Kevin and Randy Muise have a jar containing 95 coins, all of which are either quarters or nickels. The total value in the jar is $13.95. How many of each type of coin do they have?

Answer:

Graph each side of the equation. The solution is the x-value of the point of intersection.

x≈−10.37392996

Step-by-step explanation:

Answer:

9.85

that is 9.8

Step-by-step explanation:

16sin38=x

x= 9.85

Answer: 90% of people marry there 7th grade love. since u have read this, u will be told good news tonight. if u don't pass this on nine comments your worst week starts now this isn't fake. apparently if u copy and paste this on ten comments in the next ten minutes you will have the best day of your life tomorrow. you will either get kissed or asked out in the next 53 minutes someone will say i love you.

Step-by-step explanation:

Answer:

The answer is 102, good luck<3

Answer:

(-4, 2)

Step-by-step explanation:

this is a simultaneous equation;

solving by substitution method, we have;

since y= ¼x + 3 and the same y= 2x + 10

therefore we combine both equations;

¼x + 3= 2x+10

3-10=2x-¼x

to clear out fraction, multiply throughout by 4

-28= 8x-x

-28= 7x

x= -28/7

x= -4

since x is -4, therefore All we have to do is to substitute x into y, doing this, we have;

y=2x+10

y= 2(-4) + 10

y= -8 + 10

y= 2

therefore y= 2 and x= -4

so the values are (-4, 2)

if you still don't understand, check your NGM SS2 textbook or make a comment and I will explain better

Answer:

The host can only prepare 7 plates

Step-by-step explanation:

7 plates, each will contain 1 slice of cherry pie and 2 slices of pumpkin pie. (14/7 = 2)

Answer:

C=2rPi

Step-by-step explanation:

The circumference of a circle is the 2 times the radius and Pi

Answer:

B. 4

Step-by-step explanation:

∠4 is an alternate exterior angle with ∠2 because both angles are on the exterior of the lines, and lie on opposite ends on the transversal.

The angle that is an alternate exterior angle with angle 2 is angle 4.

Parallel lines are those lines that are equidistance from each other and never intersect each other.

Given that;

The figure is shown in figure.

Now,

Since, ∠4 is an alternate exterior angle with ∠2 because both angles are on the exterior of the lines, and lie on opposite ends on the transversal.

Hence, ∠4 is an alternate exterior angle with angle 2.

Thus, The angle that is an alternate exterior angle with angle 2 is angle 4.

Learn more about the parallel lines visit:

https://brainly.com/question/26961508

#SPJ2

Answer:

1. 490.6 mm²

2. 14.1 ft²

Step-by-step explanation:

1. 3.14×(12.5)²= 3.14 × 156.25 =490.6mm²

2. ½×3.14×3²= 3.14×4.5= 14.1 ft²

Answer:

70 cents

Step-by-step explanation:

Answer:

The value of x is 14.

Step-by-step explanation:

3x° = (x + 28)° [ vertically opposite angles

are equal]

3x = x + 28°

3x - x = 28°

2x = 28°

x = 28°/2

x = 14°

Answer:

144

Step-by-step explanation:

Using the formula they gave of V= l*w*h

w=4, l=3, and h=12

3*4*12=12*12=144in^3

Answer:

B. 144

Step-by-step explanation:

Using the formula given (Volume= Length x Width x Height), you will get the equation of 3 x 4 x12. 3 x 4 = 12 and 12 x 12 is 144.

Answer:

The 95% confidence interval for the population's mean length of vacation, in days, is (4.24, 4.76).

The 92% confidence interval for the population's mean length of vacation, in days, is (4.27, 4.73).

Step-by-step explanation:

We have the standard deviations for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 85 - 1 = 84

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 84 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.95}{2} = 0.975[/tex]. So we have T = 1.989.

The margin of error is:

[tex]M = T\frac{s}{\sqrt{n}} = 1.989\frac{1.2}{\sqrt{85}} = 0.26[/tex]

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 4.5 - 0.26 = 4.24 days

The upper end of the interval is the sample mean added to M. So it is 4.5 + 0.26 = 4.76 days

The 95% confidence interval for the population's mean length of vacation, in days, is (4.24, 4.76).

92% confidence interval:

Following the sample logic, the critical value is 1.772. So

[tex]M = T\frac{s}{\sqrt{n}} = 1.772\frac{1.2}{\sqrt{85}} = 0.23[/tex]

The lower end of the interval is the sample mean subtracted by M. So it is 4.5 - 0.23 = 4.27 days

The upper end of the interval is the sample mean added to M. So it is 4.5 + 0.23 = 4.73 days

The 92% confidence interval for the population's mean length of vacation, in days, is (4.27, 4.73).

Answer:

1808.64 IN^3

Step-by-step explanation:

Helium needed = volume of spherical shaped ice cream + volume of cone

Volume of a sphere = 4/3 x pi x r^3

4/3 x 3.14 x 6^3 = 904.32 ft^3

Volume of a cone 1/3 x (nr^2h)

n = 22/7

r = radius  = 12/2 = 6

the diameter is the straight line that passes through the centre of a circle and touches the two edges of the circle.

h = height

1/3 x 3.14 x 6^2 x 24 = 904.32

helium needed = 2 x 904.32 = 1808.64 in^3

Answer:

Solution given:

The given equation of a line is

ax²+2hxy+by²=0

Let y=mx be any one line of

ax²+2hxy+by²=0

Let the perpendicular line of

y=mx is

x+my=0

According to the question the line x+my=0 is one line of Ax²+2Hxy +By²=0

Substituting value of y

Ax²+2Hx [tex] \frac{ - x}{m} [/tex]+[tex]B \frac{x²}{m²} [/tex]=0

Ax²-[tex] \frac{ 2H}{m}x² [/tex]+[tex]B \frac{x²}{m²} [/tex]=0

Taking LCM

we get

Ax²m²-2mHx²+Bx²=0

x²[Am²-2Hm+B]=0..............[1]

Again.

ax²+2hx(mx)+B(mx)²=0

ax²+2hmx²+Bm²x²=0

x²[a+2hm+Bm²]=0

bm²+2hn+a=0.....................[2]

Taking coefficient of equation 1 &2.

equation 1. b 2h a b 2h

equation 2.A. -2h B A -2H

Doing criss cross multiplication

ignore first coefficient and repeat first and second

again

lines are perpendicular so

[tex] \frac{m²}{2hB} [/tex]=[tex] \frac{m}{aA-bB} [/tex]=[tex]\frac{1}{-2Hb-2hA} [/tex]

Taking 1st & 2nd ratio,we get,

m=[tex] \frac{2hB+2Ha}{aA-bB} [/tex]....[*]

Taking 3rd & 2nd ratio,we get,

m=[tex] \frac{aA-bB}{-2Hb-2hA} [/tex] ....[#]

Again

Equating equation * &# we get;

[tex] \frac{2hB+2Ha}{aA-bB} [/tex]=[tex] \frac{aA-bB}{-2Hb-2hA} [/tex]

(aA-bB)²=-4(hB+Ha)(Hb+HA)

(aA-bB)²+4(hB+Ha)(Hb+HA)=0 is a required condition.

Answer:

awnser is 36

Step-by-step explanation:

Answer:  x = 37

Step-by-step explanation:

you can use the pythagorean theorem to solve this:

[tex]a^{2} + b^{2} = c^{2}[/tex]

'a' and 'b' in the equation is referring to the legs of the right triangle, or the sides adjacent to the hypotenuse (which is the side opposite the right angle)

plug in the legs into the equation and you get :

[tex]12^{2} + 35^{2} = x^{2}[/tex]

now to solve that we just square 12 and 35 and you get

144 + 1225 = [tex]x^{2}[/tex]

1369 =  [tex]x^{2}[/tex]

x = [tex]\sqrt{1369}[/tex]

x= 37

:))

Answer:

-20

Step-by-step explanation:

cause i did the math hope this helped :D

I used a calculator, the answer is

-20

Answer:

i belive the correct awnser is A

Step-by-step explanation:

9514 1404 393

Answer:

  A  figure 1

Step-by-step explanation:

The given equation is in slope-intercept form:

  y = mx + b . . . . . . where m is the slope and b is the y-intercept

The slope and y-intercept in your equation are both 2. That means the graph crosses the y-axis at y=2 (above the x-axis), and has a positive slope (goes up to the right.

The two top graphs, figures 1 and 2, show lines with positive slope. Only figure 1 shows a graph with positive slope and a positive y-intercept.

Answer:

I believe Awnser B..  (0.5, 0)

Step-by-step explanation:

Answer:

Dwane has 25 rare edition stamps, or 5/6.

Step-by-step explanation:

The question is basically answered in the problem.

It says, "25 of the stamps are a rare edition."

And the question is, "How many rare edition stamps does Dwane have?"

So, 25 is the answer.

Hope this helps!

9514 1404 393

Answer:

  23)  35.77 in²

  25)  48.19 cm²

Step-by-step explanation:

Use the appropriate area formula with the given information.

__

23) The area of a triangle is given by the formula ...

  A = 1/2bh . . . . . base b, height h

  A = 1/2(9.8 in)(7.3 in) = 35.77 in²

__

25) The area of a parallelogram is given by the formula ...

  A = bh . . . . . . base b, height h

  A = (7.9 cm)(6.1 cm) = 48.19 cm²

_____

The height in each figure is measured perpendicular to the base. This tells you that the length 10.6 cm of the diagonal side of the triangle is not relevant to finding the area.

Answer:

23) 35.77 in2

24) 48.19cm2

Answer:

im not gonna help u~ but i can talk to u~ <3

Step-by-step explanation:

Answer:

The 95% confidence interval to determine the true proportion of all patrons who would be interested in attending the showing is (0.2121, 0.4671).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].

Of the 53 patrons, 18 said that they would come to see the film.

This means that [tex]n = 53, \pi = \frac{18}{53} = 0.3396[/tex]

95% confidence level

So [tex]\alpha = 0.05[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3396 - 1.96\sqrt{\frac{0.3396*0.6604}{53}} = 0.2121[/tex]

The upper limit of this interval is:

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3396 + 1.96\sqrt{\frac{0.3396*0.6604}{53}} = 0.4671[/tex]

The 95% confidence interval to determine the true proportion of all patrons who would be interested in attending the showing is (0.2121, 0.4671).

Answer:

d. 15

Step-by-step explanation:

Putting the values in the shift 2 function

X1 + X2 ≥ 15

where x1=  13, and x2=2

13+12≥ 15

15≥ 15

At least 15 workers must be assigned to the shift 2.

The LP model questions require that the constraints are satisfied.

The constraint for the shift 2 is that the  number of workers must be equal or greater than 15

This can be solved using other constraint functions e.g

Putting  X4= 0 in

X1 + X4 ≥ 12

gives

X1 ≥ 12

Now Putting the value X1 ≥ 12  in shift 2 constraint

X1 + X2 ≥ 15

12+ 2≥ 15

14 ≥ 15

this does not satisfy the condition so this is wrong.

Now from

X2 + X3 ≥ 16

Putting X3= 14

X2 + 14 ≥ 16

gives

X2  ≥ 2

Putting these in the shift 2

X1 + X2 ≥ 15

13+2 ≥ 15

15 ≥ 15

Which gives the same result as above.

Answer:

The limit of the (228.533, 234.735)

Step-by-step explanation:

The values are missing in the given question. Therefore, in order to attempt this question, we will make assumptions.

So, let's assume that:

sample size of the journal article that was reported = 5

which is applied for determining a 95% CI

Also, assuming that the resulting interval = (229.764, 233.504)

However, if we are to use a 99% CI which we deemed to be more appropriate;

Then, our objective is to find the limits of this particular interval in question.

To do that:

We need to first find the sample mean at 95% CI by using the formula:

[tex]\Big ( \bar {x} - t_{\alpha/2, df} \ \dfrac{s}{\sqrt{n}}}, \bar x + t_{\alpha/2, df} \ \dfrac{s}{\sqrt{n}}} \Big) = (229.764,233.504)[/tex]

Since; df = n - 1

df = 5 - 1

df = 4

Then;

[tex]\Big ( \bar {x} - t_{\alpha/2, 4} \ \dfrac{s}{\sqrt{n}}}, \bar x + t_{\alpha/2, 4} \ \dfrac{s}{\sqrt{n}}} \Big) = (233.504+229.764)[/tex]

[tex]2 \bar {x} = (463.268)[/tex]

[tex]\bar {x} =\dfrac{463.268}{2 }[/tex]

[tex]\bar {x} =231.634[/tex]

Sample mean = 231.634

Using the same formula to determine the standard deviation, we have:

[tex]\Big ( \bar {x} - t_{\alpha/2, df} \ \dfrac{s}{\sqrt{n}}}, \bar x + t_{\alpha/2, df} \ \dfrac{s}{\sqrt{n}}} \Big) = (229.764,233.504)[/tex]

[tex]\Big ( \bar {x} - t_{\alpha/2, 4} \ \dfrac{s}{\sqrt{n}}}, \bar x + t_{\alpha/2, 4} \ \dfrac{s}{\sqrt{n}}} \Big) = (233.504-229.764)[/tex]

[tex]\Big ( (\bar x + t_{\alpha/2, 4} \ \dfrac{s}{\sqrt{n}}} )- (\bar {x} - t_{\alpha/2, 4} \ \dfrac{s}{\sqrt{n}}}) \Big) = (233.504-229.764)[/tex]

[tex]\Big ( (\bar x + t_{0.05/2, 4} \ \dfrac{s}{\sqrt{4}}} )- (\bar {x} - t_{0.05/2, 4} \ \dfrac{s}{\sqrt{5}}}) \Big) = (233.504-229.764)[/tex]

At t = 0.025 and df = 4; = 2.776

[tex]2\times 2.776 \dfrac{s}{\sqrt{5}}= 3.74[/tex]

[tex]5.552 \dfrac{s}{\sqrt{5}}= 3.74[/tex]

[tex]\dfrac{s}{\sqrt{5}}= \dfrac{ 3.74}{5.552}[/tex]

[tex]\dfrac{s}{\sqrt{5}}= 0.6736[/tex]

[tex]s = 0.6736 \times \sqrt{5}[/tex]

s = 1.5063

The 99% CI is:

[tex]\implies \Big(\bar x \pm t_{\alpha/2,4} \dfrac{s}{\sqrt{n}} \Big)[/tex]

At t =0.005 and df =4; = 4.604

[tex]\implies \Big(231.634 \pm 4.604 \dfrac{1.5063}{\sqrt{5}} \Big)[/tex]

[tex]\implies \Big(231.634 \pm 4.604 (0.67364) \Big)[/tex]

[tex]\implies \Big(231.634 \pm 3.1014 \Big)[/tex]

[tex]\implies \Big((231.634 - 3.1014), (231.634 + 3.1014) \Big)[/tex]

[tex]\implies \Big( 228.533, 234.735 \Big)[/tex]