At Jefferson Middle School, eighty-two students were asked which sports they plan to participate in for the coming year. Twenty students plan to participate in track and cross country; six students in cross country and basketball; and eight students in track and basketball. Twelve students plan to participate in all three sports. A total of thirty students plan to participate in basketball, and a total of forty students plan to participate in cross country. Ten students don't plan to participate in any of the three sports. How many students plan to just participate in cross country? 2 4 40 30

Answer:

Solution : Option B

Step-by-Step Explanation:

We have the following system of equations at hand here.

{ x = 5 cot(t), y = - 3csc(t) + 4 }

Now instead of isolating the t from either equation, let's isolate cot(t) and csc(t) --- Step #1,

x = 5 cot(t) ⇒ x - 5 = cot(t),

y = - 3csc(t) + 4 ⇒ y - 4 = - 3csc(t) ⇒ y - 4 / - 3 = csc(t)

Now let's square these two equations. We know that csc²θ - cot²θ = 1, so let's subtract the equations  as well. --- Step #2

( y - 4 / - 3 )² = (csc(t))²

- ( x - 5 / 1 )² = (cot(t))²  

___________________

(y - 4)² / 9 - x² / 25 = 1

And as we are subtracting the two expressions, this is an example of a hyperbola. Therefore your solution is option b.

Answer:

x ≥-10

Step-by-step explanation:

–2.5x ≤ 25

Divide each side by -2.5, remembering to flip the inequality

–2.5x/-2.5 ≥ 25 /-2.5

x ≥-10

Answer:

[tex]x\leq -10[/tex]

Step-by-step explanation:

[tex]-2.5x\leq 25[/tex]-----> Multiply by -1:

[tex]2.5x\geq -25[/tex]-----> Divide by 2.5:

[tex]x\geq -10[/tex]

Hope this helps!

Answer:

[tex]L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ][/tex]

Step-by-step explanation:

Given that:

[tex]f(t) = 12 cos (t- \dfrac{\pi}{6})[/tex]

recall that:

cos (A-B) = cos AcosB + sin A sin B

∴

[tex]f(t) = 12 [cos\ t \ cos \dfrac{\pi}{6}+ sin \ t \ sin \dfrac{\pi}{6}][/tex]

[tex]f(t) = 12 [cos \ t \ \dfrac{3}{2}+ sin \ t \ sin \dfrac{1}{2}][/tex]

[tex]f(t) = 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t)[/tex]

[tex]L(f(t)) = L ( 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t) ][/tex]

[tex]L(f(t)) = 6 \sqrt{3} \ L [cos \ (t) ] + 6\ L [ sin \ (t) ][/tex]

[tex]L(f(t)) = 6 \sqrt{3} \dfrac{S}{S^2 + 1^2}+ 6 \dfrac{1}{S^2 +1^2}[/tex]

[tex]L(f(t)) = \dfrac{6 \sqrt{3} +6 }{S^2+1}[/tex]

[tex]L(f(t)) = \dfrac{6( \sqrt{3} \ S +1 }{S^2+1}[/tex]

[tex]L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ][/tex]

Step-by-step explanation:

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

= x/2-3 + 4x²+x+4

= ..........

Answer:

-3x - 7y = 36

Step-by-step explanation:

The given line -3x - 7y = 10 has an infinite number of parallel lines, all of the form -3x - 7y = C.

If we want the equation of a line parallel to -3x - 7y = 10 that passes through (-5, -3), we substitute -5 for x in -3x - 7y = 10 and substitute -3 for y in -3x - 7y = 10:

-3(-5) - 7(-3) = C, or

15  + 21 = C, or C = 36

Then the desired equation is -3x - 7y = 36.

Answer:

see below

Step-by-step explanation:

n < 2^n

First let n=1

1 < 2^1

1 <2  This is true

Next, assume that

(k) < 2^(k)

as we attempt to prove that

(k+1) < 2^(k+1)

.

.

.

Therefore we can conclude that

k+1 < 2^(k+1)

Answer:

Step-by-step explanation:

Hello, please consider the following.

First, assume that n equals [tex]\boxed{1}[/tex]. Therefore, [tex]\boxed{1<2^1}[/tex] is [tex]\boxed{\text{True}}[/tex]

Next, assume that [tex]\boxed{k<2^k}[/tex], as we attempt to prove [tex]\boxed{k+1<2^{k+1}}[/tex]

Since .... Therefore, we can conclude that [tex]\boxed{k+1<2^{k+1}}[/tex]

The choice for the last box is confusing. Based on your feedback, we can assume that we are still in the step 2 though.

And the last step which is not included in your question is the conclusion where we can say that we prove that for any integer [tex]n\geq 1[/tex], we have [tex]n<2^n[/tex].

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

Answer:

40 minutes

Step-by-step explanation:

If it takes her 8 minutes to mow 1/6 of it, we can find the total amount of time it  will take by multiplying 8 by 6, since 1/6 times 6 is 1 (1 represents the whole lawn mowed)

8(6) = 48

The question asks for how many more minutes it will take, so subtract 48 by 8.

48 - 8 = 40

= 40 minutes

Answer:

40 minutes

Step-by-step explanation:

We can use ratios to solve

8 minutes          x minutes

------------------- = ----------------

1/6 yard                 1 yard

Using cross products

8 * 1 = 1/6 x

Multiply each side by 6

8*6 = 1/6 * x * 6

48 = x

48 minutes total

She has already done 8 minutes

48-8 = 40 minutes

Answer:

A:$2478

B:$728

Total:$3206

Step-by-step explanation:

2.95x3000=8850

1.30x2000=2600

8850x0.28=2478

2600x0.28=728

2478+728=3206

De acordo com a disponibilidade da unidade, há apenas a seguinte dosagem: 1g/2mL - ou seja, uma grama de dipirona a cada 2mL

O enunciado está meio mal formulado, pois é dito que foram prescritos 500mg de dipirona e é essa quantidade de farmaco que a criança tem que tomar. Deseja-se saber quantos mL deverao ser administrados.

Fazendo a classica regra de 3, podemos chegar no volume desejado:

(atentar que 500mg = 0,5g)

     g               mL

     1    ---------   2

    0,5  ---------  X    

1 . X = 0,5 . 2

Answer:

16.50

Step-by-step explanation:

Constant of proportionality = no of gallons of water per 1 minute.

In the first row, we have 16.50 gallons of water per 1 minute.

In the 2nd row, we have 24.75 gallons of water in 1.5 minutes. In 1 minute, we will have 24.75 ÷ 1.5 = 16.50 gallons

In the 3rd row, we have 33 gallons in 2 minutes. In 1 minute, we will have 33 ÷ 2 = 16.50 gallons.

We can see that there seems to be the same constant of proportionality for the 2nd and 3rd row, which is 16.50.

Thus, a relationship between gallons of water (w) and time (t), considering the constant, 16.50, can be written as: [tex] w = 16.50t [/tex]

This means the constant of proportionality, 16.50, is same for all rows.

Answer:

204

Step-by-step explanation:

To simplify the shape, you can do multiple things. I've opted to shave down both prongs to take it from a 'T' shape to a rectangular prism.

For height of the prongs, take 4 from 6.

6 - 4 = 2

Divide by 2 as there are 2 prongs.

2 / 2 = 1

Remember L * W * H

6 * 3 * 1 = 18

Remember that there are two prongs!

3 + 4 = 7

6 * 7 * 4 = 168

168 + 2(18) = 204

Answer:

x=9/2

Step-by-step explanation:

Let's solve your equation step-by-step.

5x+4(−x−2)=−5x+2(x−1)+12

Step 1: Simplify both sides of the equation.

5x+4(−x−2)=−5x+2(x−1)+12

5x+(4)(−x)+(4)(−2)=−5x+(2)(x)+(2)(−1)+12 (Distribute)

5x+−4x+−8=−5x+2x+−2+12

(5x+−4x)+(−8)=(−5x+2x)+(−2+12) (Combine Like Terms)

x+−8=−3x+10

x−8=−3x+10

Step 2: Add 3x to both sides.

x−8+3x=−3x+10+3x

4x−8=10

Step 3: Add 8 to both sides.

4x−8+8=10+8

4x=18

Step 4: Divide both sides by 4.

4x/4=18/4

x=9/2

Answer:

30 cm

Step-by-step explanation:

let x be the lenght of the two sides of equal lenghts, so the other is x+3

and the perimeter is x+x +x +3

P=3x+3

P=3(x+1)

93=3(x+1)

31=x+1

x=30

so the shorter sides are of 30 centimeters and the longest is 33

Answer:

less than

Step-by-step explanation:

If the normality requirement is not satisfied​ (that is, ​np(1​ - p) is not at least​ 10), then a​ 95% confidence interval about the population proportion will include the population proportion in​ _less than__ 95% of the intervals.

The confidence interval consist of all reasonable values of a population mean. These are value for which the null hypothesis will not be rejected.

So, let assume that If the 95%  confidence interval contains the value for the hypothesized mean, then the sample mean  is reasonably close to the hypothesized mean. The effect of this is that the p- value is going to be greater than 0.05, so we fail to reject the null hypothesis.

On the other hand,

If the 95%  confidence interval do not contains the value for the hypothesized mean, then the sample mean  is far away from the hypothesized mean. The effect of this is that the p- value is going to be lesser than 0.05, so we reject the null hypothesis.

Answer:

The 99% confidence interval for estimating the population mean μ is ($60,112.60, $68087.40).

Step-by-step explanation:

The complete question is:

Salaries of 42 college graduates who took a statistics course in college have a​ mean, [tex]\bar x[/tex] of, $64, 100. Assuming a standard​ deviation, σ of ​$10​,016 construct a ​99% confidence interval for estimating the population mean μ.

Solution:

The (1 - α)% confidence interval for estimating the population mean μ is:

[tex]CI=\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]

The critical value of z for 99% confidence interval is:

[tex]z_{\alpha/2}=z_{0.01/2}=z_{0.005}=2.57[/tex]

Compute the 99% confidence interval for estimating the population mean μ as follows:

[tex]CI=\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]

     [tex]=64100\pm 2.58\times\frac{10016}{\sqrt{42}}\\\\=64100+3987.3961\\\\=(60112.6039, 68087.3961)\\\\\approx (60112.60, 68087.40)[/tex]

Thus, the 99% confidence interval for estimating the population mean μ is ($60,112.60, $68087.40).

Answer:

Divide the diameter by 2.

Step-by-step explanation:

The radius of any circle is always the end to the center.

The diameter is a point of the circle to the opposite side.

This means that the diameter is twice the size of the radius, so to find the radius from the diameter, divide the diameter by 2.

Hope this helped!

Answer:

Divide the diameter by 2. d/2=r

Step-by-step explanation:

If a diameter has been given instead of a radius, you can find the radius by dividing the diameter by 2, for example.

If the diameter was 10, the radius would 10/2=5.

Answer:

1. cups of coffee sold

2.Y = 605.7 - 5.943x

3. -0.952

4. 70.84

Step-by-step explanation:

1. the dependent variable in this question is the cups of coffee sold

2. least square estimation line

Y = a+bx

we have y as the cups of coffee sold

x as temperature.

first we will have to solve for a and then b

∑X = 450

∑Y = 960

∑XY = 61600

∑X² = 35500

∑Y² = 221800

a = ∑y∑x²-∑x∑xy/n∑x²-(∑x)²

a = 960 * 35500-450*61600/6*35500-450²

a = 6360000/10500

= 605.7

b = n∑xy - ∑x∑y/n∑x²-(∑x)²

= 6*61600 - 450*960/6*35500 - 450²

= -5.943

the regression line

Y = a + bx

Y = 605.7 - 5.943x

3. we are to find correlation coefficient

r = n∑xy - ∑x∑y multiplied by√(n∑x²-(∑x)² * (n∑y² - (∑y)²)

= 6*61600 -960*450/√(6*35500 - 450²)*(6*221800 - 960²)

=-62400/√4296600000

= -62400/65548.5

= -0.952

4. we have to predict sales of a 90 degree day fro the regression line

Y = 605.7 - 5.943x

y = 605.7 - 5.943(90)

y = 605.7 - 534.87

= 70.84

Answer:

6.25

Step-by-step explanation:

2.5 *2.5=6.25

Answer:

6.25cm^2.

Step-by-step explanation:

To find the area of a square, you multiply the two sides, 2.5✖️2.5.

This gives the area of 6.25cm^2.

Hope this helped!

Have a nice day:)

The base 1 is matched with height 2, base 2 is matched with height 3 and base 3 is matched with height 1. The base to the corresponding height is matched in the attached figure.

Triangle is the closed shaped polygon which has 3 sides and 3 interior angles. The height of the triangle is the dimension of the elevation from the opposite peak to the length of the base.

Thus, the base 1 is matched with height 2, base 2 is matched with height 3 and base 3 is matched with height 1. The base to the corresponding height is matched in the attached figure.

In the given figure, three triangles is shown with base and height. Here,

Thus, the base 1 is matched with height 2, base 2 is matched with height 3 and base 3 is matched with height 1. The base to the corresponding height is matched in the attached figure.

Learn more about the base and height of the triangle here;

https://brainly.com/question/26043588

#SPJ2

Answer:

b = 10.5

Step-by-step explanation:

2(b-9) = 3

then:

2*b + 2*-9 = 3

2b - 18 = 3

2b = 3 + 18

2b = 21

b = 21/2

b = 10.5

check:

2(10.5 - 9) = 3

2*1.5 = 3

Answer:

X is uniformly distributed.

Step-by-step explanation:

Uniform Distribution:

This is the type of distribution where all outcome of a certain event have equal likeliness of occurrence.

Example of Uniform Distribution is - tossing a coin. The probability of getting a head is the same as the probability of getting a tail. The have equal likeliness of occurrence.

Answer:

Hello the options to your question is missing below are the options

 A) if sample means were obtained for a long series of samples, approximately 95 percent of all sample means would be between 10 and 16 miles

B.the unknown population mean is definitely between 10 and 16 miles

C.if these intervals were constructed for a long series of samples, approximately 95 percent would include the unknown mean commute distance for all Californians

D.the unknown population mean is between 10 and 16 miles with probability .95

Answer : if these intervals were constructed for a long series of samples, approximately 95 percent would include the unknown mean commute distance for all Californians  ( c )

Step-by-step explanation:

95%  confidence

interval = 10 to 16 miles

To have 95% confidence signifies that if these intervals were constructed for a long series of samples, approximately 95 percent would include the unknown mean commute distance for all Californians

confidence interval covers a range of samples/values in the interval and the higher the % of the confidence interval the more precise the interval is,

Answer:

Below

Step-by-step explanation:

● 8 ÷ (3-x)

Dividing by 3-x is like multiplying by 1/(3-x)

● 8 × (1/3-x)

● 8 /(3-x)

Answer:

A.the type 1 error probability is [tex]\mathbf{\alpha = 0.0244 }[/tex]

B. β  = 0.0122

C. β  = 0.0000

Step-by-step explanation:

Given that:

Mean = 100

standard deviation = 2

sample size = 9

The null and the alternative hypothesis can be computed as follows:

[tex]\mathtt{H_o: \mu = 100}[/tex]

[tex]\mathtt{H_1: \mu \neq 100}[/tex]

A. If the acceptance region is defined as [tex]98.5 < \overline x > 101.5[/tex] , find the type I error probability [tex]\alpha[/tex] .

Assuming the critical region lies within [tex]\overline x < 98.5[/tex] or [tex]\overline x > 101.5[/tex], for a type 1 error to take place, then the sample average x will be within the critical region when the true mean heat evolved is [tex]\mu = 100[/tex]

∴

[tex]\mathtt{\alpha = P( type \ 1 \ error ) = P( reject \ H_o)}[/tex]

[tex]\mathtt{\alpha = P( \overline x < 98.5 ) + P( \overline x > 101.5 )}[/tex]

when  [tex]\mu = 100[/tex]

[tex]\mathtt{\alpha = P \begin {pmatrix} \dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} < \dfrac{\overline 98.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} + \begin {pmatrix}P(\dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} > \dfrac{101.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} }[/tex]

[tex]\mathtt{\alpha = P ( Z < \dfrac{-1.5}{\dfrac{2}{3}} ) + P(Z > \dfrac{1.5}{\dfrac{2}{3}}) }[/tex]

[tex]\mathtt{\alpha = P ( Z <-2.25 ) + P(Z > 2.25) }[/tex]

[tex]\mathtt{\alpha = P ( Z <-2.25 ) +( 1- P(Z < 2.25) })[/tex]

From the standard normal distribution tables

[tex]\mathtt{\alpha = 0.0122+( 1- 0.9878) })[/tex]

[tex]\mathtt{\alpha = 0.0122+( 0.0122) })[/tex]

[tex]\mathbf{\alpha = 0.0244 }[/tex]

Thus, the type 1 error probability is [tex]\mathbf{\alpha = 0.0244 }[/tex]

B. Find beta for the case where the true mean heat evolved is 103.

The probability of type II error is represented by β. Type II error implies that we fail to reject null hypothesis [tex]\mathtt{H_o}[/tex]

Thus;

β = P( type II error) - P( fail to reject [tex]\mathtt{H_o}[/tex] )

∴

[tex]\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }[/tex]

Given that [tex]\mu = 103[/tex]

[tex]\mathtt{\beta = P( \dfrac{98.5 -103}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-103}{\dfrac{2}{\sqrt{9}}}) }[/tex]

[tex]\mathtt{\beta = P( \dfrac{-4.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-1.5}{\dfrac{2}{3}}) }[/tex]

[tex]\mathtt{\beta = P(-6.75 \leq Z \leq -2.25) }[/tex]

[tex]\mathtt{\beta = P(z< -2.25) - P(z < -6.75 )}[/tex]

From standard normal distribution table

β  = 0.0122 - 0.0000

β  = 0.0122

C. Find beta for the case where the true mean heat evolved is 105. This value of beta is smaller than the one found in part (b) above. Why?

[tex]\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }[/tex]

Given that [tex]\mu = 105[/tex]

[tex]\mathtt{\beta = P( \dfrac{98.5 -105}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-105}{\dfrac{2}{\sqrt{9}}}) }[/tex]

[tex]\mathtt{\beta = P( \dfrac{-6.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-3.5}{\dfrac{2}{3}}) }[/tex]

[tex]\mathtt{\beta = P(-9.75 \leq Z \leq -5.25) }[/tex]

[tex]\mathtt{\beta = P(z< -5.25) - P(z < -9.75 )}[/tex]

From standard normal distribution table

β  = 0.0000 - 0.0000

β  = 0.0000

The reason why the value of beta is smaller here is that since the difference between the value for the true mean and the hypothesized value increases, the probability of type II error decreases.

Answer:

[tex](\frac{3}{4})*(\frac{8}{9})*(\frac{15}{16})*(\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49})*(\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}) = \frac{11}{20}[/tex]

Step-by-step explanation:

Given

[tex](\frac{3}{4})*(\frac{8}{9})*(\frac{15}{16})*(\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49})*(\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100})[/tex]

Required

Simplify

For clarity, group the expression in threes

[tex]((\frac{3}{4})*(\frac{8}{9})*(\frac{15}{16}))*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

Evaluate the first group [Divide 8 by 4]

[tex]((\frac{3}{1})*(\frac{2}{9})*(\frac{15}{16}))*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[Divide 9 by 3]

[tex]((\frac{1}{1})*(\frac{2}{3})*(\frac{15}{16}))*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[tex]((\frac{2}{3})*(\frac{15}{16}))*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[Divide 15 by 3]

[tex]((\frac{2}{1})*(\frac{5}{16}))*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[Divide 16 by 2]

[tex]((\frac{1}{1})*(\frac{5}{8}))*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[tex](\frac{5}{8})*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

Evaluate the second group [Divide 35 and 25 by 5]

[tex](\frac{5}{8})*((\frac{24}{5})*(\frac{7}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[Divide 49 by 7]

[tex](\frac{5}{8})*((\frac{24}{5})*(\frac{1}{3})*(\frac{4}{7}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[Divide 24 by 3]

[tex](\frac{5}{8})*((\frac{8}{5})*(\frac{1}{1})*(\frac{4}{7}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[tex](\frac{5}{8})*((\frac{8}{5})*(\frac{4}{7}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

Merge the first and second group

[tex]((\frac{5}{8})*(\frac{8}{5})*(\frac{4}{7}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[tex](1*(\frac{4}{7}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[tex](\frac{4}{7})*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

Evaluate the last group [Divide 99 by 9]

[tex](\frac{4}{7})*((\frac{63}{64})*(\frac{80}{9})*(\frac{11}{100}))[/tex]

[Divide 63 by 9]

[tex](\frac{4}{7})*((\frac{7}{64})*(\frac{80}{1})*(\frac{11}{100}))[/tex]

[Divide 64 and 80 by 8]

[tex](\frac{4}{7})*((\frac{7}{8})*(\frac{10}{1})*(\frac{11}{100}))[/tex]

[Divide 10 and 4 by 2]

[tex](\frac{4}{7})*((\frac{7}{4})*(\frac{5}{1})*(\frac{11}{100}))[/tex]

[Divide 100 by 5]

[tex](\frac{4}{7})*((\frac{7}{4})*(\frac{1}{1})*(\frac{11}{20}))[/tex]

[tex](\frac{4}{7})*((\frac{7}{4})*(\frac{11}{20}))[/tex]

[tex](\frac{4}{7})*(\frac{7}{4})*(\frac{11}{20})[/tex]

[tex]1*(\frac{11}{20})[/tex]

[tex]\frac{11}{20}[/tex]

Hence;

[tex](\frac{3}{4})*(\frac{8}{9})*(\frac{15}{16})*(\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49})*(\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}) = \frac{11}{20}[/tex]

Answer:

300.05 miles

Step-by-step explanation:

initial fee= $39.99

final bill = $ 100

cost =$ 0.20 per mile

remaining amount = $ 60.01

solution,

she drive = remaining amount / cost

=60.01/0.20

=300.05 miles

Answer:

500 miles

Step-by-step explanation:

Let us use cross multiplication to find the unknown amount.

Given:

1) Cost for 1 mile=$0.20

2)Cost for x miles=$100

Solution:

No of miles                             Cost

1) 1                                             $0.20

2)x                                             $100

By cross multiplying,

100 x 1= 0.20x

x=100/0.20

x=500 miles

Thank you!

[tex]_6C_3=\dfrac{6!}{3!3!}=\dfrac{4\cdot5\cdot6}{2\cdot3}=20[/tex]

[tex](f\circ g)(2)=\dfrac{1}{4\cdot2+9}=\dfrac{1}{17}\\\\(f+g)(2)=\dfrac{1}{2}+4\cdot2+9=\dfrac{1}{2}+17=\dfrac{1}{2}+\dfrac{34}{2}=\dfrac{35}{2}[/tex]