Does a point have a one dimension length

Answer:

[tex] z = \frac{0.11-0.14}{0.0139} = -2.156[/tex]

And we can use the normal standard distribution table and we got:

[tex] P(Z<-2.156) =0.0155[/tex]

Step-by-step explanation:

For this case we know the following info given:

[tex] p =0.14[/tex] represent the population proportion

[tex] n = 622[/tex] represent the sample size selected

We want to find the following proportion:

[tex] P(\hat p <0.11)[/tex]

For this case we can use the normal approximation since we have the following conditions:

i) np = 622*0.14 = 87.08>10

ii) n(1-p) = 622*(1-0.14) =534.92>10

The distribution for the sample proportion would be given by:

[tex] \hat p \sim N (p ,\sqrt{\frac{p(1-p)}{n}}) [/tex]

The mean is given by:

[tex] \mu_{\hat p}= 0.14[/tex]

And the deviation:

[tex]\sigma_{\hat p}= \sqrt{\frac{0.14*(1-0.14)}{622}}= 0.0139[/tex]

We can use the z score formula given by:

[tex] z=\frac{\hat p -\mu_{\hat p}}{\sigma_{\hat p}}[/tex]

And replacing we got:

[tex] z = \frac{0.11-0.14}{0.0139} = -2.156[/tex]

And we can use the normal standard distribution table and we got:

[tex] P(Z<-2.156) =0.0155[/tex]

Answer:

It would be 16!!!

The value of the exponent numerical expression (-1/2)⁻⁴ will be 16. Then the correct option is D.

When the relevant components and basic processes of a numerical method are given values, the expression's result is the result of the computation it depicts.

The definition of simplicity is making something simpler to achieve or grasp while also making it a little less difficult.

The expression is given below.

⇒ (-1/2)⁻⁴

Simplify the equation, then we have

⇒ (-1/2)⁻⁴

⇒ (-2)⁴

⇒ -2⁴

⇒ 16

The value of the exponent numerical expression (-1/2)⁻⁴ will be 16. Then the correct option is D.

More about the value of the expression link is given below.

https://brainly.com/question/23671908

#SPJ6

Answer:

Its right and scalene.

It has a right angle and all the sides are diferent.

Answer:

The 99.5% confidence interval for the proportion of all shrubs of this species that will resprout after fire is (0, 0.5745).

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 zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

For this problem, we have that:

[tex]n = 18, \pi = \frac{5}{18} = 0.2778[/tex]

99.5% confidence level

So [tex]\alpha = 0.005[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.005}{2} = 0.9975[/tex], so [tex]Z = 2.81[/tex].  

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2778 - 2.81\sqrt{\frac{0.2778*0.7222}{18}} = -0.01 = 0[/tex]

We cannot have a negative proportion, so we use 0.

The upper limit of this interval is:

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2778 + 2.81\sqrt{\frac{0.2778*0.7222}{18}} = 0.5745[/tex]

The 99.5% confidence interval for the proportion of all shrubs of this species that will resprout after fire is (0, 0.5745).

Answer:

Step-by-step explanation:

[tex]100 (0.96)^{25} =[/tex] around 36.04

Answer:

The IV will run [tex]66.67 \ ml /hr[/tex]

Step-by-step explanation:

From the question we are told that

   The mass of Magnesium sulfate is  [tex]m_g = 30 \ g[/tex]

   The volume of the Magnesium sulfate  [tex]V_R = 500ml[/tex]

   The rate at which the dose of the solution (Magnesium sulfate + Lactated Ringers. ) is infused is  [tex]R = 4g/hr[/tex]

The concentration of Magnesium sulfate in Lactated Ringers  is mathematically evaluated as

         [tex]C_m = \frac{m_g}{V_R}[/tex]

substituting values

          [tex]C_m = \frac{30}{500}[/tex]

          [tex]C_m = 0.06\ g/ ml[/tex]

This implies that

     0.06 g of Magnesium sulfate is in every 1 ml of  Lactated Ringers

So  4 g  of  Magnesium sulfate is in x ml of  Lactated Ringers

So

        [tex]x = \frac{4}{0.06}[/tex]

       [tex]x = 66.67 \ ml[/tex]

So the amount of the solution in ml that is been infused in 1 hour is  

      [tex]66.67 \ ml /hr[/tex]

Answer:

y=2/3x+1

Step-by-step explanation:

The slope is 2/3 and the y-intercept is 1.

Answer:

a) [tex]\mathbf{4 + \dfrac{x}{1!}- \dfrac{2x^2}{2!} ...}[/tex]

b)  See Below for proper explanation

Step-by-step explanation:

a) The objective here  is to Use the power series expansions for ex, sin x, cos x, and geometric series to find the first three nonzero terms in the power series expansion of the given function.

The function is [tex]e^x + 3 \ cos \ x[/tex]

The expansion is of  [tex]e^x[/tex] is [tex]e^x = 1 + \dfrac{x}{1!}+ \dfrac{x^2}{2!}+ \dfrac{x^3}{3!} + ...[/tex]

The expansion of cos x is [tex]cos \ x = 1 - \dfrac{x^2}{2!}+ \dfrac{x^4}{4!}- \dfrac{x^6}{6!}+ ...[/tex]

Therefore; [tex]e^x + 3 \ cos \ x = 1 + \dfrac{x}{1!}+ \dfrac{x^2}{2!}+ \dfrac{x^3}{3!} + ... 3[1 - \dfrac{x^2}{2!}+ \dfrac{x^4}{4!}- \dfrac{x^6}{6!}+ ...][/tex]

[tex]e^x + 3 \ cos \ x = 4 + \dfrac{x}{1!}- \dfrac{2x^2}{2!} + \dfrac{x^3}{3!}+ ...[/tex]

Thus, the first three terms of the above series are:

[tex]\mathbf{4 + \dfrac{x}{1!}- \dfrac{2x^2}{2!} ...}[/tex]

b)

The series for [tex]e^x + 3 \ cos \ x[/tex] is [tex]\sum \limits^{\infty}_{x=0} \dfrac{x^x}{n!} + 3 \sum \limits^{\infty}_{x=0} ( -1 )^x \dfrac{x^{2x}}{(2n)!}[/tex]

let consider the series; [tex]\sum \limits^{\infty}_{x=0} \dfrac{x^x}{n!}[/tex]

[tex]|\frac{a_x+1}{a_x}| = | \frac{x^{n+1}}{(n+1)!} * \frac{n!}{x^x}| = |\frac{x}{(n+1)}| \to 0 \ as \ n \to \infty[/tex]

Thus it converges for all value of x

Let also consider the series [tex]\sum \limits^{\infty}_{x=0}(-1)^x\dfrac{x^{2n}}{(2n)!}[/tex]

It also converges for all values of x

Answer:

The probability that all four received a raised when they asked for one is 0.370.

Step-by-step explanation:

Let the random variable X represent the number of business executives who received a pay raise when they asked for one.

The probability that a business executives received a pay raise when they asked for one is, p = 0.78.

A random sample of n = 4 executives was selected.

The events of any executive receiving a pay raise when they asked for one is independent of the others.

The random variable X follows a Binomial distribution with parameters n = 4 and p = 0.78.

The probability mass function of X is:

[tex]P(X=x)={4\choose x}\ (0.78)^{x}\ (1-0.78)^{4-x};\ x=0,1,2,3...[/tex]

Compute the probability that all four received a raised when they asked for one as follows:

[tex]P(X=4)={4\choose 4}\ (0.78)^{4}\ (1-0.78)^{4-4}[/tex]

                [tex]=1\times 0.37015056\times 1\\\\=0.37015056\\\\\apporx 0.370[/tex]

Thus, the probability that all four received a raised when they asked for one is 0.370.

Answer:

7.1

Step-by-step explanation:

d = sqrt(7^2 + -1^2)= sqrt(50)=7.1

Answer:

by using distance formula

putting values

d=√(-1-6)²+(-4--5)²

d=√(-7)²+(1)²

d=√49+1

d=√50

d=5√2=7.1

Answer:

5q + 9

Step-by-step explanation:

Combine like terms to simplify the expression.

Have a blessed day!

Answer:

7q+9

Step-by-step explanation:

6q+4+q+5

6q+q+4+5

=7q+9

Answer:

y = x - 2

Step-by-step explanation:

y = x + b

3 = 5 + b

y = x - 2

We can use the slope intercept form of a line.

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

y = 1x +b

Substitute the point into the equation

3 = 1*5+b

3 = 5+b

Subtract 5 from each side

3-5 = 5+b-5

-2 =b

y = x-2

Answer:

Step-by-step explanation:

hey

(f*g)(x) = f(g(x)) = f(2) = 2

second answer is correct

thanks

Answer:

Mathematics could make scientists to have a preliminary understanding of the dimensions, perimeters, areas and volumes of different craters on Mars.

Step-by-step explanation:

Martian Craters are series of craters formed on the surface of Mars. The study of a planets crater gives an understanding of the properties of matter that lies under the crater.

Mathematics can be applied to determine the dimensions, perimeter, area and volume of the features of a crater using appropriate conversions and theorems.

The Pi in the sky theorem can be applied to determine the area and perimeter, even volume of different craters on the Mars surface. Also, eingenfunction expansion theorem gives a preliminary knowledge of the craters.

By measurements and conversions processes, the features of Martian crater could be studied from images.

Answer:

There is an estimated increase in revenue of $0.2 million for each 1,000 additional clicks

Step-by-step explanation:

The slope (0.2) is the rate of change in Y-hat for each unit change in x.

In this specific case, since Y-hat is the revenue, in millions, and x is the number of clicks, in thousands, the best interpretation is that there is an estimated increase in revenue of $0.2 million for each 1,000 additional clicks

Answer:

Please read the complete answer below!

Step-by-step explanation:

You have the following function:

[tex]f(x)=x^3-3x^2-9x+4[/tex]   (1)

a) To find the interval on which f is increasing or decreasing, you first calculate the critical points of f(x).

You calculate the derivative f(x) respect to x:

[tex]\frac{df}{dx}=3x^2-6x-9[/tex]    (2)

Next, you equal the derivative to zero, and then you find the roots of the polynomial by using the quadratic formula:

[tex]3x^2-6x-9=0\\\\x_{1,2}=\frac{-(-6)\pm\sqrt{(-6)^2-4(3)(-9)}}{2(3)}\\\\x_{1,2}=\frac{6\pm12}{6}\\\\x_1=-1\\\\x_2=3[/tex]

Then, the critical points are x=-1 and x=3

Next, you calculate df/dx for a values of x to the left and to the right of the critical points x1 and x2. If df/dx < 0 the function is decreasing, if df/dx > 0 the function is increasing.

for x = -1.01

[tex]\frac{df(-1.01)}{dx}=3(-1.01)^2-6(-1.01)-9=0.12[/tex]

Then, in the interval (-∞,-1), the function is increasing

for x = -0.99

[tex]\frac{df(-0.99)}{dx}=3(-0.99)^2-6(-0.99)-9=-0.11[/tex]

In the interval (-1,3) the function is decreasing

for x = 3.01

[tex]\frac{df(3.01)}{dx}=3(3.01)^2-6(3.01)-9=0.12[/tex]

In the interval (3,+∞) the function is increasing

b) To find the local minimum and maximum you use the second derivative of the function:

[tex]\frac{d^2f}{dx^2}=6x-6[/tex]     (3)

you evaluate the second derivative for the critical points x1 and x2, if the second derivative is positive, you have a local minimum. If the second derivative is negative, you have a local maximum:

for x1 = -1

[tex]6(-1)-6=-12<0[/tex]

x=-1 is a local maximum

for x2 = 3

[tex]6(3)-6=12>0[/tex]

x=3 is a local minimum

c) upward concavity: (-1,3)

downward concavity: (-∞,-1)U(3,+∞)

The inflection points are calculated with the second derivative equal to zero:

[tex]6x-6=0\\\\x=1[/tex]

For x = 1 you have an inflection point

[tex]answer = \frac{25}{56} \\ solution \\ \frac{3}{7} + \frac{1}{56} \\ = \frac{3 \times 8 + 1}{56} \\ = \frac{24 + 1}{56} \\ = \frac{25}{56} \\ hope \: it \: helps \: \\ good \: luck \: on \: your \: assignment[/tex]

Answer:

25/56

Step-by-step explanation:

3/7 + 1/56

We have to find the L.C.M of 7 and 56

The L.C.M of 7 and 56 is 56

Now, we have to change the denominators to 56

we dont need to change the denominator of 1/56 to 56 as it is already 56

[tex]\frac{3}{7}[/tex] * [tex]\frac{8}{8}[/tex] = [tex]\frac{24}{56}[/tex]

Now we can add the fractions

[tex]\frac{24}{56} + \frac{1}{56}[/tex] [tex]= \frac{25}{56}[/tex]

Hope it helped :>

Answer:

Step-by-step explanation:

If a sequence c1,c2,c3,...has limit K then the sequence ec1,ec2,ec3,...has limit e^K. Use this fact together with l'Hopital's rule to compute the limit of the sequence given by

bn=(n)^(5.6/n).

a)

[tex]L = \lim_{n \to \infty} b_n \\\\\\L= \lim_{n \to \infty} n^{\frac{5.6}{n} }[/tex]

Log on both sides

[tex]In (L) = \lim_{n \to \infty} In (n)^{\frac{5.6}{n} }\\\\= \lim_{n \to \infty} \frac{5.6}{n} In(n)[/tex]

[tex]=5.6 \lim_{n \to \infty} \frac{d}{dn} In(n)/\frac{d}{dn} (n)\\\\=5.6 \lim_{n \to \infty} \frac{1}{n} /1 \\\\=5.6 \lim_{n \to \infty} \frac{1}{n} \\\\=5.6 \times 0\\\\In(L) =0\\\\L=e^0\\\\L=1[/tex]

[tex]\therefore \lim_{n \to \infty} (n)^{\frac{5.6}{n} =1[/tex]

The limit value of given sequece is 1.

To understand more, check below explanation.

The given sequence is,

                   [tex]b_{n}=n^{5.6/n}[/tex]

We have to find limit of above sequence.

           [tex]L=\lim_{n \to \infty} b_n \\\\L=\lim_{n \to \infty}n^{5.6/n} \\\\ln(L)=\lim_{n \to \infty}\frac{5.6}{n}ln(n) \\\\ln(L)=5.6\lim_{n \to \infty}\frac{ln(n)}{n} \\\\ln(L)=5.6\lim_{n \to \infty}\frac{1/n}{1} \\\\ln(L)=5.6*0=0\\\\L=e^{0}=1[/tex]

Therefore, the limit value of given sequece is 1.

Learn more about the limit of function here:

https://brainly.com/question/2166212

Answer:

0.45

Step-by-step explanation:

9/20 is the same as 0.45

Answer:

0.45

Step-by-step explanation:

9/20= 9*5/20*5= 45/100= 0.45

Answer:

a rigt angle is a total of 90 degrees so subtratct 51 from 90 and you get 39 degrees.

Answer:

so halve the deck is black and halve the deck is red so you have 26 black cards then divide by 5 and you get 5 with 1 card leftover

Answer:

[tex]\mathbf{\dfrac{e^{2t}}{36} + \dfrac{e^{3t}}{18} + \dfrac{e^{4t}}{12} +\dfrac{e^{5t}}{9} + \dfrac{5e^{6t}}{36} + \dfrac{7e^{7t}}{6} + \dfrac{5e^{8t}}{36} + \dfrac{e^{9t}}{9} + \dfrac{e^{10t}}{12} + \dfrac{e^{11t}}{18} + \dfrac{e^{12t}}{36} }[/tex]

Step-by-step explanation:

The objective is to find the moment generating  function of  [tex]M_{X+Y}(t) \ of \ X+Y[/tex].

We are being informed that the fair die is rolled twice;

So; X to be the value for the  first roll

Y to be the value of the second roll

The outcomes  of X are:  X = {1,2,3,4,5,6}

Where ;

[tex]P (X=x) = \dfrac{1}{6}[/tex]

The outcomes  of Y are:  y = {1,2,3,4,5,6}

Where ;

[tex]P (Y=y) = \dfrac{1}{6}[/tex]

The outcome of Z = X+Y

[tex]= \left[\begin{array}{cccccc}(1,1)&(1,2)&(1,3)&(1,4)&(1,5)&(1,6)\\ (2,1)&(2,2)&(2,3)&(2,4)&(2,5)&(2,6)\\ (3,1)&(3,2)&(3,3)&(3,4)&(3,5)&(3,6) \\ (4,1)&(4,2)&(4,3)&(4,4)&(4,5)&(4,6) \\ (5,1)&(5,2)&(5,3)&(5,4)&(5,5)&(5,6) \\ (6,1)&(6,2)&(6,3)&(6,4)&(6,5)&(6,6) \end{array}\right][/tex]

= [2,3,4,5,6,7,8,9,10,11,12]

Here;

[tex]P (Z=z) = \dfrac{1}{36}[/tex]

∴ the moment generating function [tex]M_{X+Y}(t) \ of \ X+Y[/tex]is as follows:

[tex]M_{X+Y}(t) \ of \ X+Y[/tex] = [tex]E(e^{t(X+Y)}) = E(e^{tz})[/tex]

⇒ [tex]\sum \limits^{12}_ {z=2 } et ^z \ P(Z=z)[/tex]

= [tex]\mathbf{\dfrac{e^{2t}}{36} + \dfrac{e^{3t}}{18} + \dfrac{e^{4t}}{12} +\dfrac{e^{5t}}{9} + \dfrac{5e^{6t}}{36} + \dfrac{7e^{7t}}{6} + \dfrac{5e^{8t}}{36} + \dfrac{e^{9t}}{9} + \dfrac{e^{10t}}{12} + \dfrac{e^{11t}}{18} + \dfrac{e^{12t}}{36} }[/tex]

Answer:

[tex] P(A) = 0.21[/tex]

We want to find the probability that a randomly selected freshman from this college does not take an introductory statistics class, so then we can use the complement rule given by:

[tex] P(A') = 1-P(A)[/tex]

Where A is the event of interest (a freshman at a certain college takes an introductory statistics class) and A' the complement (a freshman at a certain college NOT takes an introductory statistics class) and then replacing we got:

[tex] P(A')=1-0.21= 0.79[/tex]

Step-by-step explanation:

For this problem we know that the probability that a freshman at a certain college takes an introductory statistics class is 0.21, let's define of interest as A and we can set the probability like this:

[tex] P(A) = 0.21[/tex]

We want to find the probability that a randomly selected freshman from this college does not take an introductory statistics class, so then we can use the complement rule given by:

[tex] P(A') = 1-P(A)[/tex]

Where A is the event of interest (a freshman at a certain college takes an introductory statistics class) and A' the complement (a freshman at a certain college NOT takes an introductory statistics class) and then replacing we got:

[tex] P(A')=1-0.21= 0.79[/tex]

Answer:

The average rate of change is 12x=12.0x.

Description:

Function: x= 1x = 3  convert to short form: x 1x 3

Interval:  x= 1  ,       x 3

Steps:

Input:  Find the average rate of change of f(x)=3x2 on the interval [x,3x].

We have that a=x, b=3x, f(x)=3x2

Thus, f(b)−f(a)b−a=3((3x))2−(3(x)2)3x−(x)=12x.

Answer: the average rate of change is 12x=12.0x.

Please mark brainliest

Hope this helps.

Answer:

3

Step-by-step explanation:

A P E X

Answer:

The area of the rectangle increasing at the rate of 140 cm²/s

Step-by-step explanation:

Rectangle area:

A rectangle has two dimensions, length l and width w.

It's area is:

A = l*w.

When the length is 20 cm and the width is 10 cm, how fast is the area of the rectangle increasing?

We apply implicit differentiation to solve this question:

[tex]A = l*w[/tex]

So

[tex]\frac{dA}{dt} = l\frac{dw}{dt} + w\frac{dl}{dt}[/tex]

Length is 20, so [tex]l = 20[/tex].

Width is 10, so [tex]w = 10[/tex]

The length of a rectangle is increasing at a rate of 8 cm/s and its width is increasing at a rate of 3 cm/s.

This means that [tex]\frac{dl}{dt} = 8, \frac{dw}{dt} = 3[/tex]

So

[tex]\frac{dA}{dt} = l\frac{dw}{dt} + w\frac{dl}{dt}[/tex]

[tex]\frac{dA}{dt} = 20*3 + 10*8 = 140[/tex]

Area in cm².

So

The area of the rectangle increasing at the rate of 140 cm²/s

Answer:

x= -3    x=8

Step-by-step explanation:

(x+3)(x-8)=0

We can use the zero product property to solve

x+3 =0   x-8 =0

x= -3    x=8

Answer:

x=8

Step-by-step explanation:

There is a missing content in the question.

After the statements and before the the options given; there is an omitted content which says:

Referring to Table 16-5, in testing the coefficient of X in the regression equation (0.117) the results were a t-statistic of 9.08 and an associated p-value of 0.0000. Which of the following is the best interpretation of this result?

Answer:

C. The quarterly growth rate in the number of contracts is significantly different from 0% (? = 0.05).

Step-by-step explanation:

From the given question:

The resulting regression equation can be represented as:

[tex]\hat Y = 3.37 + 0.117 X - 0.083 Q_1 + 1.28 Q_2 + 0.617Q_3[/tex]

where;

the estimated number of contracts in a quarter X is the coded quarterly value with X = 0

the first quarter of 2010 Q1 is a dummy variable equal to 1 in the first quarter of a year and 0 otherwise

Q2 is a dummy variable equal to 1 in the second quarter of a year and 0 otherwise

Q3 is a dummy variable equal to 1 in the third quarter of a year and 0 otherwise

Our null and alternative hypothesis can be stated as;

Null hypothesis :

[tex]H_0 :[/tex]  The quarterly growth rate in the number of contracts is not  significantly different from 0% (? = 0.05)

[tex]H_a:[/tex]  The quarterly growth rate in the number of contracts is significantly different from 0% (? = 0.05)

The decision rule is to reject the null hypothesis if the p-value is less than 0.05.

From the missing omitted part we added above; we can see that the   t-statistics value = 9.08 and the p-value = 0.000 .

Conclusion:

Thus; we reject the null hypothesis and accept the alternative hypothesis. i.e

The quarterly growth rate in the number of contracts is significantly different from 0% (? = 0.05)

Answer:

D = 0 , Dx = 4 , Dy = -6 , Dz = 2

Step-by-step explanation:

As per cramer's rule,

D = |   7    6    4  |  = 0

      |   3    3    3  |

      |   4    4    4  |

Dx =  |  10    6    4  | = 4

        |   1    3    3    |

        |   2    4    4   |

Dy = |   7    10    4  | = -6

       |   3    1     3   |

       |   4    2    4   |

Dz =  |  7    6    10 | = 2

        |   3    3    1   |

        |   4    4    2  |