The ratio of boys to girls in a high school is found to be three to five. If a class has 12boys, how many girls would you expect to be in the class?

Answer:

D

Step-by-step explanation:

fro the diagram below there line has no slope

Answer: B) The slope is zero

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Explanation:

Any horizontal line will always have a slope of 0. This is because there is no change in y (aka the rise is 0).

So we could say something like

slope = rise/run = 0/1 = 0

The run can be anything we want, and we'd still get 0 every time.

------------

Another way to see this is to pick two points from this line. Whichever points are selected, they are plugged into the slope formula

m = (y2-y1)/(x2-x1)

You'll find that the y2-y1 expression turns into 0. Why? Because y1 and y2 are the same, so they subtract to 0. It doesn't matter what x2-x1 turns into.

Answer:

The standard error of a proportion p in a sample of size n is given by: [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]

In this question:

The standard error of a proportion p in a sample of size n is given by: [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]

Hi there!  

»»————-　★　————-««

I believe your answer is:  

[tex]n = t + r[/tex]

»»————-　★　————-««  

Here’s why:  

⸻⸻⸻⸻

[tex]\boxed{\text{Solving for 'n'...}}\\\\t = n - r\\----------\\\rightarrow t + r = n -r + r\\\\\rightarrow t+r = n\\\\\rightarrow \boxed{n=t+r}[/tex]

⸻⸻⸻⸻

»»————-　★　————-««  

Hope this helps you. I apologize if it’s incorrect.  

Answer:

Step-by-step explanation:

Use the coordinates (30, 1) and (150, 5) to solve this. Time is always an x thing, while things like distance and weight and value are y things. Put them into the slope formula:

[tex]m(\frac{miles}{sec})=\frac{5-1}{150-30}=\frac{4}{120}=\frac{1}{30}[/tex]  This translates to:

The rocket is ascending at a constant rate of 1 mile every 30 seconds; or, conversely, for every 30 seconds the rocket is flying, it is traveling 1 mile.

Expanding each square on the left side, you have

(x cos(A) + y sin(A))² = x² cos²(A) + 2xy cos(A) sin(A) + y² sin²(A)

(x sin(A) - y cos(A))² = x² sin²(A) - 2xy sin(A) cos(A) + y² cos²(A)

so that adding them together eliminates the identical middle terms and reduces to the sum to

x² cos²(A) + y² sin²(A) + x² sin²(A) + y² cos²(A)

Collecting terms to factorize gives us

(y² + x²) sin²(A) + (x² + y²) cos²(A)

(x² + y²) (sin²(A) + cos²(A))

and sin²(A) + cos²(A) = 1 for any A, so we end up with

x² + y²

as required.

Answer:

<1 = 33

Step-by-step explanation:

The sum of the angle of a triangle is 180

31+116+x = 180

x+147=180

x = 180-147

x = 33

Answer:

a. approximately 0.0187

b. 0.047

Step-by-step explanation:

q = 1-p

= 1-0.4

q = 0.6

a. the probability that the first arrival will occur during seventh one-second interval

probability(7) = 0.6⁷⁻¹ x 0.4

= 0.6⁶ x 0.4

= 0.046656 x 0.4

= 0.0186624

approximately 0.0187

b. probability that the first arrival will not occur until at least the seventh one second interval

p(y≥7) = 1-p(x<7)

= 1-[(0.4)(0.6)⁰ + (0.4)(0.6)¹ +(0.4)(0.6)²+(0.4)(0.6)³+(0.4)(0.6)⁴+(0.4)(0.6)⁵]

= 1-(0.4+0.24+0.144+0.0864+0.05184+0.031104

= 1-0.95334

= 0.04667

= 0.047

Answer:

14 hours

Step-by-step explanation:

Take any two consecutive high tides and to find their x coordinatey and sub them..

Answer:

b I think!!!!!!!!!!!##$

Answer:

4

Step-by-step explanation:

Pick two points on the line

(0,-5)  and (1,-1)

We can find the slope using

m = (y2-y1)/(x2-x1)

   = ( -1 - -5)/(1 - 0)

  (-1+5)/(1-0)

    4/1

  = 4

Answer:

a) The 90% confidence interval for the mean amount owed in student loans of graduates of four-year business colleges is ($13,600, $15,162), having a margin of error of $781.

b) We are 90% sure that the mean amount owed in student loans of graduates of  all four-year business colleges is between $13,600 and $15,162.

Step-by-step explanation:

Question a:

We have the standard deviation 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,which is the sample size subtracted by 1. So

df = 25 - 1 = 24

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 24 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.9}{2} = 0.95[/tex]. So we have T = 2.0639

The margin of error is:

[tex]M = T\frac{s}{\sqrt{n}} = 2.0639\frac{1892}{\sqrt{25}} = 781[/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 14381 - 781 = $13,600

The upper end of the interval is the sample mean added to M. So it is 14381 + 781 = $15,162

The 90% confidence interval for the mean amount owed in student loans of graduates of four-year business colleges is ($13,600, $15,162), having a margin of error of $781.

b) Interpret the confidence interval you have computed.

We are 90% sure that the mean amount owed in student loans of graduates of  all four-year business colleges is between $13,600 and $15,162.

Answer:

$150

Step-by-step explanation:

0.2 X 750 = 150

hope this helps

Answer <1 = 145

< 2= 35

<3 = 35

Step-by-step explanation:

<1 + <4 = 180 (supplementary since you can find corresponding side and straight angle)

<1 = 180-35

=145 degrees

<2 =<4 corresponding angles

<2 = 35 degrees

<3 = 35 degree (corresponding to <2)

Answer:

(a)

Number of cars with defective turn signals = 65

Number of cars with no defective turn signals = 410 - 65 = 345

Required probability:

(b)

Number of cars with defects = 65 + 35 = 100

Number of cars with no defects = 410 - 100 = 310

Required probability:

Answer:

1250000cm

Step-by-step explanation:

1:500000

1x2.5 : 500000x2.5

2.5:1250000

YEsStep-by-step explanation:

Answer:

22hrs

Step-by-step explanation:

hope it is well understood?

Given:

d = 2

f = 4

To find:

Value of  [tex]\frac{14(7)-d}{2f}[/tex]

Steps:

we need to substitute and then find the value,

[tex]= \frac{14(7)-2}{2(4)}\\ \\=\frac{98-2}{8} \\\\=\frac{96}{8}\\\\=12[/tex]

Therefore, the answer is option C) 12

Happy to help :)

If you need help, feel free to ask

Answer:

377.19 (rounded off to 2dp)

Step-by-step explanation:

since its a right angled triangle, we can use tangent

tan(x) =opp/adj

tan(5) =33/AC

AC =33/tan(5)

Answer:

35 cm

Step-by-step explanation:

is the correct answer

Answer:

Step-by-step explanation:

If A is 170 less than B, than the equation for that is:

A = B - 170 (1) where the word "is" means equals and less than is subtraction.

If the total of A + B is 1520, then

A + B = 1520 (2). Sub equation (1) into equation (2):

(B - 170) + B = 1520 and

2B - 170 = 1520 and

2B = 1690 so

B = 845. Building B is 845 feet tall and Building A is

A = 845 - 170 (this is equation (1) with the height of B subbed in) so

A = 675 feet

675 + 845 should equal 1520 according to our equation. And of course it does.

Answer: 675 + 845 should equal 1520 according to our equation. And of course it does.

It looks like the limit you want to find is

[tex]\displaystyle \lim_{x\to\infty} \left(\frac{x+4}{x-1}\right)^{x+4}[/tex]

One way to compute this limit relies only on the definition of the constant e and some basic properties of limits. In particular,

[tex]e = \displaystyle\lim_{x\to\infty}\left(1+\frac1x\right)^x[/tex]

The idea is to recast the given limit to make it resemble this definition. The definition contains a fraction with x as its denominator. If we expand the fraction in the given limand, we have a denominator of x - 1. So we rewrite everything in terms of x - 1 :

[tex]\left(\dfrac{x+4}{x-1}\right)^{x+4} = \left(\dfrac{x-1+5}{x-1}\right)^{x-1+5} \\\\ = \left(1+\dfrac5{x-1}\right)^{x-1+5} \\\\ =\left(1+\dfrac5{x-1}\right)^{x-1} \times \left(1+\dfrac5{x-1}\right)^5[/tex]

Now in the first term of this product, we substitute y = (x - 1)/5 :

[tex]\left(\dfrac{x+4}{x-1}\right)^{x+4} = \left(1+\dfrac1y\right)^{5y} \times \left(1+\dfrac5{x-1}\right)^5[/tex]

Then use a property of exponentiation to write this as

[tex]\left(\dfrac{x+4}{x-1}\right)^{x+4} = \left(\left(1+\dfrac1y\right)^y\right)^5 \times \left(1+\dfrac5{x-1}\right)^5[/tex]

In terms of end behavior, (x - 1)/5 and x behave the same way because they both approach ∞ at a proportional rate, so we can essentially y with x. Then by applying some limit properties, we have

[tex]\displaystyle \lim_{x\to\infty} \left(\frac{x+4}{x-1}\right)^{x+4} = \lim_{x\to\infty} \left(\left(1+\dfrac1x\right)^x\right)^5 \times \left(1+\dfrac5{x-1}\right)^5 \\\\ = \lim_{x\to\infty}\left(\left(1+\dfrac1x\right)^x\right)^5 \times \lim_{x\to\infty}\left(1+\dfrac5{x-1}\right)^5 \\\\ =\left(\lim_{x\to\infty}\left(1+\dfrac1x\right)^x\right)^5 \times \left(\lim_{x\to\infty}\left(1+\dfrac5{x-1}\right)\right)^5[/tex]

By definition, the first limit is e and the second limit is 1, so that

[tex]\displaystyle \lim_{x\to\infty} \left(\frac{x+4}{x-1}\right)^{x+4} = e^5\times1^5 = \boxed{e^5}[/tex]

You can also use L'Hopital's rule to compute it. Evaluating the limit "directly" at infinity results in the indeterminate form [tex]1^\infty[/tex].

Rewrite

[tex]\left(\dfrac{x+4}{x-1}\right)^{x+4} = \exp\left((x+4)\ln\dfrac{x+4}{x-1}\right)[/tex]

so that

[tex]\displaystyle \lim_{x\to\infty} \left(\frac{x+4}{x-1}\right)^{x+4} = \lim_{x\to\infty}\exp\left((x+4)\ln\dfrac{x+4}{x-1}\right) \\\\ = \exp\left(\lim_{x\to\infty}(x+4)\ln\dfrac{x+4}{x-1}\right) \\\\ =\exp\left(\lim_{x\to\infty}\frac{\ln\dfrac{x+4}{x-1}}{\dfrac1{x+4}}\right)[/tex]

and now evaluating "directly" at infinity gives the indeterminate form 0/0, making the limit ready for L'Hopital's rule.

We have

[tex]\dfrac{\mathrm d}{\mathrm dx}\left[\ln\dfrac{x+4}{x-1}\right] = -\dfrac5{(x-1)^2}\times\dfrac{1}{\frac{x+4}{x-1}} = -\dfrac5{(x-1)(x+4)}[/tex]

[tex]\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1{x+4}\right]=-\dfrac1{(x+4)^2}[/tex]

and so

[tex]\displaystyle \exp\left(\lim_{x\to\infty}\frac{\ln\dfrac{x+4}{x-1}}{\dfrac1{x+4}}\right) = \exp\left(\lim_{x\to\infty}\frac{-\dfrac5{(x-1)(x+4)}}{-\dfrac1{(x+4)^2}}\right) \\\\ = \exp\left(5\lim_{x\to\infty}\frac{x+4}{x-1}\right) \\\\ = \exp(5) = \boxed{e^5}[/tex]

Answer:

0.0036

Step-by-step explanation:

Given that :

Proportion of population that are math phobic = 6% = 6/100 = 0.06

P(math phobic) = 0.06

If two selections are made ; probability that both are math phobic ;

P1 = selection 1 = 0.06

P2 = selection 2 = 0.06

Probability that both are math phobic :

P1 * P2 = (0.06 * 0.06) = 0.0036

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Explanation:

Let y be the length of the vertical dashed red line in the drawing. More specifically, this dashed line is an altitude.

Along the top, the entire segment is 400 units long. The right piece is 144 units long, so the left piece is 400-144 = 256 units long.

The triangles are similar, allowing us to set up a proportion like so:

144/y = y/256

144*256 = y*y

36864 = y^2

y^2 = 36864

y = sqrt(36864)

y = 192

So this is the length of that vertical dashed red line.

--------------------------------

Now shift your attention solely on the smaller triangle on the right side. It is a right triangle with legs 144 and 192. The hypotenuse is x.

We can use the pythagorean theorem to find x.

a^2 + b^2 = c^2

c = sqrt( a^2 + b^2 )

x = sqrt( 144^2 + 192^2 )

x = 240

240.

Let y be the length of the vertical dashed red line in the drawing. More specifically, this dashed line is an altitude.

Along the top, the entire segment is 400 units long. The right piece is 144 units long, so the left piece is 400-144 = 256 units long.

The triangles are similar, allowing us to set up a proportion like so:

144/y = y/256

144*256 = y*y

36864 = y^2

y^2 = 36864

y = sqrt(36864)

y = 192

So this is the length of that vertical dashed red line.

Now shift your attention solely to the smaller triangle on the right side. It is a right triangle with legs 144 and 192. The hypotenuse is x.

We can use the Pythagorean theorem to find x.

a^2 + b^2 = c^2

c = sqrt( a^2 + b^2 )

x = sqrt( 144^2 + 192^2 )

x = 240

Pythagorean theorem, the well-known geometric theorem that the sum of the squares on the legs of a right triangle is equal to the square on the hypotenuse (the side opposite the right angle)—or, in familiar algebraic notation, a2 + b2 = c2.

Learn more about the Pythagorean theorem at

https://brainly.com/question/343682

#SPJ2

Answer:

12

Step-by-step explanation:

You're looking for a solution in the form

[tex]y(x) = \displaystyle \sum_{n=0}^\infty a_nx^n[/tex]

Differentiating, we get

[tex]y'(x) = \displaystyle \sum_{n=0}^\infty na_nx^{n-1} = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1)a_{n+1}x^n[/tex]

[tex]y''(x) = \displaystyle \sum_{n=0}^\infty (n+1)na_{n+1}x^{n-1} = \sum_{n=1}^\infty (n+1)na_{n+1}x^{n-1} = \sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n[/tex]

Substitute these for y' and y'' in the differential equation:

[tex]\displaystyle \sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n - \sum_{n=0}^\infty (n+1)a_{n+1}x^n = 0[/tex]

[tex]\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1)a_{n+2}-(n+1)a_{n+1}\bigg)x^n = 0[/tex]

Then the coefficients of y are given by the recurrence

[tex]\begin{cases}a_0=y(0)\\a_1=y'(0)\\a_{n+2}=\frac{a_{n+1}}{n+2}&\text{for }n\ge0\end{cases}[/tex]

or

[tex]a_n = \dfrac{a_{n-1}}n[/tex]

But we cannot assume that [tex]a_0[/tex] and [tex]a_1[/tex] depend on each other; we can only guarantee that the recurrence holds for n ≥ 1, so that

[tex]a_2=\dfrac{a_1}2 \\\\ a_3=\dfrac{a_2}3=\dfrac{a_1}{3\times2} \\\\ a_4=\dfrac{a_3}4=\dfrac{a_1}{4\times3\times2} \\\\ \vdots \\\\ a_n=\dfrac{a_1}{n!}[/tex]

So in the power series solution, we split off the constant term and we're left with

[tex]y(x) = a_0 + a_1 \displaystyle \sum_{n=1}^\infty \frac{x^n}{n!}[/tex]

so that the fundamental solutions are

[tex]y_1=1[/tex]

and

[tex]y_2=x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\cdots[/tex]

Answer:

[11.906 ; 12.894]

Step-by-step explanation:

Given :

Sample mean, xbar = 12.4

Sample standard deviation, s = 1.7

Sample size, n = 48

We use the T distribution since we are using the sample standard deviation;

α - level = 95% ; df = n - 1 = 48 - 1 = 47

Tcritical = T(1 - α/2), 47 = 2.012

Using the confidence interval for one sample mean

Xbar ± Tcritical * s/√n

12.4 ± (2.012 * 1.7/√48)

12.4 ± 0.4936922

C. I = [11.906 ; 12.894]