In 2006, there were 160 teachers in College A, and three fourth of them had their own vehicles. In 2007, 20 new teachers came to the school and 6 of them had own vehicles. Calculate the percentage increase in the numbers of teacher who had own vehicles.​

Hi there!  

»»————-　★　————-««

I believe your answer is:  

330

»»————-　★　————-««  

Here’s why:  

⸻⸻⸻⸻

We are rounding the number 326 to the nearest tens.

The '2' is in the tens place, and the '6' is to the right of the digit.

Since the '6' is greater than or equal to 5, then we would round up.

326 ≈ 330

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

Answer:

8 months 11 days 1 year

Answer:

Does the answer help you?

To solve this question, the normal distribution and the central limit theorem are used.

Doing this, there is an 0.0129 = 1.29% probability that the mean installation time for 31 computers is less than 43 minutes.

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First, these concepts are presented, then we identify mean, standard deviation and sample size, and then, we find the desired probability.

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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.

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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

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

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Mean of 45, standard deviation of 5:

This means that [tex]\mu = 45, \sigma = 5[/tex]

Sample size of 31:

This means that [tex]n = 31, s = \frac{5}{\sqrt{31}}[/tex]

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Probability the sample mean is less than 43:

This is the p-value of Z when X = 43, so:

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

By the Central Limit Theorem

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

[tex]Z = \frac{43 - 45}{\frac{5}{\sqrt{31}}}[/tex]

[tex]Z = -2.23[/tex]

[tex]Z = -2.23[/tex] has a p-value of 0.0129.

Thus, 0.0129 = 1.29% probability that the mean installation time for 31 computers is less than 43 minutes.

A similar question is given at: https://brainly.com/question/15020228

The 2 triangles are congruent

Answer:

es1433

Step-by-step explanation:

Hi

let's call X the initial deposit.

the interest rate is 5.2 %

so each year X increase by 1.052.

so we have : X *1.052^7 =2300

X = 2300/1.052^7

X = 1612,94

please note that the deposit was rounded to the next cent. as the result would be 1612,937...

Answer:

1686.22

Step-by-step explanation:

1686.22

 2300=P(1+(0.052x7))  

2300=P1.364  

P=2300/1.364  

=1686.217...  

=1686.22..

=1686 for the nearest dollar

Answer:

184.8

Step-by-step explanation:

y =kx

k=y/x

k=42/5=8.4

y=8.4*22

Answer:

a

Step-by-step explanation:

3x tables

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Answer:

a

Step-by-step explanation:

Has pattern 3x tables. Answer is 15 A.

Step-by-step explanation:

Let the husband's income be x

Wife's income be x + 9000

X + X + 9000 = 81000

2X + 9000 – 9000 = 81000 – 9000

2X= 72000

X = 36000

Husband, 36000,

Wife, 9000+36000, 45000

Answer:

Hello

Step-by-step explanation:

[tex]Formula: \\\\\boxed{\Large a^3\pm b^3=(a \pm b)(a^2 \mp ab+b^2)}\\\\8x^3+1=(2x)^3+1^3=(2x+1)(4x^2-2x+1)\\\\8x^3-1=(2x)^3-1^3=(2x-1)(4x^2+2x+1)\\[/tex]

The factors of the expression 8x³ + 1 and 8x³ - 1³ will be (2x + 1) & (4x² – 2x + 1) and (2x – 1) & (4x² + 2x + 1), respectively.

A polynomial expression is an algebraic expression with variables and coefficients. Unknown variables are what they're termed. We can use addition, subtraction, and other mathematical operations. However, a variable is not divisible.

The expression is given below.

8x³ + 1 and 8x³ - 1³

(2x)³ + 1³ and (2x)³ - 1³

We know that the formula is given as,

a³ + b³ = (a + b) (a² – ab + b²)

a³ – b³ = (a – b) (a² + ab + b²)

Then the expression is written as,

(2x)³ + 1³ = (2x + 1) [(2x)² – 2x + 1²]

(2x)³ + 1³ = (2x + 1) (4x² – 2x + 1)

(2x)³ – 1³ = (2x – 1) [(2x)² + 2x + 1²]

(2x)³ – 1³ = (2x – 1) (4x² + 2x + 1)

The factors of the expression 8x³ + 1 and 8x³ - 1³ will be (2x + 1) & (4x² – 2x + 1) and (2x – 1) & (4x² + 2x + 1), respectively.

More about the polynomial link is given below.

https://brainly.com/question/17822016

#SPJ2

Answer:

2) t distribution with 58 degrees of freedom.

Step-by-step explanation:

Population standard deviations not known:

This means that the t-distribution is used to solve this question.

The sample sizes are n1 = 25 and n2 = 35.

The number of degrees of freedom is the sum of the sample sizes subtracted by the number of samples, in this case 2. So

25 + 35 - 2 = 58 df.

Thus the correct answer is given by option 2.

Answer:

A. 0; 1

Step-by-step explanation:

Required

Mean and standard deviation of a standardized normal distribution

A standardized normal distribution is represented as:

[tex](\mu,\sigma) = (0,1)[/tex]

This implies that:

[tex]\mu = 0[/tex] -- mean

[tex]\sigma = 1[/tex] --- standard deviation

Hence, (a) is true

Answer:

40 per minute.

Step-by-step explanation:

360/9=40

Answer:

OD) 16 cubic centimeters

Hi there!

To find the Trigonometric Equation, we have to isolate sin, cos, tan, etc. We are also given the interval [0,2π).

First Question

What we have to do is to isolate cos first.

[tex] \displaystyle \large{ cos \theta = - \frac{1}{2} }[/tex]

Then find the reference angle. As we know cos(π/3) equals 1/2. Therefore π/3 is our reference angle.

Since we know that cos is negative in Q2 and Q3. We will be using π + (ref. angle) for Q3. and π - (ref. angle) for Q2.

Find Q2

[tex] \displaystyle \large{ \pi - \frac{ \pi}{3} = \frac{3 \pi}{3} - \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{2 \pi}{3} }[/tex]

Find Q3

[tex] \displaystyle \large{ \pi + \frac{ \pi}{3} = \frac{3 \pi}{3} + \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{4 \pi}{3} }[/tex]

Both values are apart of the interval. Hence,

[tex] \displaystyle \large \boxed{ \theta = \frac{2 \pi}{3} , \frac{4 \pi}{3} }[/tex]

Second Question

Isolate sin(4 theta).

[tex] \displaystyle \large{sin 4 \theta = - \frac{1}{ \sqrt{2} } }[/tex]

Rationalize the denominator.

[tex] \displaystyle \large{sin4 \theta = - \frac{ \sqrt{2} }{2} }[/tex]

The problem here is 4 beside theta. What we are going to do is to expand the interval.

[tex] \displaystyle \large{0 \leqslant \theta < 2 \pi}[/tex]

Multiply whole by 4.

[tex] \displaystyle \large{0 \times 4 \leqslant \theta \times 4 < 2 \pi \times 4} \\ \displaystyle \large \boxed{0 \leqslant 4 \theta < 8 \pi}[/tex]

Then find the reference angle.

We know that sin(π/4) = √2/2. Hence π/4 is our reference angle.

sin is negative in Q3 and Q4. We use π + (ref. angle) for Q3 and 2π - (ref. angle for Q4.)

Find Q3

[tex] \displaystyle \large{ \pi + \frac{ \pi}{4} = \frac{ 4 \pi}{4} + \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{5 \pi}{4} }[/tex]

Find Q4

[tex] \displaystyle \large{2 \pi - \frac{ \pi}{4} = \frac{8 \pi}{4} - \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{7 \pi}{4} }[/tex]

Both values are in [0,2π). However, we exceed our interval to < 8π.

We will be using these following:-

[tex] \displaystyle \large{ \theta + 2 \pi k = \theta \: \: \: \: \: \sf{(k \: \: is \: \: integer)}}[/tex]

Hence:-

For Q3

[tex] \displaystyle \large{ \frac{5 \pi}{4} + 2 \pi = \frac{13 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 4\pi = \frac{21 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 6\pi = \frac{29 \pi}{4} }[/tex]

We cannot use any further k-values (or k cannot be 4 or higher) because it'd be +8π and not in the interval.

For Q4

[tex] \displaystyle \large{ \frac{ 7 \pi}{4} + 2 \pi = \frac{15 \pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 4 \pi = \frac{23\pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 6 \pi = \frac{31 \pi}{4} }[/tex]

Therefore:-

[tex] \displaystyle \large{4 \theta = \frac{5 \pi}{4} , \frac{7 \pi}{4} , \frac{13\pi}{4} , \frac{21\pi}{4} , \frac{29\pi}{4}, \frac{15 \pi}{4} , \frac{23\pi}{4} , \frac{31\pi}{4} }[/tex]

Then we divide all these values by 4.

[tex] \displaystyle \large \boxed{\theta = \frac{5 \pi}{16} , \frac{7 \pi}{16} , \frac{13\pi}{16} , \frac{21\pi}{16} , \frac{29\pi}{16}, \frac{15 \pi}{16} , \frac{23\pi}{16} , \frac{31\pi}{16} }[/tex]

Let me know if you have any questions!

Answer:  [tex]\frac{x^2}{16}+\frac{y^2}{49} = 1\\\\[/tex]

This is the same as writing (x^2)/16 + (y^2)/49 = 1

===========================================================

Explanation:

The general equation for an ellipse is

[tex]\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2} = 1\\\\[/tex]

where

Notice how 'a' pairs with the x term, so that's why 'a' describes the horizontal width along the x axis. The horizontal width is 8 ft, which cuts in half to 4 ft. So a = 4.

The vertical length is 14 ft, which cuts in half to 7 ft. So b = 7.

The center isn't mentioned (other than the fact that the actor is located here), but I'm assuming by default it's at the origin (0,0).

With that all in mind, we then get the following:

[tex]\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2} = 1\\\\\frac{(x-0)^2}{4^2}+\frac{(y-0)^2}{7^2} = 1\\\\\frac{x^2}{16}+\frac{y^2}{49} = 1\\\\[/tex]

The graph is below. I used GeoGebra to make the graph.

From the graph, we can see that the horizontal width spans from x = -4 to x = 4. This is a total distance of |-4-4| = 8 feet. Similarly, the vertical length spans from y = -7 to y = 7 which is a distance of 14 feet.

Answer:

The answer is 72.

I am right .

Answer:

total number of boys in skool is 216

And

ratio of girls to boys is a 3:4

Step-by-step explanation:

let total no. of boys be X

ACCORDING TO QUESTION

ratio of boys to girls = 4:3

total number of girls = 162

now

4/3=X/162

or, 4×162= 3x

or, 4×162/3= X

hence

X = 216

therefore total number of boys is 216

and

ratio of girls to boys is =162/216

=3/4

=3:4

Answer:

Step-by-step explanation:

-24

-55

84

20

-60

Answer:

Option C

We need to make the negative exponent positive.

The rule is: [tex]A^{-B} =\frac{1}{A^{B} }[/tex]

Here, A=x, B= 12

[tex]so,[/tex] [tex]x^{-12}[/tex]

[tex]=\frac{1}{x^{12} }[/tex]

OAmalOHopeO

(2^-1)^2/2×2^0

2^(-1×2)/2^1

2^-2/2^1

2^(-2-1)

2^(-3)

(1/2)^3

Properties used (m^n)^a = m^na

(m)^-n = (1/m)^n

m^0 = 1

m^n/m^a = m^(n-a)

Must click thanks and mark brainliest

Answer:

60

Step-by-step explanation:

Use similar triangles or the ratios from the right triangle altitude theorem.

x/36 = (64 + 36)/x

x² = 3600

x = 60

Cost of 17 inches of wire = 68 cents

Cost of 1 inch of wire

= 68 cents/17

= 4 cents

Cost of 39 inches of wire

= 4 cents × 39

= 156 cents

= $1.56

Answer:

17 inches of wire costs 68 cents,

Step-by-step explanation:

x=234

b=456

Answer:

I think its C:37

Step-by-step explanation:

And if im wrong sorry :/

Answer:

-3

Step-by-step explanation:

Slope is equal to (-5-7)/(2-(-2)=-12/4=-3

Answer:

1/52

Step-by-step explanation:

9514 1404 393

Answer:

  x = π

Step-by-step explanation:

You want f(g(x)) = g(f(x)):

  3 +cos(2x) = 2(3 +cos(x))

  cos(2x) -2cos(x) = 3 . . . . . . . rearrange

  2cos(x)²-1 -2cos(x) = 3 . . . . . use an identity for cos(2x)

  2(c² -c -2) = 0 . . . . . . . . . . . . substitute c = cos(x)

  (c -2)(c +1) = 0 . . . . . . . . . . . factor

  c = 2 (not possible)

  c = -1 = cos(x) . . . . . true for x = π

The value of x that makes f(g(x)) = g(f(x)) is x = π.

_____

Additional comment

The substitution c=cos(x) just makes the equation easier to write and the form of it easier to see. There is really no other reason for making any sort of substitution. In the end, the equation is quadratic in cos(x), so is solved by any of the usual methods of solving quadratics.

Answer:

Hello,

Step-by-step explanation:

[tex]\displaystyleV=2*\pi*\int\limits^4_2 {(\dfrac{1}{x}-\dfrac{1}{x^3} )*x } \, dx \\=2*\pi*\int\limits^4_2 {(1-\dfrac{1}{x^2} ) } \, dx \\=2*\pi* [x+\dfrac{1}{x}]^4_2\\\\\boxed{V=\dfrac{7\pi}{2}}\\[/tex]