Which is the equation of the line for the points in the given table

Answer:

1.9375.

Step-by-step explanation:

To solve this, we must use PEMDAS.

The first things we take care of are parentheses and exponents.

Since there are no parentheses, we do exponents.

4^-1+8^-1÷1/2/3^3​

= [tex]\frac{1}{4} +\frac{1}{8} / 1/ 2/ 27[/tex]

= 1/4 + (1/8) / 1 * (27 / 2)

= 1/4 + (27 / 8) / 2

= 1/4 + (27 / 8) * (1 / 2)

= 1/4 + (27 / 16)

= 4 / 16 + 27 / 16

= 31 / 16

= 1.9375.

Hope this helps!

Answer:

Step-by-step explanation:

Together, they have 4 times what Linda has, plus $10. So, Linda has 1/4 of $60 = $15.

  Linda has $15

  Reuben has $25 . . . . . . $10 more than Linda

  Manuel has $30 . . . . . . twice what Linda has

Answer:

A sample of at least 541 is needed if we wish to be 98% confident that our sample mean will be within 4 hours of the true mean.

Step-by-step explanation:

We are given that an electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed with a standard deviation of 40 hours.

We have to find a sample such that we are 98% confident that our sample mean will be within 4 hours of the true mean.

As we know that the Margin of error formula is given by;

The margin of error =  [tex]Z_(_\frac{\alpha}{2}_) \times \frac{\sigma}{\sqrt{n} }[/tex]

where, [tex]\sigma[/tex] = standard deviation = 40 hours

            n = sample size

            [tex]\alpha[/tex] = level of significance = 1 - 0.98 = 0.02 or 2%

Now, the critical value of z at ([tex]\frac{0.02}{2}[/tex] = 1%) level of significance n the z table is given as 2.3263.

So, the margin of error =  [tex]Z_(_\frac{\alpha}{2}_) \times \frac{\sigma}{\sqrt{n} }[/tex]

                 [tex]4=2.3263 \times \frac{40}{\sqrt{n} }[/tex]

                 [tex]\sqrt{n}= \frac{40 \times 2.3263}{ 4}[/tex]

                  [tex]\sqrt{n}=23.26[/tex]

                   n = [tex]23.26^{2}[/tex] = 541.03 ≈ 541

Hence, a sample of at least 541 is needed if we wish to be 98% confident that our sample mean will be within 4 hours of the true mean.

Answer:

z(c)  = - 1,64

We reject the null hypothesis

Step-by-step explanation:

We need to solve a proportion test ( one tail-test ) left test

Normal distribution

p₀ = 63 %

proportion size  p = 51 %

sample size  n = 114

At 5% level of significance   α = 0,05, and with this value we find in z- table z score of z(c) = 1,64  ( critical value )

Test of proportion:

H₀     Null Hypothesis                        p = p₀

Hₐ    Alternate Hypothesis                p < p₀

We now compute z(s) as:

z(s) =  ( p - p₀ ) / √ p₀q₀/n

z(s) =( 0,51 - 0,63) / √0,63*0,37/114

z(s) =  - 0,12 / 0,045

z(s) = - 2,66

We compare z(s) and z(c)

z(s) < z(c)      - 2,66 < -1,64

Therefore as z(s) < z(c)  z(s) is in the rejection zone we reject the null hypothesis

Answer:

The variance of the data is 15.

σ² = 15

Step-by-step explanation:

The mean is given as

X = 8

х        |    (x - X)    |    (x - X) ²

12       |        4         |    16

9        |        1         |     1    

11        |        3         |    9

5       |        -3        |    9

3       |        -5        |    25

The variance is given by

[tex]\sigma^2 = \frac{1}{n-1} \sum (x - X)^2[/tex]

[tex]\sigma^2 = \frac{1}{5 - 1} (16 + 1 + 9 + 9 +25) \\\\\sigma^2 = \frac{1}{4} ( 16 + 1 + 9 + 9 +25) \\\\\sigma^2 = \frac{1}{4} (60) \\\\\sigma^2 = 15[/tex]

Therefore, the variance of the data is 15.

Answer:

The answer is below

Step-by-step explanation:

A sugar refinery has three processing plants, all receiving raw sugar in bulk. The amount of raw sugar (in tons) that one plant can process in one day can be modelled using an exponential distribution with mean of 4 tons for each of three plants. If each plant operates independently,a.Find the probability that any given plant processes more than 5 tons of raw sugar on a given day.b.Find the probability that exactly two of the three plants process more than 5 tons of raw sugar on a given day.c.How much raw sugar should be stocked for the plant each day so that the chance of running out of the raw sugar is only 0.05?

Answer: The mean (μ) of the plants is 4 tons. The probability density function of an exponential distribution is given by:

[tex]f(x)=\lambda e^{-\lambda x}\\But\ \lambda= 1/\mu=1/4 = 0.25\\Therefore:\\f(x)=0.25e^{-0.25x}\\[/tex]

a) P(x > 5) = [tex]\int\limits^\infty_5 {f(x)} \, dx =\int\limits^\infty_5 {0.25e^{-0.25x}} \, dx =-e^{-0.25x}|^\infty_5=e^{-1.25}=0.2865[/tex]

b) Probability that exactly two of the three plants process more than 5 tons of raw sugar on a given day can be solved when considered as a binomial.

That is P(2 of the three plant use more than five tons) = C(3,2) × [P(x > 5)]² × (1-P(x > 5)) = 3(0.2865²)(1-0.2865) = 0.1757

c) Let b be the amount of raw sugar should be stocked for the plant each day.

P(x > a) = [tex]\int\limits^\infty_a {f(x)} \, dx =\int\limits^\infty_a {0.25e^{-0.25x}} \, dx =-e^{-0.25x}|^\infty_a=e^{-0.25a}[/tex]

But P(x > a) = 0.05

Therefore:

[tex]e^{-0.25a}=0.05\\ln[e^{-0.25a}]=ln(0.05)\\-0.25a=-2.9957\\a=11.98[/tex]

a  ≅ 12

Answer:

a[tex]\geq[/tex]-3

Step-by-step explanation:

Answer:

-3  ≤  a

Step-by-step explanation:

6≥ -6(a+2)

Divide each side by -6, remembering to flip the inequality

6/-6 ≤ -6/-6(a+2)

-1 ≤ (a+2)

Subtract 2 from each side

-1 -2  ≤  a+2-2

-3  ≤  a

Answer:

In Table C, y vary inversely with x.

1×18 = 18

2×9 = 18

3×6 = 18

18 = 18 = 18

Step-by-step explanation:

We are given four tables and asked to find out in which table y vary inversely with x.

We know that an inverse relation has a form given by

y = k/x

xy = k

where k must be a constant

Table A:

x     |      y

1     |      3

2     |     9

3     |    27

1×3 = 3

2×9 = 18

3×27 = 81

3 ≠ 18 ≠ 81

Hence y does not vary inversely with x.

Table B:

x     |      y

1     |     -5

2     |     5

3     |    15

1×-5 = -5

2×5 = 10

3×15 = 45

-5 ≠ 10 ≠ 45

Hence y does not vary inversely with x.

Table C:

x     |      y

1     |      18

2     |     9

3     |     6

1×18 = 18

2×9 = 18

3×6 = 18

18 = 18 = 18

Hence y vary inversely with x.

Table D:

x     |      y

1     |      4

2     |     8

3     |    12

1×4 = 4

2×8 = 16

3×12 = 36

4 ≠ 16 ≠ 36

Hence y does not vary inversely with x.

Answer:

The measure of each angle:

152.8°   and     27.2°

Step-by-step explanation:

Supplementary angles sum 180°

then:

a + b = 180°

a - b = 125.6°

then:

a = 180 - b

a = 125.6 + b

180 - b = 125.6 + b

180 - 125.6 = b + b

54.4 = 2b

b = 54.4/2

b = 27.2°

a = 180 - b

a = 180 - 27.2

a = 152.8°

Check:

152.8 + 27.2 = 180°

Answers:

Step-by-step explanation:

Let x and y be the measures of each angle.

x + y = 180°

x - y = 125.6°

180 - 125.6 = 54.4

Now we divide 54.4 evenly to get y.

y = 27.2°

To get x, we substitute y into the equation.

x = 27.2 + 125.6

x = 152.8°

To check, we plug these in to see if they equal 180°.

27.2 + 152.8 = 180° ✅

Answer:

a=6, b=5.5

Step-by-step explanation:

By looking at the sides of the triangles it can easily be seen that some of the sides match up. Side b is similar to the side of 11 and same with side a and the side of 3. Since one side is 16 and the other side on the smaller triangle is 8, the bigger triangle is twice as large than the smaller one. So 3 x 2 = 6 and 11 / 2 = 5.5

Answer:

Hey there!

The common ratio is 5, because you multiply by 5 to get from one term to the next.

Hope this helps :)

Answer:

5

Step-by-step explanation:

To find the common ratio take the second term and divide by the first term

55/11 = 5

The common ratio would be 5

Answer:

Step-by-step explanation:

The class interval is simply the difference between the lower or upper class boundary or limit  of a class and the lower or upper class boundary or limit of the next class.

In this case for the class

$4 up to $7 18 and

$7 up to $10 36

The lower class boundary of the first class is $4 and the lower class boundary of the second class is $7

The area of the surface above the region R is 4096π square units.

Given that:

The function: [tex]f(x, y) = 64 + x^2 - y^2[/tex]

The region R is the disk with a radius of 8 units [tex]x^2 + y^2 \le 64[/tex].

To find the area of the surface given by z = f(x, y) that lies above the region R, to calculate the double integral over the region R of the function f(x, y) with respect to dA.

The integral for the area is given by:

[tex]Area = \int\int_R f(x, y) dA[/tex]

To evaluate this integral, we need to set up the limits of integration for x and y over the region R, which is the disk cantered at the origin with a radius of 8 units.

Using polar coordinates, we can parameterize the region R as follows:

x = rcos(θ)

y = rsin(θ)

where r goes from 0 to 8, and θ goes from 0 to 2π.

Now, rewrite the integral in polar coordinates:

[tex]Area =\int\int_R f(x, y) dA\\Area = \int_0 ^{2\pi} \int_0^8(64 + r^2cos^2(\theta) - r^2sin^2(\theta)) \times r dr d \theta[/tex]

Now, we can integrate with respect to r first and then with respect to θ:

[tex]Area = \int_0^{2\pi} \int_0^8] (64r + r^3cos^2(\theta) - r^3sin^2(\theta)) dr d \theta[/tex]

Integrate with respect to r:

[tex]Area = \int_0^{2\pi}[(32r^2 + (1/4)r^4cos^2(\theta) - (1/4)r^4sin^2(\theta))]_0^8 d \theta\\Area = \int_0^{2\pi} (2048 + 256cos^2(\theta) - 256sin^2(\theta)) d \theta[/tex]

Now, we can integrate with respect to θ:

[tex]Area = [2048\theta + 128(sin(2\theta) + \theta)]_0 ^{2\pi}[/tex]

Area = 2048(2π) + 128(sin(4π) + 2π) - (2048(0) + 128(sin(0) + 0))

Area = 4096π + 128(0) - 0

Area = 4096π square units

So, the area of the surface above the region R is 4096π square units.

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Answer:

49.923 mph

Step-by-step explanation:

we know that the car traveled 133 miles in h hours at an average speed of x mph.

That is, xh = 133.

We can also write this in terms of hours driven: h = 133/x.

If x was 30 mph faster, then h would be one hour less.

That is, (x + 30)(h - 1) = 133, or h - 1 = 133/(x + 30).

We can rewrite the latter equation as h = 133/(x + 30) + 1

We can then make a system of equations using the formulas in terms of h to find x:

h = 133/x = 133/(x + 30) + 1

133/x = 133/(x + 30) + (x + 30)/(x + 30)

133/x = (133 + x + 30)/(x + 30)

133 = x*(133 + x + 30)/(x + 30)

133*(x + 30) =  x*(133 + x + 30)

133x + 3990 = 133x + x^2 + 30x

3990 = x^2 + 30x

x^2 + 30x - 3990 = 0

Using the quadratic formula:

x = [-b ± √(b^2 - 4ac)]/2a  

= [-30 ± √(30^2 - 4*1*(-3990))]/2(1)  

= [-30 ± √(900 + 15,960)]/2

= [-30 ± √(16,860)]/2

= [-30 ± 129.846]/2

= 99.846/2  -----------  x is miles per hour, and a negative value of x is neglected, so we'll use the positive value only)

= 49.923

Check if the answer is correct:

h = 133/49.923 = 2.664, so the car took 2.664 hours to drive 133 miles at an average speed of 49.923 mph.

If the car went 30 mph faster on average, then h = 133/(49.923 + 30) = 133/79.923 = 1.664, and 2.664 - 1 = 1.664.

Thus, we have confirmed that a car driving 133 miles at about 49.923 mph would have arrive precisely one hour earlier by going 30 mph faster

Answer:

11/36

Step-by-step explanation:

Find the probability that neither dice shows a 5 (also means the dice can show any number except 5- where there are 5 possible choices out of 6):

= 5/6 x 5/6

=25/36

If we subtract the probability that neither dice shows a 5, we can obtain the probability that at least 1 dice shows a 5- (either one of them is 5, or both of them is 5)

1- 25/36

=11/36

Answer:

To verify the Cauchy-Bunyakovsky-Schwarz Inequality, (p,q) must be less than (or equal to) ||p|| • ||q||

(1,1,1) is not equal to (-10,5)

Step-by-step explanation:

a°b° + a^1b^1 + a^2b^2 < 5x (-2x^2 + 1)

Any algebra raised to the power of zero is equal to 1.

a°b° = 1 × 1 = 1

1 + ab + a^2b^2 < -10x^3 + 5x

The vectors:

(1,1,1) < (-10,5)

This verifies the Cauchy-Schwarz Inequality

Triangle Inequality states that for any triangle, the sum of the lengths of two sides must be greater than or equal to the length of the third side.

Answer:

1-{0.2(m-3)+¼}=0

1{0.2m-0.6+¼}=0

1-{(0.8m-2.4+1)/4}=0

1-(0.8m-1.4)/4=0

lcm

(4-0.8m-1.4)/4=0

(2.6-0.8m)/4=0

cross multiply

2.6-0.8m=0

m=2.6/0.8

m=3.25

The solution of the expression are,

⇒ m = 3.25

Mathematical expression is defined as the collection of the numbers variables and functions by using operations like addition, subtraction, multiplication, and division.

Given that;

Expression is,

⇒ 1 - | 0.2 (m - 3) + 1/4 | = 0

Now, We can simplify as;

⇒ 1 - | 0.2m - 0.6 + 1/4| = 0

⇒ 1 - |0.2m - 0.6 + 0.25| = 0

⇒ 1 - |0.2m - 0.35| = 0

⇒ 1 = 0.2m + 0.35

⇒ 1 - 0.35 = 0.2m

⇒ 0.2m = 0.65

⇒ m = 3.25

Thus, The solution of the expression are,

⇒ m = 3.25

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I hope this will help uh.....

Answer:

the probability that one parachute of the  five parachute is damaged is 0.156

Step-by-step explanation:

From the given information;

Let consider X to be the altitude above the  ground that a parachute opens

Then; we can posit that the probability that the parachute is damaged is:

P(X ≤ 100 )

Given that the population mean μ = 155

the standard deviation σ = 30

Then;

[tex]P(X \leq 100 ) = ( \dfrac{X- \mu}{\sigma} \leq \dfrac{100- \mu}{\sigma})[/tex]

[tex]P(X \leq 100 ) = ( \dfrac{X- 155}{30} \leq \dfrac{100- 155}{30})[/tex]

[tex]P(X \leq 100 ) = (Z \leq \dfrac{- 55}{30})[/tex]

[tex]P(X \leq 100 ) = (Z \leq -1.8333)[/tex]

[tex]P(X \leq 100 ) = \Phi( -1.8333)[/tex]

From standard normal tables

[tex]P(X \leq 100 ) = 0.0334[/tex]

Hence; the probability of the given parachute damaged is 0.0334

Let consider Q to be the dropped parachute

Given that the number of parachute be n= 5

The probability that the parachute opens in each trail be  p = 0.0334

Now; the random variable Q follows the binomial distribution with parameters n= 5 and p = 0.0334

The probability mass function is:

Q [tex]\sim[/tex] B(5, 0.0334)

Similarly; the event that one parachute is damaged is :

Q ≥ 1

P( Q ≥ 1 ) = 1 - P( Q < 1 )

P( Q ≥ 1 ) = 1 - P( Y = 0 )

P( Q ≥ 1 ) = 1 - b(0;5; 0.0334 )

P( Q ≥ 1 ) = [tex]1 -(^5_0)* (0.0334)^0*(1-0.0334)^5[/tex]

P( Q ≥ 1 ) = [tex]1 -( \dfrac{5!}{(5-0)!}) * (0.0334)^0*(1-0.0334)^5[/tex]

P( Q ≥ 1 ) = 1 -  0.8437891838

P( Q ≥ 1 ) = 0.1562108162

P( Q ≥ 1 ) [tex]\approx[/tex] 0.156

Therefore; the probability that one parachute of the  five parachute is damaged is 0.156

Answer:

The required probability = 0.144

Step-by-step explanation:

Since the probability of making money is 60%, then the probability of losing money will be 100-60% = 40%

Now the probability we want to calculate is the probability of making money in the first two days and losing money on the third day.

That would be;

P(making money) * P(making money) * P(losing money)

Kindly recollect;

P(making money) = 60% = 60/100 = 0.6

P(losing money) = 40% = 40/100 = 0.4

The probability we want to calculate is thus;

0.6 * 0.6 * 0.4 = 0.144

Answer:

Step-by-step explanation:

From the summary of the given data;

After the second dose, 137 of 452 subjects in the experimental group (group 1) experienced drowsiness as a side effect.

Let consider [tex]p_1[/tex] to be the probability of those that experience the drowsiness in group 1

[tex]p_1[/tex] = [tex]\dfrac{137}{452}[/tex]

[tex]p_1[/tex] = 0.3031

After the second dose, 31 of 99 subjects in the control group (group 2) experienced drowsiness as a side effect.

Let consider [tex]p_2[/tex] to be the probability of those that experience the drowsiness in group 1

[tex]p_2[/tex] = [tex]\dfrac{31}{99}[/tex]

[tex]p_2[/tex] = 0.3131

The objective is to be able to determine if the evidence suggest that a lower proportion of subjects in group 1 experienced drowsiness as a side effect than subjects in group 2 at the α=0.05 level of significance.

In order to do that; we have to state the null and alternative hypothesis; carry out our test statistics and make conclusion based on it.

So; the null and the  alternative hypothesis can be computed as:

[tex]H_o :p_1 =p_2[/tex]

[tex]H_a= p_1<p_2[/tex]

The test statistics is computed as follows:

[tex]Z = \dfrac{p_1-p_2}{\sqrt{p_1 *\dfrac{1-p_1}{n_1} +p_2 *\dfrac{1-p_2}{n_2}} }[/tex]

[tex]Z = \dfrac{0.3031-0.3131}{\sqrt{0.3031 *\dfrac{1-0.3031}{452} +0.3131 *\dfrac{1-0.3131}{99}} }[/tex]

[tex]Z = \dfrac{-0.01}{\sqrt{0.3031 *\dfrac{0.6969}{452} +0.3131 *\dfrac{0.6869}{99}} }[/tex]

[tex]Z = \dfrac{-0.01}{\sqrt{0.3031 *0.0015418 +0.3131 *0.0069384} }[/tex]

[tex]Z = \dfrac{-0.01}{\sqrt{4.6731958*10^{-4}+0.00217241304} }[/tex]

[tex]Z = \dfrac{-0.01}{0.051378 }[/tex]

Z = - 0.1946

At the level of significance ∝ = 0.05

From the standard normal table;

the critical value for Z(0.05) = -1.645

Decision Rule: Reject the null hypothesis if Z-value is lesser than the critical value.

Conclusion: We do not reject the null hypothesis because the Z value is greater than the critical value. Therefore, we cannot conclude that a lower proportion of subjects in group 1 experienced drowsiness as a side effect than subjects in group 2

Answer:

a. uses a related sample - repeated measures

c. uses a related sample (matched subjects)

Step-by-step explanation:

A) You are interested in a potential treatment for compulsive hoarding. You treat a group of 50 compulsive hoarders and compare their scores on the Hoarding Severity scale before and after the treatment. You want to see if the treatment will lead to lower hoarding scores.  

The design described uses a related sample - repeated measures because the scores were compared on the Hoarding Severity scale before and after the treatment.

B) John Caccioppo was interested in possible mechanisms by which loneliness may have deterious effects of health. He compared the sleep quality of a random sample of lonely people to the sleep quality of a random sample of nonlonely people.

The design described uses a related sample (matched subjects)

Answer:

√3/2

Explanation:

The directional derivative at the given point is gotten using the formula;

∇f(x,y)•u where u is the unit vector in that direction.

∇f(x,y) = f/x i + f/y j

Given the function f(x, y) = y cos(xy),

f/x = -y²sin(xy) and

f/y = -xysin(xy)+cos(xy)

∇f(x,y) = -y²sin(xy) i + (cos(xy)-xysin(xy)) j

∇f(x,y) at (0,1) will give;

∇f(0,1) = -0sin0 i + cos0j

∇f(0,1) = 0i+j

The unit vector in the direction of angle θ is given as u = cosθ i + sinθ j

u = cos(π/3)i+ sin(π/3)j

u = 1/2 i + √3/2 j

Taking the dot product of both vectors;

∇f(x,y)•u = (0i+j)•(1/2 i + √3/2 j)

Note that i.i = j.j = 1 and i.j = 0

∇f(x,y)•u = 0 + √3/2

∇f(x,y)•u = √3/2

The directional derivative of [tex]f[/tex] at the given point in the direction indicated is [tex]\frac{\sqrt{3}}{2}[/tex].

The directional derivative is represented by the following formula:

[tex]\nabla_{\vec v} f = \nabla f(x_{o},y_{o}) \cdot \vec v[/tex]    (1)

Where:

The gradient of [tex]f[/tex] is calculated below:

[tex]\nabla f (x_{o},y_{o}) = \left[\begin{array}{cc}\frac{\partial f}{\partial x} (x_{o}, y_{o}) \\\frac{\partial f}{\partial y} (x_{o}, y_{o})\end{array}\right][/tex] (2)

Where [tex]\frac{\partial f}{\partial x}[/tex] and [tex]\frac{\partial f}{\partial y}[/tex] are the partial derivatives with respect to [tex]x[/tex] and [tex]y[/tex], respectively.

If we know that [tex](x_{o}, y_{o}) = (0, 1)[/tex], then the gradient is:

[tex]\nabla f(x_{o}, y_{o}) = \left[\begin{array}{cc}-y^{2}\cdot \sin xy\\\cos xy -x\cdot y\cdot \sin xy\end{array}\right][/tex]

[tex]\nabla f (x_{o}, y_{o}) = \left[\begin{array}{cc}-1^{2}\cdot \sin 0\\\cos 0-0\cdot 1\cdot \sin 0\end{array}\right][/tex]

[tex]\nabla f (x_{o}, y_{o}) = \left[\begin{array}{cc}0\\1\end{array}\right][/tex]

If we know that [tex]\vec v = \cos \frac{\pi}{3}\,\hat{i} + \sin \frac{\pi}{3} \,\hat{j}[/tex], then the directional derivative is:

[tex]\Delta_{\vec v} f = \left[\begin{array}{cc}0\\1\end{array}\right]\cdot \left[\begin{array}{cc}\cos \frac{\pi}{3} \\\sin \frac{\pi}{3} \end{array}\right][/tex]

[tex]\nabla_{\vec v} f = (0)\cdot \cos \frac{\pi}{3} + (1)\cdot \sin \frac{\pi}{3}[/tex]

[tex]\nabla_{\vec v} f = \frac{\sqrt{3}}{2}[/tex]

The directional derivative of [tex]f[/tex] at the given point in the direction indicated is [tex]\frac{\sqrt{3}}{2}[/tex]. [tex]\blacksquare[/tex]

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Answer:

115 miles

Step-by-step explanation:

First find the distance at 50 mph

d = 50 mph * .5 hours

   = 25 miles

Then find the distance at 60 mph

d = 60 mph * 1.5 hours

   = 90 miles

Add the distances together

25+90

115 miles

Answer:

he drives a 115 miles

Step-by-step explanation:

if he drives 50 mph for half an hour he drove 25 miles then if he drives 60 mph for 1 hour and 30 minutes he would of drove 90 miles. 60 + 30=90

90+25=115 so he drove 115 miles.

Answer:

y = -3.5x² + 2.7x -8.2

Step-by-step explanation:

the quadratic equation is set up as a² + bx + c, so just plug in the values

Answer:

[tex]-3.5x^2 + 2.7x -8.2[/tex]

Step-by-step explanation:

Quadratic functions are always formatted in the form [tex]ax^2+bx+c[/tex].

So, we can use your values of a, b, and c, and plug them into the equation.

A is -3.5, so the first term becomes [tex]-3.5x^2[/tex].

B is 2.7, so the second term is [tex]2.7x[/tex]

And -8.2 is the C, so the third term is [tex]-8.2[/tex]

So we have [tex]-3.5x^2+2.7x-8.2[/tex]

Hope this helped!

Answer:

H0: μ=2; Ha: μ>2

Step-by-step explanation:

The null hypothesis is the default hypothesis while the alternative hypothesis is the opposite of the null and is always tested against the null hypothesis.

In this case study, the null hypothesis is that adults were eating, on average, the recommended 2 cups of vegetables a day: H0: μ=2 while the alternative hypothesis is adults were eating, on average, more than the recommended 2 cups of vegetables a day Ha: μ>2.

Answer:

The probability that all three people on the subcommittee are men

= 20%

Step-by-step explanation:

Number of members in the committee = 15

= 8 men + 7 women

The probability of selecting a man in the committee

= 8/15

= 53%

The probability of selecting three men from eight men

= 3/8

= 37.5%

The probability that all three people on the subcommittee are men

= probability of selecting a man multiplied by the probability of selecting three men from eight men

= 53% x 37.5%

= 19.875%

= 20% approx.

This is the same as:

The probability of selecting 3 men from the 15 member-committee

= 3/15

= 20%

Answer: length = 1, width = 1, height = 3, volume = 5

Step-by-step explanation:

Degree is the biggest exponent for the variables in the expression

Length = 4x - 1. The exponent for x is 1 --> degree = 1

Width = x          The exponent for x is 1  --> degree = 1

Height = x³       The exponent for x³ is 3  --> degree = 3

Volume = 4x⁵ - x⁴.  The biggest exponent for x is 5  --> degree = 5

Answer:

- First answer: 1

- Second answer: 1

- Third answer: 3

- Last answer: 5

Step-by-step explanation:

Correct on E2020

Answer:

The answer is 5th angle = [tex]\bold{42^\circ}[/tex]

Step-by-step explanation:

Given that pizza is divided into six unequal slices.

Largest slice has an angle of [tex]90^\circ[/tex].

He eats the pizza from largest to smallest.

Let the difference in angles in each slice = [tex]d^\circ[/tex]

1st angle = [tex]90^\circ[/tex]

2nd angle = 90-d

3rd angle = 90-d-d = 90 - 2d

4th angle = 90-2d-d = 90 - 3d

5th angle = 90-3d-d = 90 - 4d

6th angle = 90-4d -d = 90 - 5d

We know that the sum of all the angles will be equal to [tex]360^\circ[/tex] (The sum of all the angles subtended at the center).

i.e.

[tex]90+90-d+90-2d+90-3d+90-4d+90-5d=360\\\Rightarrow 540 - 15d = 360\\\Rightarrow 15d = 540 -360\\\Rightarrow 15d = 180\\\Rightarrow d = 12^\circ[/tex]

So, the angles will be:

1st angle = [tex]90^\circ[/tex]

2nd angle = 90- 12 = 78

3rd angle = 78-12 = 66

4th angle = 66-12 = 54

5th angle = 54-12 = 42

6th angle = 42 -12 = 30

So, the answer is 5th angle = [tex]\bold{42^\circ}[/tex]

Answer:

8lb of the cheaper Candy

17.5lb of the expensive candy

Step-by-step explanation:

Let the cheaper candy be x

let the costly candy be y

X+y = 25.5....equation one

2.2x +7.3y = 25.5(5.7)

2.2x +7.3y = 145.35.....equation two

X+y = 25.5

2.2x +7.3y = 145.35

Solving simultaneously

X= 25.5-y

Substituting value of X into equation two

2.2(25.5-y) + 7.3y = 145.35

56.1 -2.2y +7.3y = 145.35

5.1y = 145.35-56.1

5.1y = 89.25

Y= 89.25/5.1

Y= 17.5

X= 25.5-y

X= 25.5-17.5

X= 8