-10 degrees Celsius is what Fahrenheit

Answer:

t = 13.56915448 hrs.

Step-by-step explanation:

500 = 300 [tex]e^{2k}[/tex]

5/3 = [tex]e^{2k}[/tex]

ln(5/3) = 2k ln(e)

k = ln(5/3)/2

k= 0.255412812

~~~~~~~~~~~~

9600 = 300 [tex]e^{0.255412812 t}[/tex]

32= [tex]e^{0.255412812 t}[/tex]

ln(32) = [tex]0.255412812 t[/tex] ln(e)

t = ln(32)/0.255412812

t = 13.56915448 hrs.

Answer:

1,108 in²

Step-by-step explanation:

SA = (12×20) + (2×20×5 + 2×12×5) + (2×½×12×9)

+ (2×20×11)

= 240+320+108+440

= 1,108 in²

Answer:

sin(x)

Step-by-step explanation:

sec x tanx(1 - sin^2 x)

1 - sin^2 x = cos^2 x

sec(x)tan(x)cos^2(x)

[tex]\frac{1}{cos(x)}[/tex] * [tex]\frac{sin(x)}{cos(x)}[/tex] * cos^2(x)

[tex]\frac{sin(x)cos^2(x)}{cos^2(x)}[/tex]

sin(x)

The requirement is fulfilled when 10 bananas and 2 apples are kept in the basket.

The equation can be given as

10y + 2x = 80

Here,

y means 7 ounces (weight of banasas)

x means 5 ounces (weight of apples)

Answer:

[tex]180 - 2x + 180 - 4x + x = 180 \\ 360 - 5x = 180 \\ 180 = 5x \\ x = 36[/tex]

Answer:

9:2

Step-by-step explanation:

Answer

y = 3x - 2

Step-by-step explanation

The slope-intercept form of the equation for this line is,

⇒ y  = x + 2

The equation of line in point-slope form passing through the points

(x₁ , y₁) and (x₂, y₂) with slope m is defined as;

⇒ y - y₁ = m (x - x₁)

Where, m = (y₂ - y₁) / (x₂ - x₁)

Given that;

The point-slope form of the equation of the line that passes through points (-9, -2) and (1, 3) is,

⇒ y - 3 = (x - 1)

Now, We can simplify as;

⇒ y - 3 = (x - 1)

⇒ y - 3 + 3 = (x - 1) + 3

⇒ y  = x + 2

Thus, The slope-intercept form of the equation for this line is,

⇒ y  = x + 2

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Answer:

See attachment showing the rise and run

Slope = 1

Step-by-step explanation:

In the diagram attached below, the rise is represented by the blue line, while the run is represented by the red line.

Rise = 4 units

Run = 4 units

It's a positive slope because the line slopes upwards from left to right

Slope = rise/run = 4/4

Slope = 1

9514 1404 393

Answer:

  48 +j64 ohms

Step-by-step explanation:

The impedance of a series circuit is the sum ...

  R + jXl -jXc

  = 48 +j100 -j36

  = 48 +j64 . . . . ohms

_____

Additional comment

"j" is the electrical engineer's name for √-1, because "i" is used to represent current.

Answer:

yes it is

Step-by-step explanation:

Answer:

x = 128.472

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The number of diners each day has a mean of 107 and a standard deviation of 60.

This means that [tex]\mu = 107, \sigma = 60[/tex]

Distribution of the daily average:

Over a month of 30 days, so [tex]n = 30, s = \frac{60}{\sqrt{30}} = 10.955[/tex]

The probability that a daily average over a given month is greater than x is 2.5%. Calculate x.

This is X when Z has a p-value of 1 - 0.025 = 0.975, so X when Z = 1.96. Then

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]1.96 = \frac{X - 107}{10.955}[/tex]

[tex]X - 107 = 1.96*10.955[/tex]

[tex]X = 128.472[/tex]

So x = 128.472

Answer:

[tex]\begin{array}{cc}{Class}& {Frequency} & 138 - 202 & 14 & 203 - 267 & 5 & 268 - 332 & 3 & 333 - 397 & 1 & 398 - 462 & 3 \ \end{array}[/tex]

The class with the greatest is 138- 202 and the class with the least relative frequency is 333 - 397

Step-by-step explanation:

Solving (a): The frequency distribution

Given that:

[tex]Lowest = 138[/tex] --- i.e. the lowest class value

[tex]Class = 5[/tex] --- Number of classes

From the given dataset is:

[tex]Highest = 459[/tex]

So, the range is:

[tex]Range = Highest - Lowest[/tex]

[tex]Range = 459 - 138[/tex]

[tex]Range = 321[/tex]

Divide by the number of class (5) to get the class width

[tex]Width = 321 \div 5[/tex]

[tex]Width = 64.2[/tex]

Approximate

[tex]Width = 64[/tex]

So, we have a class width of 64 in each class;

The frequency table is as follows:

[tex]\begin{array}{cc}{Class}& {Frequency} & 138 - 202 & 14 & 203 - 267 & 5 & 268 - 332 & 3 & 333 - 397 & 1 & 398 - 462 & 3 \ \end{array}[/tex]

Solving (b) The relative frequency histogram

First, we calculate the relative frequency by dividing the frequency of each class by the total frequency

So, we have:

[tex]\begin{array}{ccc}{Class}& {Frequency} & {Relative\ Frequency} & 138 - 202 & 14 & 0.53 & 203 - 267 & 5 & 0.19 & 268 - 332 & 3 & 0.12 & 333 - 397 & 1 & 0.04 & 398 - 462 & 3 & 0.12 \ \end{array}[/tex]

See attachment for histogram

The class with the greatest is 138- 202 and the class with the least relative frequency is 333 - 397

Answer:

B. (-r, 0)

Step-by-step explanation:

W is on the x axis, so the y coordinate is 0.

It is also r away from the origin, so the x coordinate is -r

Hope this helps!

9514 1404 393

Answer:

  ∠CAB = 86°

  ∠ABC = 43°

  ∠BCA = 51°

Step-by-step explanation:

This can be done a couple of different ways (as with most math problems). We can use the distance formula to find the side lengths, then the law of cosines to find the angles. Or, we could use the dot product. In the end, the math is about the same.

The lengths of the sides are given by the distance formula.

  AB² = (4-1)² +(-3-0)² +(0-(-1)) = 16 +9 +1 = 26

  BC² = (1-4)² +(2-(-3))³ +(3-0)² = 9 +25 +9 = 43

  CA² = (1-1)² +(0-2)² +(-1-3)² = 4 +16 = 20

From the law of cosines, ...

  ∠A = arccos((AB² +CA² -BC²)/(2·AB·CA)) = arccos((26 +20 -43)/(2√(26·20)))

  ∠A = arccos(3/(4√130)) ≈ 86°

  ∠B = arccos((AB² +BC² -AC²)/(2·AB·BC)) = arccos((26 +43 -20)/(2√(26·43)))

  ∠B = arccos(49/(2√1118)) ≈ 43°

  ∠C = arccos((BC² +CA² -AB²)/(2·BC·CA)) = arccos((43 +20 -26)/(2√(43·20)))

  ∠C = arccos(37/(4√215)) ≈ 51°

The three angles are ...

  ∠CAB = 86°

  ∠ABC = 43°

  ∠BCA = 51°

_____

Additional comment

This sort of repetitive arithmetic is nicely done by a spreadsheet.

Answer:

Fill in the blanks with the correct verbs given below.

(croak, is ,brush, writes plays, chirp, combines, mows, are look , keeps rises ,has,

reflects leaves, walks, is)

1. Marry ...keeps rises...... to office every morning .Walking is good exercise.

2. We...brush..... our teeth after dinner. It .....makes.....them clean.

3. Salim....wakes..... for school everyday at 7:00 a.m.

4. The sun ...rises.......at 5:00 a.m. these days. The days are getting longer.

5. The tiger chirp......an endangered animal. There are only 1410 tigers left.

6. Vikram Seth.....combines........ poetry, fiction and travelogues.

7. Sania Mirza....plays..... tennis. She plays both singles and doubles.

8. Frogs....croak... ...and birds.....chirp......

9. Please ....look......for my red shirt. I cannot find it.

10. The gardener .....mows....... the loan once a week.

11. The moon ...keep rises.....the light of the sun.

12. The times of India......writes....... a national newspaper.

13. The UN headquarters.....is..... in Geneva.

14. Maya ...has........ a pet dog .Its name is Lily.

15. This book....writes..... projects and activities

Answer:

Cuboid = width*height*length

Cuboid = 24 cm^2

Answer:

[tex]\frac{-2x+11}{(x-4)(x+1)}[/tex]

Step-by-step explanation:

I don't think we can factor this so we'll have to multiply to make the denominators the same

[tex]\frac{3(x+1)}{(x^2-3x-4)(x+1)}-\frac{2(x^2-3x-4)}{(x+1)(x^2-3x-4)}\\\\\frac{3x+3-(2x^2-6x-8)}{(x^2-3x-4)(x+1)}=\frac{-2x^2+9x+11}{(x^2-3x-4)(x+1)}\\-2x^2+9x+11=(x+1)(-2x+11)\\\\x^2-3x-4=(x+1)(x-4)\\\frac{(x+1)(-2x+11)}{(x+1)(x-4)(x+1)}=\frac{-2x+11}{(x-4)(x+1)}[/tex]

Answer:

Step-by-step explanation:

                          Statements                                 Reasons

1). CD is an altitude of ΔABC                1). Given

2). ΔACD and ΔBCD are right              2). Definition of right triangles.

    triangles.

3). a² = (c - x)² + h²                                 3). Pythagoras theorem

4). a² = c² + x² - 2cx + h²                       4). Square the binomial.

5). b² = x² + h²                                       5). Pythagoras theorem.

6). cos(x) = [tex]\frac{x}{a}[/tex]                                           6). definition of cosine ratio for an angle

7). bcos(A) = x                                        7). Multiplication property of equality.

8). a² = c² - 2c(bcosA) + b²                    8). Substitution property

9). a² = b² + c² - 2bc(cosA)                    9). Commutative properties of

                                                                    addition and multiplication.

Answer:

Hope it is helpful and useful

Hello,

[tex]f(x)=2x^4-x^3-18x+9x\\\\=x(2x^3-x^2-18x+9)\\\\=x(x^2(2x-1)-9(2x-1))\\\\=x(2x-1)(x^2-9)\\=x(2x-1)(x-3)(x+3)\\[/tex]

Zeros are : 0;  1/2; -3; 3.

The zeros of the function are -3, 0, 1/2 and 3.

The given function is [tex]f(x)=2x^{4} -x^{3} -18x^{2} +9x[/tex].

Zeros of a function are the points where the graph of the function meets the X-axis i.e., at the solutions of f(x) = 0.

Now, factorise the given function, that is f(x)=x(2x³-x²-18x+9).

=x[x²(2x-1)-9x(2x-1)]

=x(2x-1)(x²-9)

=x(2x-1)(x+3)(x-3)

= -3, 0, 1/2, 3

Therefore, the zeros of the function are -3, 0, 1/2 and 3.

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Answer:

False

Step-by-step explanation:

Each f(x) increases by 8 therefore this equation is a linear function. If you where to graph it would be a straight line

Hope this helped :)

Answer:

False

Step-by-step explanation:

The slope is the same between all pounts which means the function is linear.

Hope this helps!

Answer:

8

Step-by-step explanation:

5 = 40

1 = x

Then we multiply by the rule of crisscrossing

5 x X = 40 x 1

5x = 40 then divide both by 5

X = 8

Answer:

D. Critical value =± 0.707; there is not sufficient evidence to support the claim of a linear correlation between heights of mothers and heights of their daughters.

Step-by-step explanation:

Given that :

Correlation Coefficient, r = 0.693

The sample size, n = 8

The degree of freedom used for linear correlation :

df = n - 2

df = 8 - 2 = 6

Using a critical value calculator for correlation Coefficient at α = 0.05

The critical value obtained is : 0.707

The test statistic :

T = r / √(1 - r²) / (n - 2)

T = 0.693 / √(1 - 0.693²) / (8 - 2)

T = 0.693 / 0.2943215

T = 2.354

Since ;

Test statistic < Critical value ; we fail to reject the null and conclude that there is not sufficient evidence to support the claim of a linear correlation between heights of mothers and heights of their daughters.

Answer:

Frumpyton

Step-by-step explanation:

Since the standard deviation of Frumpyton is a lower number, this means a higher percentage of outcomes (job salaries) will be within a closer range to the mean salary. Since Frumpyton's standard deviation is $2,000 and the window your looking for is $32,000 to $36,000, if you go one interval up or down from the mean of $34,000, it falls in that range. Whereas, Dirtballville's standard deviation is $3,000 so it's more likely to fall outside of that range.

Answer:

Neither one. They will both result in the same price.

Step-by-step explanation:

To discount an item 10%, you charge 90% of the price of the item. To find 90% of a price, you multiply the price by 0.9.

To discount an item 15%, you charge 85% of the price of the item. To find 85% of a price, you multiply the price by 0.85.

Since multiplication is commutative, multiplying a number by 0.9 and then by 0.85 is the same as multiplying the number by 0.85 first and then by 0.9.

Let's say the item costs x.

Take off the 10% discount first: 0.9x

Now take off the 15% discount: 0.85 * (0.9x)

Now do it the other way.

Take off the 15% discount first: 0.85x

Now take off the 10% discount: 0.9 * (0.85x)

Since 0.85 * 0.9 * x = 0.9 * 0.85 * x, the results are the same.

Answer: neither

Answer:

-4 < x < 6

Step-by-step explanation:

[tex]\bar{x} = 0[/tex]

[tex]\bar{y} =\dfrac{136}{125}[/tex]

Step-by-step explanation:

Let's define our functions [tex]f(x)\:\text{and}\:g(x)[/tex] as follows:

[tex]f(x) = x^2 + 1[/tex]

[tex]g(x) = 6x^2[/tex]

The two functions intersect when [tex]f(x)=g(x)[/tex] and that occurs at [tex]x = \pm\frac{1}{5}[/tex] so they're going to be the limits of integration. To solve for the coordinates of the centroid [tex]\bar{x}\:\text{and}\:\bar{y}[/tex], we need to solve for the area A first:

[tex]\displaystyle A = \int_a^b [f(x) - g(x)]dx[/tex]

[tex]\displaystyle \:\:\:\:\:\:\:=\int_{-\frac{1}{5}}^{+\frac{1}{5}}[(x^2 + 1) - 6x^2]dx[/tex]

[tex]\displaystyle \:\:\:\:\:\:\:=\int_{-\frac{1}{5}}^{+\frac{1}{5}}(1 - 5x^2)dx[/tex]

[tex]\displaystyle \:\:\:\:\:\:\:=\left(x - \frac{5}{3}x^3 \right)_{-\frac{1}{5}}^{+\frac{1}{5}}[/tex]

[tex]\:\:\:\:\:\:\:= \dfrac{28}{75}[/tex]

The x-coordinate of the centroid [tex]\bar{x}[/tex] is given by

[tex]\displaystyle \bar{x} = \dfrac{1}{A}\int_a^b x[f(x) - g(x)]dx[/tex]

[tex]\displaystyle \:\:\:\:\:\:\:= \frac{75}{28}\int_{-\frac{1}{5}}^{+\frac{1}{5}} (x - 5x^3)dx[/tex]

[tex]\:\:\:\:\:\:\:=\dfrac{75}{28}\left(\dfrac{1}{2}x^2 -\dfrac{5}{4}x^4 \right)_{-\frac{1}{5}}^{+\frac{1}{5}}[/tex]

[tex]\:\:\:\:\:\:\:= 0[/tex]

The y-coordinate of the centroid [tex]\bar{y}[/tex] is given by

[tex]\displaystyle \bar{y} = \frac{1}{A}\int_a^b \frac{1}{2}[f^2(x) - g^2(x)]dx[/tex]

[tex]\displaystyle \:\:\:\:\:\:\:=\frac{75}{28}\int_{-\frac{1}{5}}^{+\frac{1}{5}} \frac{1}{2}(-35x^4 + 2x^2 + 1)dx[/tex]

[tex]\:\:\:\:\:\:\:=\frac{75}{56} \left[-7x^5 + \frac{2}{3}x^3 + x \right]_{-\frac{1}{5}}^{+\frac{1}{5}}[/tex]

[tex]\:\:\:\:\:\:\:=\dfrac{136}{125}[/tex]