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Setler [38]
1 year ago
14

As Evin is driving her car, she notices that after 1 hour her gas tank has 7.25 gallons left and after 4 hours of driving, it ha

s 3.5 gallons of gas left in it. What is the rate at which her car is using gas.
Mathematics
1 answer:
kompoz [17]1 year ago
5 0

Answer:

7.25 per hour

Step-by-step explanation:

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6 tractors take 10days to collect the harvest. How long would it take 18 tractors to do the same amount of work?
Nutka1998 [239]
This is a question of proportionality. However, it is not direct proportionality as it is expected that as the number of tractors increases, the work is finished faster as opposed to fewer number of tractors. This is referred to as inverse proportionality.

Therefore;
6 tractors ----- 10 days
18 tractors --- x days

Then,
x = (6*10)/18 = 10/3 days = 3 days, and 8 hours.

This means that 18 tractors will take 3 days and 8 hours to collect the same harvest.
6 0
2 years ago
which shows the correct substitution of the values a, b, and c from the equation 0 = 4x2 2x – 1 into the quadratic formula below
irinina [24]

0 = 4x^2+2x- 1

Quadratic formula is

x = \frac{-b+-\sqrt{b^2-4ac}}{2a}

'a' is the coefficient of x^2 = 4

'b' is the coefficient of x = 2

'c' is the constant = -1

Now we plug in all the values in quadratic formula

x = \frac{-b+-\sqrt{b^2-4ac}}{2a}

x = \frac{-2+-\sqrt{2^2-4(4)(-1)}}{2(4)}

x = \frac{-2+-\sqrt{18}}{8}

The above one is the substitution of values of a,b,c in quadratic formula.

5 0
2 years ago
Read 2 more answers
If r(x) = 3x – 1 and s(x) = 2x + 1, which expression is equivalent to (StartFraction r Over s EndFraction) (6)?
Airida [17]

Answer:

\dfrac{r}{s}(6)=\dfrac{3(6)-1}{2(6)+1}

Step-by-step explanation:

We know that for any two function f(x) and g(x) ,

\dfrac{f}{g}(x)=\dfrac{f(x)}{g(x)}

Given functions : r(x)=3x-1  and s(x)=2x+1

Then, \dfrac{r}{s}(x)=\dfrac{r(x)}{s(x)}

\Rightarrow\ \dfrac{r}{s}(x)=\dfrac{3x-1}{2x+1}

At x= 6 , we get

\dfrac{r}{s}(6)=\dfrac{3(6)-1}{2(6)+1}

The , The expression is equivalent to \dfrac{r}{s}(6)=\dfrac{3(6)-1}{2(6)+1}

When we further simplify it , we get \dfrac{r}{s}(6)=\dfrac{18-1}{12+1}=\dfrac{17}{13}

5 0
2 years ago
In the diagram of circle o, what is the measure of ABC?<br> 27°<br> 54°<br> 108°<br> 120°
Vlad1618 [11]

<u>Given</u>:

Given that a circle O with two tangents BA and BC.

The major arc AC is 234°

The minor arc AC is 126°

We need to determine the measure of ∠ABC

<u>Measure of ∠ABC:</u>

We know the property that, "if a tangent and a secant, two tangents or two secants intersect in the interior of the circle, then the measure of angle formed is one half the difference of the measures of the intercepted arcs."

Hence, applying the above property, we have;

\angle ABC=\frac{1}{2}( major \widehat{AC} - minor \widehat{AC})

Substituting the values, we get;

\angle ABC=\frac{1}{2}( 234^{\circ} -126^{\circ})

\angle ABC=\frac{1}{2}( 108^{\circ})

\angle ABC=54^{\circ}

Thus, the measure of ∠ABC is 54°

Hence, Option b is the correct answer.

5 0
2 years ago
In a study by Peter D. Hart Research Associates for the Nasdaq Stock Market, it was determined that 20% of all stock investors a
solong [7]

Answer:

The answer to the questions are;

a. The probability that exactly six are retired people is 0.1633459.

b. The probability that 9 or more are retired people is 0.04677.

c. The number of expected retired people in a random sample of 25 stock investors is 0.179705.

d. In a random sample of 20 U.S. adults the probability that exactly eight adults invested in mutual funds is 0.179705.

e. The probability that fewer than five adults invested in mutual funds out of a random sample of 20 U.S. adults is 5.095×10⁻².

f. The probability that exactly one adult invested in mutual funds out of a random sample of 20 U.S. adults is 4.87×10⁻⁴.

g. The probability that 13 or more adults out of a random sample of 20 U.S. adults invested in mutual funds is 2.103×10⁻².

h. 4, 1, 13. They tend to converge to the probability of the expected value.

Step-by-step explanation:

To solve the question, we note that the binomial distribution probability mass function is given by

f(n,p,x) = \left(\begin{array}{c}n&x&\end{array}\right) × pˣ × (1-p)ⁿ⁻ˣ = ₙCₓ × pˣ × (1-p)ⁿ⁻ˣ

Also the mean of the Binomial distribution is given by

Mean = μ = n·p = 25 × 0.2 = 5

Variance = σ² = n·p·(1-p) = 25 × 0.2 × (1-0.2) = 4

Standard Deviation = σ = \sqrt{n*p*(1-p)}

Since the variance < 5 the normal distribution approximation is not appropriate to sole the question

We proceed as follows

a. The probability that exactly six are retired people is given by

f(25, 0.2, 6) = ₂₅C₆ × 0.2⁶ × (1-0.2)¹⁹ = 0.1633459.

b. The probability that 9 or more are retired people is given by

P(x>9) = 1- P(x≤8) = 1- ∑f(25, 0.2, x where x = 0 →8)

Therefore we have

f(25, 0.2, 0) = ₂₅C₀ × 0.2⁰ × (1-0.2)²⁵ = 3.78×10⁻³

f(25, 0.2, 1) = ₂₅C₁ × 0.2¹ × (1-0.2)²⁴ = 2.36 ×10⁻²

f(25, 0.2, 2) = ₂₅C₂ × 0.2² × (1-0.2)²³ = 7.08×10⁻²

f(25, 0.2, 3) = ₂₅C₃ × 0.2³ × (1-0.2)²² = 0.135768

f(25, 0.2, 4) = ₂₅C₄ × 0.2⁴ × (1-0.2)²¹ = 0.1866811

f(25, 0.2, 5) = ₂₅C₅ × 0.2⁵ × (1-0.2)²⁰ = 0.1960151

f(25, 0.2, 6) = ₂₅C₆ × 0.2⁶ × (1-0.2)¹⁹ = 0.1633459

f(25, 0.2, 7) = ₂₅C₇ × 0.2⁷ × (1-0.2)¹⁸ = 0.11084187

f(25, 0.2, 8) = ₂₅C₈ × 0.2⁸ × (1-0.2)¹⁷ = 6.235×10⁻²

∑f(25, 0.2, x where x = 0 →8) = 0.953226

and P(x>9) = 1- P(x≤8)  = 1 - 0.953226 = 0.04677.

c. The number of expected retired people in a random sample of 25 stock investors is given by

Proportion of retired stock investors × Sample count

= 0.2 × 25 = 5.

d. In a random sample of 20 U.S. adults the probability that exactly eight adults invested in mutual funds is given by

Here we have p = 0.4 and n·p = 8 while n·p·q = 4.8 which is < 5 so we have

f(20, 0.4, 8) = ₂₀C₈ × 0.4⁸ × (1-0.4)¹² = 0.179705.

e. The probability that fewer than five adults invested in mutual funds out of a random sample of 20 U.S. adults is

P(x<5) = ∑f(20, 0.4, x, where x = 0 →4)

Which gives

f(20, 0.4, 0) = ₂₀C₀ × 0.4⁰ × (1-0.4)²⁰ = 3.66×10⁻⁵

f(20, 0.4, 1) = ₂₀C₁ × 0.4¹ × (1-0.4)¹⁹ = 4.87×10⁻⁴

f(20, 0.4, 2) = ₂₀C₂ × 0.4² × (1-0.4)¹⁸ = 3.09×10⁻³

f(20, 0.4, 3) = ₂₀C₃ × 0.4³ × (1-0.4)¹⁷ = 1.235×10⁻²

f(20, 0.4, 4) = ₂₀C₄ × 0.4⁴ × (1-0.4)¹⁶ = 3.499×10⁻²

Therefore P(x<5) = 5.095×10⁻².

f. The probability that exactly one adult invested in mutual funds out of a random sample of 20 U.S. adults is given by

f(20, 0.4, 1) = ₂₀C₁ × 0.2¹ × (1-0.2)¹⁹ = 4.87×10⁻⁴.

g. The probability that 13 or more adults out of a random sample of 20 U.S. adults invested in mutual funds is

P(x≥13) =  ∑f(20, 0.4, x where x = 13 →20) we have

f(20, 0.4, 13) = ₂₀C₁₃ × 0.4¹³ × (1-0.4)⁷ = 1.46×10⁻²

f(20, 0.4, 14) = ₂₀C₁₄ × 0.4¹⁴ × (1-0.4)⁶ = 4.85×10⁻³

f(20, 0.4, 15) = ₂₀C₁₅ × 0.4¹⁵ × (1-0.4)⁵ = 1.29×10⁻³

f(20, 0.4, 16) = ₂₀C₁₆ × 0.4¹⁶ × (1-0.4)⁴ = 2.697×10⁻⁴

f(20, 0.4, 17) = ₂₀C₁₇ × 0.4¹⁷ × (1-0.4)³ = 4.23×10⁻⁵

f(20, 0.4, 18) = ₂₀C₁₈ × 0.4¹⁸ × (1-0.4)² = 4.70×10⁻⁶

f(20, 0.4, 19) = ₂₀C₁₉ × 0.4¹⁹ × (1-0.4)⁴ = 3.299×10⁻⁷

f(20, 0.4, 20) = ₂₀C₂₀ × 0.4²⁰ × (1-0.4)⁰ = 1.0995×10⁻⁸

P(x≥13) = 2.103×10⁻².

h.  For part e we have exactly 4 with a probability of 3.499×10⁻²

For part f the  probability for the one adult is 4.87×10⁻⁴

For part g, we have exactly 13 with a probability of 1.46×10⁻²

The expected number is 8 towards which the exact numbers with the highest probabilities in parts e to g are converging.

5 0
2 years ago
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