10^2x+3=5
Solve, expressing your answer in an exact form involving a common logarithm and showing your steps.

Help please

Answer: distributive property

Step-by-step explanation: the 4 is multiplied by everting in the parenthesis

Answer:

3 meters.

Step-by-step explanation:

400 rev / minute = 400 × 60 rev / 60 minutes

= 24,000 rev / hour

24,000 × C = 72,000 m : C is the circumference

C = 3 meters

Answer:

3 meters

Step-by-step explanation:

72 km / hour * 1 hour/ 60 min  * 1000m/ 1 km

72000 meters /60 minute

1200 meters / minute

velocity = radius * w

Where w is 2*pi * the revolutions per minute

1200 = r * 2 * pi *400

1200 / 800 pi = r

1.5 /pi = r meters

We want to find the circumference

C = 2 * pi *r

C = 2* pi ( 1.5 / pi)

C = 3 meters

Answer:

I think around $36

Hope it helps!

Answer:

It depends...

Step-by-step explanation:

It depends how much weeks are in the month if there are three weeks and no extra days then you would have an answer of about 1093 (exact: 1093.33333333). just divide the number of weeks by the number of money.

Answer:

Option D

Step-by-step explanation:

correct answer on edge :)

Answer:

D <3

Step-by-step explanation:

Answer:

$170.40

Step-by-step explanation:

cost of policy=$8.52

Groups of $1000=20,00÷1000

=20

20,000 worth=$8.52×20

$170.40

Answer:  

a) h(x; 6, 9, 17).

b) P[X=2] = 0.2036

P[X ≤ 2] = 0.2466

P[X ≥ 2] = 0.9570.

c) Mean  = 3.176.

Variance = 1.028.

Standard deviation = 1.014.

Step-by-step explanation:

From the given details K=6, n=9, N=-17.

We conclude that it is the hypergeometric distribution:  

a) h(x; 6, 9, 17).

b)

[tex]P[X=2]=\frac{(^{g}C_{2})^{17-9}C_{6-2}}{^{17}C_{6}\textrm{}}[/tex]

P[X=2] = 0.2036

P[X ≤ 2] = P(x=0)+ P(x=1) + P(x=2)

P[X ≤ 2] = 0.2466

P[X ≥ 2] = 1-[P(x=0)+P(x=1)]

P[X ≥ 2] = 0.9570.

c)

Mean= [tex]n\frac{K}{N}[/tex]

            = 3.176.

Variance = [tex]n\frac{K}{N}( \frac{N-K}{N})(\frac{N-n}{n-1} )[/tex]

               = 2.824 x 0.6471 x 0.5625

               = 1.028.

Standard deviation = [tex]\sqrt{1.028}[/tex] = 1.014.

[tex]\boxed{\large{\bold{\blue{ANSWER~:) }}}}[/tex]

For each subset it can either contain or not contain an element. For each element, there are 2 possibilities. Multiplying these together we get 27 or 128 subsets. For generalisation the total number of subsets of a set containing n elements is 2 to the power n.

total subsets

Answer:

x=3

Step-by-step explanation:

7(x - 7) = -28

x - 7 = -4

x = 3

Answer:

x = 3

Step-by-step explanation:

Your goal is to isolate the x from the other numbers.

-28 = 7(x - 7)

Distribute the 7 to the (x - 7)

You will end up with:

-28 = 7x - 49

Add 49 to both sides of the equation to further isolate the x

21 = 7x

Finally, divide both sides by 7 so x is by itself

x = 3

Answer:

0.1423 = 14.23% probability that, in any hour, more than 5 customers will arrive.

Step-by-step explanation:

We have the mean, which means that the Poisson distribution is used to solve this question.

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

In which

x is the number of sucesses

e = 2.71828 is the Euler number

[tex]\mu[/tex] is the mean in the given interval.

A mean of 3.5 customers arrive hourly at the drive-through window.

This means that [tex]\mu = 3.5[/tex]

What is the probability that, in any hour, more than 5 customers will arrive?

This is:

[tex]P(X > 5) = 1 - P(X \leq 5)[/tex]

In which

[tex]P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)[/tex]

Then

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

[tex]P(X = 0) = \frac{e^{-3.5}*3.5^{0}}{(0)!} = 0.0302[/tex]

[tex]P(X = 1) = \frac{e^{-3.5}*3.5^{1}}{(1)!} = 0.1057[/tex]

[tex]P(X = 2) = \frac{e^{-3.5}*3.5^{2}}{(2)!} = 0.1850[/tex]

[tex]P(X = 3) = \frac{e^{-3.5}*3.5^{3}}{(3)!} = 0.2158[/tex]

[tex]P(X = 4) = \frac{e^{-3.5}*3.5^{4}}{(4)!} = 0.1888[/tex]

[tex]P(X = 5) = \frac{e^{-3.5}*3.5^{5}}{(5)!} = 0.1322[/tex]

Finally

[tex]P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0302 + 0.1057 + 0.1850 + 0.2158 + 0.1888 + 0.1322 = 0.8577[/tex]

[tex]P(X > 5) = 1 - P(X \leq 5) = 1 - 0.8577 = 0.1423[/tex]

0.1423 = 14.23% probability that, in any hour, more than 5 customers will arrive.

Answer:

im pretty sure the degree is 4.

Step-by-step explanation:

Answer:

7x10 ^-10

Step-by-step explanation:

Answer:

C. (-4099)^1/5

Step-by-step explanation:

[tex]x^{\frac{1}{2} } = \sqrt{x}[/tex]

you can not take roots (real roots) of a negative number if the exponent is

even ... A,B,D have even exponents (in the denominator of the exponent.. in other words the index of the radical is even)...

the only odd index is in "B" (the 5 in the 1/5)

Step-by-step explanation:

-27 okay 3^-3 its same as 3^3

Answer: A)

[tex]3^{-3}[/tex]

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

[tex]=\frac{1}{3^3}[/tex]

[tex]3^3=27[/tex]

[tex]=\frac{1}{27}[/tex]

OAmalOHopeO

Answer:

268.40

Step-by-step explanation:

We can write a ratio to solve

2.5 meters        22 meters

-----------------  = --------------

30.5 dollars       x dollars

Using cross products

2.5 * x = 30.5 * 22

2.5x =671

Divide each side by 2.5

2.5x / 2.5 = 671/2.5

x =268.4

Answer:

[tex]\pi[/tex]

Step-by-step explanation:

1. [tex]c = 2\pi r[/tex]

2. [tex]c=2\pi \frac{1}{2}[/tex]

3. [tex]c=\pi[/tex]

Answer:

8×1000 milligrams

8000 milligrams

Answer:

Length (L) = 72 feet

Step-by-step explanation:

From the question given above, the following data were obtained:

Period (T) = 9.42 s

Pi (π) = 3.14

Length (L) =?

The length of the pendulum can be obtained as follow:

T = 2π √(L/32)

9.42 = (2 × 3.14) √(L/32)

9.42 = 6.28 √(L/32)

Divide both side by 6.28

√(L/32) = 9.42 / 6.28

Take the square of both side

L/32 = (9.42 / 6.28)²

Cross multiply

L = 32 × (9.42 / 6.28)²

L = 72 feet

Thus, the Lenght is 72 feet

(a) Data with the eight day's measurement.

Raw data:      [60,58,64,64,68,50,57,82],  

Sorted data:  [50,57,58,60,64,64,68,82]

Sample size = 8 (even)

mean            = 62.875

median         = (60+64)/2 = 62

1st quartile   = (57+58)/2 = 57.5

3rd quartile  = (64+68)/2 =  66

IQR = 66 - 57.5 = 8.5

(b) Data without the eight day's measurement.

Raw data:      [60,58,64,64,68,50,57]

Sorted data:  [50,57,58,60,64,64,68]

Sample size = 7 (odd)

mean            = 60.143

median         = 60

1st quartile   = 57

3rd quartile = 64

IQR = 64 -57 = 7

Answers:

1. The average is the same with or without the 8th day's data.  FALSE

2. The median is the same with or without the 8th day's data.  FALSE

3. The IQR decreases when the 8th day is included.                  FALSE

4. The IQR increases when the 8th day is included.                   TRUE

5. The median is higher when the 8th day is included.              TRUE

Answer:

The Interquartile range increases when the 8th day is included.          

The median is higher when the 8th day is included.

Step-by-step explanation:

i got it right. Hope this helps.

The length of wire used for the equilateral triangle is approximately 5.61 meters.

The remaining length of wire used for the circle will be 9 - 5.61 ≈ 3.39 meters.

Here,

To minimize the total area of both figures, we need to find the optimal cut point for the wire.

Let's assume the length of the wire used for the equilateral triangle is x meters, and the remaining length of the wire used for the circle is (9 - x) meters.

For the equilateral triangle:

An equilateral triangle has all three sides equal in length.

Let's call each side of the triangle s meters. Since the total length of the wire is x meters, each side will be x/3 meters.

The formula to find the area of an equilateral triangle with side length s is:

Area = (√(3)/4) * s²

Substitute s = x/3 into the area formula:

Area = (√(3)/4) * (x/3)²

Area = (√(3)/4) * (x²/9)

Now, for the circle:

The circumference (perimeter) of a circle is given by the formula:

Circumference = 2 * π * r

Since the remaining length of wire is (9 - x) meters, the circumference of the circle will be 2π(9 - x) meters.

The formula to find the area of a circle with radius r is:

Area = π * r²

To find the area of the circle, we need to find the radius.

Since the circumference is equal to 2πr, we can set up the equation:

2πr = 2π(9 - x)

Now, solve for r:

r = (9 - x)

Now, substitute r = (9 - x) into the area formula for the circle:

Area = π * (9 - x)²

Now, we want to minimize the total area, which is the sum of the areas of the triangle and the circle:

Total Area = (√(3)/4) * (x²/9) + π * (9 - x)²

To find the optimal value of x that minimizes the total area, we can take the derivative of the total area with respect to x, set it to zero, and solve for x.

d(Total Area)/dx = 0

Now, find the critical points and determine which one yields the minimum area.

Taking the derivative and setting it to zero:

d(Total Area)/dx = (√(3)/4) * (2x/9) - 2π * (9 - x)

Setting it to zero:

(√(3)/4) * (2x/9) - 2π * (9 - x) = 0

Now, solve for x:

(√(3)/4) * (2x/9) = 2π * (9 - x)

x/9 = (8π - 2πx) / (√(3))

Now, isolate x:

x = 9 * (8π - 2πx) / (√(3))

x(√(3)) = 9 * (8π - 2πx)

x(√(3) + 2π) = 9 * 8π

x = (9 * 8π) / (√(3) + 2π)

Now, we can calculate the value of x:

x ≈ 5.61 meters

So, the length of wire used for the equilateral triangle is approximately 5.61 meters.

The remaining length of wire used for the circle will be 9 - 5.61 ≈ 3.39 meters.

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Answer:

23) No

24) No

25) Yes

Step-by-step explanation:

Question 23)

We want to determine if a zero exists between 1 and 2 for the function:

[tex]f(x)=x^2-4x-5[/tex]

Find the zeros of the function. We can factor:

[tex]\displaystyle 0 = (x-5)(x+1)[/tex]

Zero Product Property:

[tex]x-5=0\text{ or } x+1=0[/tex]

Solve for each case. Hence:

[tex]\displaystyle x = 5\text{ or } x=-1[/tex]

Therefore, our zeros are at x = 5 and x = -1.

In conclusion, a zero does not exist between 1 and 2.

Question 24)

We have the function:

[tex]f(x)=2x^2-7x+3[/tex]

And we want to determine if a zero exists between 1 and 2.

Factor. We want to find two numbers that multiply to (2)(3) = 6 and that add to -7.

-6 and -1 suffice. Hence:

[tex]\displaystyle \begin{aligned} 0 & = 2x^2-7x + 3 \\ & = 2x^2 -6x -x + 3 \\ &= 2x(x-3) - (x-3) \\ &= (2x-1)(x-3) \end{aligned}[/tex]

By the Zero Product Property:

[tex]2x-1=0\text{ or } x-3=0[/tex]

Solve for each case:

[tex]\displaystyle x=\frac{1}{2} \text{ or } x=3[/tex]

Therefore, our zeros are at x = 1/2 and x = 3.

In conclusion, a zero does not exist between 1 and 2.

Question 25)

We have the function:

[tex]f(x)=3x^2-2x-5[/tex]

And we want to determine if a zero exists between -2 and 3.

Factor. Again, we want to find two numbers that multiply to 3(-5) = -15 and that add to -2.

-5 and 3 works perfectly. Hence:

[tex]\displaystyle \begin{aligned} 0&= 3x^2 -2x -5 \\ &= 3x^2 +3x - 5x -5 \\ &= 3x(x+1)-5(x+1) \\ &= (3x-5)(x+1)\end{aligned}[/tex]

By the Zero Product Property:

[tex]\displaystyle 3x-5=0\text{ or } x+1=0[/tex]

Solve for each case:

[tex]\displaystyle x = \frac{5}{3}\text{ or } x=-1[/tex]

In conclusion, there indeed exists a zero between -2 and 3.

Answer:

B

Step-by-step explanation:

Let me know if you need an explanation.

Answer:

B

Step-by-step explanation:

4x^3+x^2+5x+2

4x^3 cannot cancel with others= 4x^3

4x^2-3x^2= x^2

5x cannot cancel with others= 5x

-3+5= 2

4x^3+x^2+5x+2

Answer:

The inverse function is [tex]g(x) = \frac{x - 2300}{170}[/tex]

The value of g(x) represents the number of applicants to the freshman class based on the number of years since 2010.

Step-by-step explanation:

Number of applicants in x years after 2010:

Is given by the following function:

[tex]f(x) = 2300 + 170x[/tex]

Inverse function:

We exchange the values of y = f(x) and x in the original function, and then find y. So

[tex]x = 2300 + 170y[/tex]

[tex]170y = x - 2300[/tex]

[tex]y = \frac{x - 2300}{170}[/tex]

[tex]g(x) = \frac{x - 2300}{170}[/tex]

The inverse function is [tex]g(x) = \frac{x - 2300}{170}[/tex]

Meaning of g:

f(x): Number of students in x years:

g(x): Inverse of f(x), is the number of years it takes for there to be x applicants, so the answer is:

The value of g(x) represents the number of applicants to the freshman class based on the number of years since 2010.

Answer:

y + 1 = 7(x + 2).

Step-by-step explanation:

Point slope form:

y - y1 =m(x - x1).

Here m = 7 and (x1,  y1) = (-2,  -1)

So the answer is;

y - (-1) = 7(x - (-2))

Answer:

1+i

Step-by-step explanation:

I do believe i to be the imaginary unit.

Let's write out some partial sums from power=0 to power=7 or whatever we need to see a pattern.

i^0=1

i^0+i^1=1+i

i^0+i^1+i^2=1+i+-1=i

i^0+i^1+i^2+i^3=i+i^3=i+-i=0

i^0+i^1+i^2+i^3+i^4=0+i^4=0+1=1

Hmmm.... we might see 1+i, then i, then 0 again.... let's see.

i^0+i^1+i^2+i^3+i^4+i^5=1+i

Coolness so we should see a pattern

Sum from power=0 to power=multiples of 4 will give us 1.

Sum from power=0 to power=remainder of 1 when final power is divided by 4 gives us 1+i.

Sum from power=0 to power=remainder of 2 when final power is divided by 4 gives us i.

Sum from power=0 to power=remainder of 3 when final power is divided by 4 gives us 1

0.

So 2021 divided by 4....

Since 2020 is a multiple of 4, then 2021 has a remainder of 1 when divided by 4.

So the answer is 1+i.

[tex] \sf \: d \: = 3.7m \\ \sf \: r \: = \frac{3.7}{2} = 1.85 \: m\\ \\ \sf \: c \: = \pi {r}^{2} \\ \\ \sf \: c \: = \pi ({1.85})^{2} \\ \sf c = 1.85 \times 1.85 \times \pi \\ \sf \: c = \boxed {\underline{ \bf a. \: 3.4225\pi \: m ^{2} }}[/tex]

Answer:

a)  The joint probability distribution

P(0,0) = 0.36, P(1,0) = 0.24,   P(2,0) = 0,   P(0,1) = 0,  P(1,1) = 0.24,  P(2,1)= 0.16

b)  P( W = 0 ) = 0.36,    P(W = 1 ) = 0.48,  P(W = 2 ) = 0.16

c) P ( z = 0 ) = 0.6

  P ( z = 1 ) = 0.4

Step-by-step explanation:

Number of head on first toss = Z

Total Number of heads on 2 tosses = W

% of head occurring = 40%

% of tail occurring = 60%

P ( head ) = 2/5 ,    P( tail ) = 3/5

a) Determine the joint probability distribution of W and Z

P( W =0 |Z = 0 ) = 0.6         P( W = 0 | Z = 1 ) = 0

P( W = 1 | Z = 0 ) = 0.4        P( W = 1 | Z = 1 ) = 0.6

P( W = 1 | Z = 0 ) = 0           P( W = 2 | Z = 1 ) = 0.4

The joint probability distribution

P(0,0) = 0.36, P(1,0) = 0.24,   P(2,0) = 0,   P(0,1) = 0,  P(1,1) = 0.24,  P(2,1)= 0.16

B) Marginal distribution of W

P( W = 0 ) = 0.36,    P(W = 1 ) = 0.48,  P(W = 2 ) = 0.16

C) Marginal distribution of Z ( pmf of Z )

P ( z = 0 ) = 0.6

P ( z = 1 ) = 0.4

Part(a): The required joint probability of W and Z is ,

[tex]P(0,0)=0.36,P(1,0)=0.24,P(2,0)=0,P(0,1)=0,P(1,1)=0.24,\\\\P(2,1)=0.16[/tex]

Part(b): The pmf (marginal distribution) of W is,

[tex]P(w=0)=0.36,P(w=1)=0.48,P(w=2)=0.16[/tex]

Part(c): The pmf (marginal distribution) of Z is,

[tex]P(z=0)=0.6,P(z=1)=0.4[/tex]

Part(a):

The joint distribution is,

[tex]P(w=0\z=0)=0.6,P(w=1|z=0)=0.4,P(w=2|z=0)=0[/tex]

Also,

[tex]P(w=0\z=1)=0,P(w=1|z=1)=0.6,P(w=2|z=1)=0.4[/tex]

Therefore,

[tex]P(0,0)=0.36,P(1,0)=0.24,P(2,0)=0,P(0,1)=0,P(1,1)=0.24,\\\\P(2,1)=0.16[/tex]

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