100[/tex]

[tex]29 - 29 + q = 100 - 29[/tex]

[tex]q = 100 - 29[/tex]

[tex]q = 71[/tex]

The proportion of p is calculate by dividing p by the total number of new cars (Same process is done for q)

For proportion of p

[tex]Proportion,\ p = \frac{p}{new\ cars}[/tex]

[tex]Proportion,\ p = \frac{29}{100}[/tex]

[tex]Proportion,\ p = 0.29[/tex]

For proportion of q

[tex]Proportion,\ q = \frac{q}{new\ cars}[/tex]

[tex]Proportion,\ q = \frac{71}{100}[/tex]

[tex]Proportion,\ q = 0.71[/tex]