In a class of 40 students, 2/5
of the total number of students like to study English and 3/5
of the

total students like to study Mathematics.
a) How many students like to study English?
b)How many students like to study Mathematics?

Answer:

B. [tex] 4x^2 + \frac{3}{2}x - 7 [/tex]

Step-by-step explanation:

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

[tex] g(x) = 4x^2 + x - 4 [/tex]

(f + g)(x) = f(x) - g(x)

= [tex] \frac{x}{2} - 3 + 4x^2 + x - 4 [/tex]

Add like terms

[tex] = 4x^2 + \frac{x}{2} + x - 3 - 4 [/tex]

[tex] = 4x^2 + \frac{3x}{2} - 7 [/tex]

[tex] = 4x^2 + \frac{3}{2}x - 7 [/tex]

Answer:

The appropriate answer is "21".

Step-by-step explanation:

Given:

Afraid percentage,

p = 21%

or,

  = 0.21

Sample size,

n = 100

As we know,

⇒ [tex]X=np[/tex]

By putting the values, we get

        [tex]=0.21\times 100[/tex]

        [tex]=21[/tex]

Answer:

Z +7

Step-by-step explanation:

Z-4/4+8

Divide first

Z -1 +8

Add and subtract

Z +7

Answer:

[tex]15,120[/tex]

Step-by-step explanation:

For the first symbol, there are 9 options to choose from. Then 8, then 7, and so on. Since each player chooses 5 symbols, they will have a total of [tex]9\cdot 8 \cdot 7 \cdot 6\cdot 5=\boxed{15,120}[/tex] permutations possible. Since the order of which they choose them matters (as a different order would be a completely different password), it's unnecessary to divide by the number of ways you can rearrange 5 distinct symbols. Therefore, the desired answer is 15,120.

Answer:15,120

Step-by-step explanation:

Answer:

-7

Step-by-step explanation:

We are given the following functions:

[tex]g(x) = x^2 + 4 + 2x[/tex]

[tex]h(x) = -3x + 2[/tex]

(g•h) (1)

The multiplication is:

[tex](g \times h)(1) = g(1) \times h(1)[/tex]

So

[tex]g(x) = 1^2 + 4 + 2(1) = 7[/tex]

[tex]h(1) = -3(1) + 2 = -3 + 2 = -1[/tex]

Then

[tex]g(1) \times h(1) = 7(-1) = -7[/tex]

So -7 is the answer.

Answer:

[tex](g + f)(8) =133[/tex]

Step-by-step explanation:

Given

[tex]g(n) = 3n - 5[/tex]

[tex]f(n) = n^2 + 50[/tex]

Required

[tex](g + f)(8)[/tex]

This is calculated as:

[tex](g + f)(n) =g(n) + f(n)[/tex]

So, we have:

[tex](g + f)(n) =3n - 5 + n^2 +50[/tex]

[tex]Substitute[/tex] 8 for n

[tex](g + f)(8) =3*8 - 5 + 8^2 +50[/tex]

[tex](g + f)(8) =24 - 5 + 64 +50[/tex]

[tex](g + f)(8) =133[/tex]

Answer:

12

Step-by-step explanation:

The amount of children that do like ice cream are 14/26 so the children that do not like ice cream 14/26, and the numerator is 12

Solution :

Given :

mean weight of the dogs, [tex]$\overline x$[/tex] = 69

Number of dogs, n = 37

Standard deviation, σ = 9.2

Confidence interval = 90%

At 90% confidence interval for the true population mean dog weight is given by :

[tex]$= \overline x \pm \frac{\sigma}{\sqrt n} \times z_{0.05}$[/tex]

[tex]$= 69\pm \frac{9.2}{\sqrt_{37}}} \times 1.64485$[/tex]

[tex]$=69 \pm 2.487799$[/tex]

= (66.5122, 71.4878)

Answer: 10

Step-by-step explanation: 15 - 5 = 10

AA~ as both the triangles are congruent because of vertically opposite angles and alternative interior angles .

Answer:

y > 0.

Step-by-step explanation:

A.

Answer:

If we define W as the width:

12m  ≤ W  ≤  22m

Step-by-step explanation:

We have a rectangle with length L and width W.

We know that:

"The length of a rectangle should be 9 meters longer than 7 times the width"

Then:

L = 9m + 7*W

We also know that the length must be between 93 and 163 meters long, so:

93m ≤ L ≤ 163m

Now we want to find the restrictions for the width W.

We start with:

93m ≤ L ≤ 163m

Now we know that L = 9m + 7*W, then we can replace that in the above inequality:

93m ≤  9m + 7*W ≤ 163m

Now we need to isolate W.

First, we can subtract 9m in the 3 sides of the inequality

93m - 9m  ≤  9m + 7*W  -9m ≤ 163m -9m

84m ≤ 7*W ≤ 154m

Now we can divide by 7 in the 3 sides, so we get:

84m/7 ≤ 7*W/7 ≤ 154m/7

12m  ≤ W  ≤  22m

Then we can conclude that the width is between 12 and 22 meters long.

Answer:

49.87

Step-by-step explanation:

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.

Suppose the mean percentage in Algebra 2B is 70% and the standard deviation is 8%.

This means that [tex]\mu = 70, \sigma = 8[/tex]

What percentage of students receive between a 70% and 94%

The proportion is the p-value of Z when X = 94 subtracted by the p-value of Z when X = 70. So

X = 94

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

[tex]Z = \frac{94 - 70}{8}[/tex]

[tex]Z = 3[/tex]

[tex]Z = 3[/tex] has a p-value of 0.9987.

X = 70

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

[tex]Z = \frac{70 - 70}{8}[/tex]

[tex]Z = 0[/tex]

[tex]Z = 0[/tex] has a p-value of 0.5.

0.9987 - 0.5 = 0.4987.

0.4987*100% = 49.87%.

So the percentage is 49.87%, and the answer, without the percent sign, is 49.87.

Answer:

x = 1 , y = -4

Step-by-step explanation:

2x - 5y = 22 ------- ( 1 )

y = 3x - 7      ------- ( 2 )

Substitute ( 2 ) in ( 1 ) :

2x - 5 (3x - 7) = 22

2x - 15x + 35 = 22

- 13x = 22 - 35

- 13x =  - 13

 x = 1

Substitute x in ( 1 ) :

2x - 5y = 22

2 ( 1 ) - 5y = 22

- 5y = 22 - 2

-5y = 20

y = - 4

Answer:

parrel line never meet

Answer:

[tex]Cov(X,Y) = -\frac{ 25}{9}[/tex]

Step-by-step explanation:

Given

[tex]Interval =[0,10][/tex]

[tex]X + Y < 10[/tex]

Required

[tex]Cov(X,Y)[/tex]

First, we calculate the joint distribution of X and Y

Plot [tex]X + Y < 10[/tex]

So, the joint pdf is:

[tex]f(X,Y) = \frac{1}{Area}[/tex] --- i.e. the area of the shaded region

The shaded area is a triangle that has: height = 10; width = 10

So, we have:

[tex]f(X,Y) = \frac{1}{0.5 * 10 * 10}[/tex]

[tex]f(X,Y) = \frac{1}{50}[/tex]

[tex]Cov(X,Y)[/tex] is calculated as:

[tex]Cov(X,Y) = E(XY) - E(X) \cdot E(Y)[/tex]

Calculate E(XY)

[tex]E(XY) =\int\limits^X_0 {\int\limits^Y_0 {\frac{XY}{50}} \, dY} \, dX[/tex]

[tex]X + Y < 10[/tex]

Make Y the subject

[tex]Y < 10 - X[/tex]

So, we have:

[tex]E(XY) =\int\limits^{10}_0 {\int\limits^{10 - X}_0 {\frac{XY}{50}} \, dY} \, dX[/tex]

Rewrite as:

[tex]E(XY) =\frac{1}{50}\int\limits^{10}_0 {\int\limits^{10 - X}_0 {XY}} \, dY} \, dX[/tex]

Integrate Y

[tex]E(XY) =\frac{1}{50}\int\limits^{10}_0 {\frac{XY^2}{2}}} }|\limits^{10 - X}_0 \, dX[/tex]

Expand

[tex]E(XY) =\frac{1}{50}\int\limits^{10}_0 {\frac{X(10 - X)^2}{2} - \frac{X(0)^2}{2}}} }\ dX[/tex]

[tex]E(XY) =\frac{1}{50}\int\limits^{10}_0 {\frac{X(10 - X)^2}{2}}} }\ dX[/tex]

Rewrite as:

[tex]E(XY) =\frac{1}{100}\int\limits^{10}_0 X(10 - X)^2\ dX[/tex]

Expand

[tex]E(XY) =\frac{1}{100}\int\limits^{10}_0 X*(100 - 20X + X^2)\ dX[/tex]

[tex]E(XY) =\frac{1}{100}\int\limits^{10}_0 100X - 20X^2 + X^3\ dX[/tex]

Integrate

[tex]E(XY) =\frac{1}{100} [\frac{100X^2}{2} - \frac{20X^3}{3} + \frac{X^4}{4}]|\limits^{10}_0[/tex]

Expand

[tex]E(XY) =\frac{1}{100} ([\frac{100*10^2}{2} - \frac{20*10^3}{3} + \frac{10^4}{4}] - [\frac{100*0^2}{2} - \frac{20*0^3}{3} + \frac{0^4}{4}])[/tex]

[tex]E(XY) =\frac{1}{100} ([\frac{10000}{2} - \frac{20000}{3} + \frac{10000}{4}] - 0)[/tex]

[tex]E(XY) =\frac{1}{100} ([5000 - \frac{20000}{3} + 2500])[/tex]

[tex]E(XY) =50 - \frac{200}{3} + 25[/tex]

Take LCM

[tex]E(XY) = \frac{150-200+75}{3}[/tex]

[tex]E(XY) = \frac{25}{3}[/tex]

Calculate E(X)

[tex]E(X) =\int\limits^{10}_0 {\int\limits^{10 - X}_0 {\frac{X}{50}}} \, dY} \, dX[/tex]

Rewrite as:

[tex]E(X) =\frac{1}{50}\int\limits^{10}_0 {\int\limits^{10 - X}_0 {X}} \, dY} \, dX[/tex]

Integrate Y

[tex]E(X) =\frac{1}{50}\int\limits^{10}_0 { (X*Y)|\limits^{10 - X}_0 \, dX[/tex]

Expand

[tex]E(X) =\frac{1}{50}\int\limits^{10}_0 ( [X*(10 - X)] - [X * 0])\ dX[/tex]

[tex]E(X) =\frac{1}{50}\int\limits^{10}_0 ( [X*(10 - X)]\ dX[/tex]

[tex]E(X) =\frac{1}{50}\int\limits^{10}_0 10X - X^2\ dX[/tex]

Integrate

[tex]E(X) =\frac{1}{50}(5X^2 - \frac{1}{3}X^3)|\limits^{10}_0[/tex]

Expand

[tex]E(X) =\frac{1}{50}[(5*10^2 - \frac{1}{3}*10^3)-(5*0^2 - \frac{1}{3}*0^3)][/tex]

[tex]E(X) =\frac{1}{50}[5*100 - \frac{1}{3}*10^3][/tex]

[tex]E(X) =\frac{1}{50}[500 - \frac{1000}{3}][/tex]

[tex]E(X) = 10- \frac{20}{3}[/tex]

Take LCM

[tex]E(X) = \frac{30-20}{3}[/tex]

[tex]E(X) = \frac{10}{3}[/tex]

Calculate E(Y)

[tex]E(Y) =\int\limits^{10}_0 {\int\limits^{10 - X}_0 {\frac{Y}{50}}} \, dY} \, dX[/tex]

Rewrite as:

[tex]E(Y) =\frac{1}{50}\int\limits^{10}_0 {\int\limits^{10 - X}_0 {Y}} \, dY} \, dX[/tex]

Integrate Y

[tex]E(Y) =\frac{1}{50}\int\limits^{10}_0 { (\frac{Y^2}{2})|\limits^{10 - X}_0 \, dX[/tex]

Expand

[tex]E(Y) =\frac{1}{50}\int\limits^{10}_0 ( [\frac{(10 - X)^2}{2}] - [\frac{(0)^2}{2}])\ dX[/tex]

[tex]E(Y) =\frac{1}{50}\int\limits^{10}_0 ( [\frac{(10 - X)^2}{2}] )\ dX[/tex]

[tex]E(Y) =\frac{1}{50}\int\limits^{10}_0 [\frac{100 - 20X + X^2}{2}] \ dX[/tex]

Rewrite as:

[tex]E(Y) =\frac{1}{100}\int\limits^{10}_0 [100 - 20X + X^2] \ dX[/tex]

Integrate

[tex]E(Y) =\frac{1}{100}( [100X - 10X^2 + \frac{1}{3}X^3]|\limits^{10}_0)[/tex]

Expand

[tex]E(Y) =\frac{1}{100}( [100*10 - 10*10^2 + \frac{1}{3}*10^3] -[100*0 - 10*0^2 + \frac{1}{3}*0^3] )[/tex]

[tex]E(Y) =\frac{1}{100}[100*10 - 10*10^2 + \frac{1}{3}*10^3][/tex]

[tex]E(Y) =10 - 10 + \frac{1}{3}*10[/tex]

[tex]E(Y) =\frac{10}{3}[/tex]

Recall that:

[tex]Cov(X,Y) = E(XY) - E(X) \cdot E(Y)[/tex]

[tex]Cov(X,Y) = \frac{25}{3} - \frac{10}{3}*\frac{10}{3}[/tex]

[tex]Cov(X,Y) = \frac{25}{3} - \frac{100}{9}[/tex]

Take LCM

[tex]Cov(X,Y) = \frac{75- 100}{9}[/tex]

[tex]Cov(X,Y) = -\frac{ 25}{9}[/tex]

Answer:

[tex]A * B = \left[\begin{array}{ccc}10&-6\\8&5\\19&-10\end{array}\right][/tex]

Step-by-step explanation:

Given

[tex]A =\left[\begin{array}{cc}2&-2\\3&4\\4&-3\end{array}\right][/tex]

[tex]B = \left[\begin{array}{cc}4&-1\\-1&2\end{array}\right][/tex]

Required

[tex]AB[/tex]

To do this, we simply multiply the rows of A by the column of B;

So, we have:

[tex]A * B = \left[\begin{array}{ccc}2*4 + -2*-1&2*-1+-2*2\\3*4+4*-1&3*-1+4*2\\4*4-3*-1&4*-1-3*2\end{array}\right][/tex]

[tex]A * B = \left[\begin{array}{ccc}10&-6\\8&5\\19&-10\end{array}\right][/tex]

Answer:

The median weight for shelter A is greater than that for shelter B.

The data for shelter B are a symmetric data set.

The interquartile range of shelter A is greater than the interquartile range of shelter B.

Step-by-step explanation:

The median weight for shelter A is greater than that for shelter B.

The median of A = 21 and the median of B = 18  true

The median weight for shelter B is greater than that for shelter A.

The median of A = 21 and the median of B = 18   false

The data for shelter A are a symmetric data set.

False, looking at the box it is not symmetric

The data for shelter B are a symmetric data set.

true, looking at the box it is  symmetric

The interquartile range of shelter A is greater than the interquartile range of shelter B.

IQR = 28 - 17 = 11 for A

IQR for B = 20 -16 = 4  True

Answer:

Whe we have two functions, f(x) and g(x), the composite function:

(f°g)(x)

is just the first function evaluated in the second one, or:

f( g(x))

And the domain of a function is the set of inputs that we can use as the variable x, we usually start by thinking that the domain is the set of all real numbers, unless there is a given value of x that causes problems, like a zero in the denominator, for example:

f(x) = 1/(x + 1)

where for x = -1 we have a zero in the denominator, then the domain is the set of all real numbers except x = -1.

Now, we have:

f(x) = x^2

g(x) = x + 9

then:

(f ∘ g)(x) = (x + 9)^2

And there is no value of x that causes problems here, so the domain is the set of all real numbers, that, in interval notation, is written as:

x ∈ (-∞, ∞)

(g ∘ f)(x)

this is g(f(x)) = (x^2) + 9 = x^2 + 9

And again, here we do not have any problem with a given value of x, so the domain is again the set of all real numbers:

x ∈ (-∞, ∞)

(f ∘ f)(x) = f(f(x)) = (f(x))^2 = (x^2)^2 = x^4

And for the domain, again, there is no value of x that causes a given problem, then the domain is the same as in the previous cases:

x ∈ (-∞, ∞)

(g ∘ g)(x) = g( g(x) ) = (g(x) + 9) = (x + 9) +9 = x + 18

And again, there are no values of x that cause a problem here, so the domain is:

x ∈ (-∞, ∞)

Answer:

2 staff members

Step-by-step explanation:

Given

See attachment for missing details

Let

[tex]s \to staff\ member[/tex]

[tex]p \to participant[/tex]

[tex]e \to puzzle[/tex]

Required

Staff members for 18 participants

From the attachment, we have:

[tex]1s \to 8p[/tex] ---- 1 staff member to 8 participants

[tex]s \to 8p[/tex]

Multiply both sides by 1

[tex]s * 2 \to 8p * 2[/tex]

[tex]2s \to 16p[/tex]

This means that 2 staff members are required for 16 participants

Answer:

answer is 0 in the ones column

9514 1404 393

Answer:

  15

Step-by-step explanation:

(2 loaves)(5/2 cups/loaf)/(1/3 cup/scoop) = (2)(5/2)(3/1) scoops = 15 scoops

The baker will need 15 scoops of flour.

Answer:

Number of 10 cents = 18

Number of  20 cents = 10

Step-by-step explanation:

Let number of 10 cents be = x

Let number 20 cents be = y

Total number of coins = x + y = 28  -------- ( 1 )

Total amount in the box = 0.10 x + 0.20y = 3.80 ---------- ( 2 )

Solve the equations to find x and y

( 1 ) => x + y = 28

         x = 28 - y

Substitute x in ( 2 )

( 2 ) => 0.10(28 - y) + 0.20y = 3.80

          2.80 - 0.10y + 0.20y = 3.80

            0.10 y = 3.80 - 2.80

            0.10 y = 1.00

                [tex]y = \frac{1}{0.10} = 10[/tex]

                y = 10

Substitute y in ( 1 ) => x + y = 28

                                x + 10 = 28

                                x = 28 - 10

                                x = 18

Answer:

Step-by-step explanation:

Answer:

24

Step-by-step explanation:

Find the prime factorization of the number 3,600. Factor Tree.

2|3,600.

2|1,800.

2|900.

2|450

5|225

5|45

3|9

3|3

|1

Setup the equation for determining the number of factors or divisors.

3600=2x2x2x2x3x3x5

Sum factors=2+2+2+2+2+3+3+5=24

Answer:

D.80

Step-by-step explanation:

You need to divide thus

16m/0.2m=80m

Answer:

Well it all started by drawing some equilateral triangles so that they made a regular hexagon: hexagon from unit length triangles. Then we ...

9514 1404 393

Answer:

  see attached

Step-by-step explanation:

The reflection across y=-x swaps the coordinates and negates both of them. The first-quadrant figure becomes a third-quadrant figure.

  (x, y) ⇒ (-y, -x)

Answer:

60 kg of 40% and 90 kg of 60%

Step-by-step explanation:

Let the amount of 40% chocolate be x.

Let the amount of 60% chocolate be y.

x + y = 150

0.4x + 0.6y = 0.52 * 150

x + y = 150

4x + 6y = 780

     -4x - 4y = -600

(+)   4x + 6y = 780

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

               2y = 180

y = 90

x + y = 150

x + 90 = 150

x = 60

Answer: 60 kg of 40% and 90 kg of 60%

[tex]m(2m+2n)+3mn+2m^2\implies \stackrel{\textit{distributing}}{2m^2+2mn}+3mn+2m^2 \\\\\\ 2m^2+2m^2+2mn+3mn\implies \stackrel{\textit{adding like-terms}}{4m^2+5mn}[/tex]