What equation can I use to pick numbers 1-70 if they're picked randomly

Answer:

7 : 8

Step-by-step explanation:

that is the procedure above

Answer:

x = 20

Step-by-step explanation:

 3x -10 = 2x +10

x = 20

Answer:

b and f

Step-by-step explanation:

Answer:

Paedtricians claim isn't true.

Step-by-step explanation:

The hypothesis :

H0 : σ = 7

H0 : σ > 7

The test statistic ; χ² :

χ² = [(n - 1) * s²] ÷ σ²

n = 25 ; s = 4.3, σ = 7

χ² = [(25 - 1) * 4.3²] ÷ 7²

χ² = [(24 * 4.3²] ÷ 49

χ² = 443.76 / 49

χ² = 9.056

At α = 0.01 ; critical value = 42.980

Since critical value > test statistic, we fail to reject the null, H0.

tan graph and its tax because tax=0

Hello!

[tex]\large\boxed{(x + 2)(x + 3)}[/tex]

x² + 5x + 6

Find two numbers that add up to 5 and multiply to 6. We get:

2, 3

Therefore:

(x + 2)(x + 3)

Answer:

i cant see the image

Step-by-step explanation:

Answer:

36

Step-by-step explanation:

(h · f)(x) = h(f(x))

h(f(x)) = h(2x+8)

h(f(x))= 3(2x+8) - 6

h(f(x)) = 6x + 24 - 6

h(f(x))= 6x + 18

If x = 3

h(f(x))= 6(3) + 18

h(f(x))= 18 + 18

h(f(x))= 36

Hence (h · f)(3) = 36

Answer:

2.5, 2.05, 0, -5, -15

Step-by-step explanation:

for negative numbers the bigger is worth less

Answer:

8/15

Step-by-step explanation:

Answer:

8/15

Step-by-step explanation:

when you divide 8/15 its 0.53

Answer:

[tex]-5 \to -24[/tex]

[tex]-1 \to -4[/tex]

[tex]2 \to 11[/tex]

[tex]3 \to 16[/tex]

[tex]4 \to 21[/tex]

Step-by-step explanation:

Given

[tex]f(x) = 5x + 1[/tex]

Required

Complete the table (see attachment)

When x = -5

[tex]f(-5) = 5 * -5 + 1 = -24[/tex]

When x = -1

[tex]f(-1) = 5 * -1 + 1 = -4[/tex]

When x = 2

[tex]f(2) = 5 * 2 + 1 = 11[/tex]

When x = 3

[tex]f(3) = 5 * 3 + 1 = 16[/tex]

When x = 4

[tex]f(4) = 5 * 4 + 1 = 21[/tex]

So, the table is:

[tex]-5 \to -24[/tex]

[tex]-1 \to -4[/tex]

[tex]2 \to 11[/tex]

[tex]3 \to 16[/tex]

[tex]4 \to 21[/tex]

9514 1404 393

Answer:

  D.  f^-1(x) = 3 -7x

Step-by-step explanation:

Solve x = f(y) for y to find the inverse function.

  x = f(y)

  x = (3 -y)/7 . . . . . . use the function definition

  7x = 3 -y . . . . . . . .multiply by 7

  y = 3 -7x . . . . . . . add y-7x to both sides

Then the inverse function is ...

  [tex]\boxed{f^{-1}(x)=3-7x}[/tex]

Answer:

Step-by-step explanation:

Um where is the diagrahm

Answer:

m>-7

Step-by-step explanation:

expand

-10m+15-4<81

-10m+11<81

collect like terms

-10m<81-11

-10m<70

m>-7

Answer:

She had 8 doubles.

Step-by-step explanation:

This question is solved by a system of equations.

I am going to say that:

x is the number of singles.

y is the number of doubles

z is the number of triples.

46 hits

This means that [tex]x + y + z = 46[/tex]

46 hits totaled 66 bases

This means that:

[tex]x + 2y + 3z = 66[/tex]

4 times as many singles as doubles

This means that [tex]x = 4y[/tex]

So

[tex]x + 2y + 3z = 66[/tex]

[tex]4y + 2y + 3z = 66[/tex]

[tex]6y + 3z = 66[/tex]

And

[tex]x + y + z = 46[/tex]

[tex]4y + y + z = 46[/tex]

[tex]5y + z = 46 \rightarrow z = 46 - 5y[/tex]

Then

[tex]6y + 3z = 66[/tex]

[tex]6y + 3(46 - 5y) = 66[/tex]

[tex]6y + 138 - 15y = 66[/tex]

[tex]9y = 72[/tex]

[tex]y = \frac{72}{9}[/tex]

[tex]y = 8[/tex]

She had 8 doubles.

Answer:

The right answer is "[tex]x\simeq 254[/tex]".

Step-by-step explanation:

Let the number of earlier case will be "x".

Now,

⇒ [tex]x+x\times \frac{26.3}{100}=321[/tex]

or,

⇒ [tex]x+x\times 0.263=321[/tex]

By taking "x" common, we get

⇒   [tex]x(1+0.263)=321[/tex]

⇒                    [tex]x=\frac{321}{1.263}[/tex]

⇒                       [tex]=254.15[/tex]

or,

⇒                    [tex]x\simeq 254[/tex]

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

Explanation:

We have an octagon because there are n = 8 sides. The diagram below shows one way to number the sides so you can count them efficiently (without missing any or double counting any).

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

Plug n = 8 into the formula below

S = 180(n-2)

S = 180(8-2)

S = 180(6)

S = 1080

The 8 interior angles add up to 1080 degrees.

Answer:

-  [tex]\frac{4}{\sqrt{29} }[/tex]

Step-by-step explanation:

The equations for the 1st object :

x₁ = 10 cos(2t),  and  y₁ = 6 sin(2t)

2nd object :

x₂ = 4 cos(t), y₂ = 4 sin(t)

Determine rate at which distance between objects will continue to change

solution Attached below

Distance( D )  = [tex]\sqrt{(10cos2(t) - 4cos(t))^2 + (6sin2(t) -4sin(t))^2}[/tex]

hence; dD/dt = - [tex]\frac{4}{\sqrt{29} }[/tex]

Answer:

Step-by-step explanation:

Here,

<6 + 125° = 180° [Co-interior angles]

=> <6 = 180 - 125

=> <6 = 55° (Ans)

Answer:

There are 256 pairs in all.

9514 1404 393

Answer:

  12

Step-by-step explanation:

In base-5 arithmetic, ...

  41 -24 = 12

_____

If we use : to separate columns with different place value, this can be looked at a couple of ways.

Subtraction by addition

  2 : 4 + 0 : 2 = 3 : 1 . . . . . make the 1s place match

  3 : 1 + 1 : 0 = 4 : 1 . . . . . . make the 5s place match

The total amount added was 0:2 +1:0 = 1:2.

Subtraction using borrowing

  4 : 1 - 2 : 4 = (4-1) : (5+1) - 2 : 4

  = (4-1-2) : (5+1)-4 = 1:2

Answer: [tex]n=13[/tex]

Step-by-step explanation:

Given

Sum of the first 10 terms is equal to sum of 20, 21, and 22 term

[tex]\Rightarrow \dfrac{10}{2}[2a+(10-1)d]=[a+19d]+[a+20d]+[a+21d]\\\\\Rightarrow 5[2a+9d]=3a+60d\\\Rightarrow 10a+45d=3a+60d\\\Rightarrow 7a=15d[/tex]

No of terms to give a sum of 960

[tex]\Rightarrow 960=\dfrac{n}{2}[2a+(n-1)d]\\\\\Rightarrow 1920=n[2a+(n-1)\cdot \dfrac{7}{15}a]\\\\\Rightarrow 28,800=n[30a+7a(n-1)]\\\\\Rightarrow a=\dfrac{28,800}{n[30+7n-7]}\\\\\Rightarrow a=\dfrac{28,800}{n[23+7n]}[/tex]

Value of first term is less than 20

[tex]\therefore \dfrac{28,800}{n[23+7n]}<20\\\\\Rightarrow 28,800<20n[23+7n]\\\Rightarrow 0<460n+140n^2-28,800\\\Rightarrow 140n^2+460n-28,800>0\\\\\Rightarrow n>12.79\\\\\text{For integer value }\\\Rightarrow n=13[/tex]

Answer:

15

Step-by-step explanation:

In the previous answer halfway through they used the equation: 960 = (n÷2)×(2a+(n-1)×(7a÷15))

Using this equation we can substitute an number to replace n, the higher the number is the smaller a would be.

When we substitute 15 into a, then it leaves us with the answer to be a = 15 which is a positive integer and also is smaller than 20, this then let’s us know that 15 is how many terms can be summed up to make 960.

To double check this answer you can find that d = 7 by changing the a into 15 in the formula 7a/15 (found in the previous answer.

Then in the expression: (n÷2)×(2a+(n-1)×d)

substitute:

n = 14 (must be an even number for the equation to work)

a = 15

d = 7

This will give you an answer of 847, but this is only 14 terms as we changed n into 14. To add the final term you need to complete the following equation: 847+(a+(n-1)×d)

substituting:

n = 15

a = 15

d = 7

This will give you the answer of 960, again proving that it takes 15 terms to sum together to make the number 960.

I hope this has helped you.

P.S. Everything in the previous solution was right apart from the start of the last section and the answer

Answer:

Step-by-step explanation:

a.

(1.36 L)/(10 hr) = (0.136 L)/(hr)

Flow rate = (0.136 L)/(hr) × (1000 mL)/L = (136 mL)/(hr)

136 mL × (27 mg)/(100 mL) = 36.72 mg

Delivery rate = (36.72 mg)/(hr)

b.

(136 mL)/(hr) × (13 gtt)/(mL) = (1868 gtt)/(hr)

c.

10 hr × (36.72 mg)/)hr) = 367.2 mg

Step-by-step explanation:

√19200cm²

=138.56cm

then the highest possible volume

=(138.56)³

=2660195.926cm³

The largest possible volume of the box is; V = 25600 cm³

Let us denote the following of the square box;

Length = x

Width = y

height = h

Formula for volume of a box is;

V = length * width * height

Thus; V = xyh

but we are dealing with a square box and as such, the base sides are all equal and so; x = y. Thus;

V = x²h

The box has an open top and as such, the surface are of the box is;

S = x² + 4xh

We are given S = 19200 cm². Thus;

19200 = x² + 4xh

h = (19200 - x²)/4x

Put (19200 - x²)/4x for h in volume equation to get;

V = x²(19200 - x²)/4x

V = 4800x - 0.25x³

To get largest possible volume, it will be dimensions when dV/dx = 0. Thus;

dV/dx = 4800 - 0.75x²

At dV/dx = 0, we have;

4800 - 0.75x² = 0

0.75x² = 4800

x² = 4800/0.75

x² = 6400

x = √6400

x = 80 cm

From h = (19200 - x²)/4x;

h = (19200 - 80²)/(4 × 80)

h = (19200 - 6400)/3200

h = 4 cm

Largest possible volume = 80² × 4

Largest possible volume = 25600 cm³

Read more at; https://brainly.com/question/19053087

Step-by-step explanation:

x²-xy-xy+y²

x²+2xy+y²

hope it helps

Answer:

42%

Step-by-step explanation:

to find percentages, you move the decimal point twice to the right

Answer:

√3 x^5y

First, let's do √3

√3=1.7

1.7 • x^5 • y

if you want

1.7 • X^4• x• y.

There are tons of equivalent's!

Answer:

[tex]P = (\frac{1}{3}t^\frac{3}{2} + \sqrt 2 - \frac{1}{3})^2[/tex]

Step-by-step explanation:

Given

[tex]\frac{dP}{dt} = \sqrt{Pt[/tex]

[tex]P(1) = 2[/tex]

Required

The solution

We have:

[tex]\frac{dP}{dt} = \sqrt{Pt[/tex]

[tex]\frac{dP}{dt} = (Pt)^\frac{1}{2}[/tex]

Split

[tex]\frac{dP}{dt} = P^\frac{1}{2} * t^\frac{1}{2}[/tex]

Divide both sides by [tex]P^\frac{1}{2}[/tex]

[tex]\frac{dP}{ P^\frac{1}{2}*dt} = t^\frac{1}{2}[/tex]

Multiply both sides by dt

[tex]\frac{dP}{ P^\frac{1}{2}} = t^\frac{1}{2} \cdot dt[/tex]

Integrate

[tex]\int \frac{dP}{ P^\frac{1}{2}} = \int t^\frac{1}{2} \cdot dt[/tex]

Rewrite as:

[tex]\int dP \cdot P^\frac{-1}{2} = \int t^\frac{1}{2} \cdot dt[/tex]

Integrate the left hand side

[tex]\frac{P^{\frac{-1}{2}+1}}{\frac{-1}{2}+1} = \int t^\frac{1}{2} \cdot dt[/tex]

[tex]\frac{P^{\frac{-1}{2}+1}}{\frac{1}{2}} = \int t^\frac{1}{2} \cdot dt[/tex]

[tex]2P^{\frac{1}{2}} = \int t^\frac{1}{2} \cdot dt[/tex]

Integrate the right hand side

[tex]2P^{\frac{1}{2}} = \frac{t^{\frac{1}{2} +1 }}{\frac{1}{2} +1 } + c[/tex]

[tex]2P^{\frac{1}{2}} = \frac{t^{\frac{3}{2}}}{\frac{3}{2} } + c[/tex]

[tex]2P^{\frac{1}{2}} = \frac{2}{3}t^\frac{3}{2} + c[/tex] ---- (1)

To solve for c, we first make c the subject

[tex]c = 2P^{\frac{1}{2}} - \frac{2}{3}t^\frac{3}{2}[/tex]

[tex]P(1) = 2[/tex] means

[tex]t = 1; P =2[/tex]

So:

[tex]c = 2*2^{\frac{1}{2}} - \frac{2}{3}*1^\frac{3}{2}[/tex]

[tex]c = 2*2^{\frac{1}{2}} - \frac{2}{3}*1[/tex]

[tex]c = 2\sqrt 2 - \frac{2}{3}[/tex]

So, we have:

[tex]2P^{\frac{1}{2}} = \frac{2}{3}t^\frac{3}{2} + c[/tex]

[tex]2P^{\frac{1}{2}} = \frac{2}{3}t^\frac{3}{2} + 2\sqrt 2 - \frac{2}{3}[/tex]

Divide through by 2

[tex]P^{\frac{1}{2}} = \frac{1}{3}t^\frac{3}{2} + \sqrt 2 - \frac{1}{3}[/tex]

Square both sides

[tex]P = (\frac{1}{3}t^\frac{3}{2} + \sqrt 2 - \frac{1}{3})^2[/tex]