Consider the two savings plans below. Compare the balances in each year after 14 years. Which person deposited more money in the plan? Which of the two investment strategies is better?

Yolanda deposits $250 per month in an account with an APR of 4%, while Zach deposits $3000 at the end of each year in an account with an APR of 4%

Answer:

y=2x+16 is the slope intercept form

Step-by-step explanation:

sorry if its wrong tried!!!

answer:

w = (-m+y)/y

Step-by-step explanation:

y-wy = m

-wy = m-y

y = -(m-y)/w

wy= -m + y

w = (-m+y)/y

.

Answer:

-4*(x+1)

Step-by-step explanation:

Simplify

⇒ 8x2 + 8x/-2x

2:  Pulling out like terms

4.1     Pull out like factors :

  8x2 + 8x  =   8x • (x + 1)

Canceling Out :

4.2    Canceling out x as it appears on both sides of the fraction line

Final result :

 -4 • (x + 1)

Answer:

-4(x+1)

Step-by-step explanation:

factor out -2x

-2x(-6x + 2x - 4) / -2x

the -2x in numerator and denominator cancel out

-6x + 2x - 4

-4x - 1

-4(x+1)

Answer:

Sonu is 15years and Monu is 21 years

Answer:

y = -t

Step-by-step explanation:

Parametric Equation = Type of equation that employs an independent variable called a parameter (often denoted by t) and in which dependent variables are defined as continuous functions of the parameter and are not dependent on another existing variable.

⇒ parametric equation of line passing through a point (a₁, b₁, c₁)

   and parallel to a vector <a, b, c> is given by :

   x =a₁ + at ,   y = b₁ + bt   ,   z = c₁ + ct

now according to question:

given -

point, P(1, 0, -3)

line,   x = −1 + 2t , y = 2−t, and z = 3+3t.

so from the line the vector is=  <2, -1, 3>

now using above formula,

equation of line is =   x = 1 + 2t , y = −t, and z = -3+3t.

we have to solve for 'y' only,

⇒ y = -t    (answer)

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Answer:

-59

Step-by-step explanation:

x = -2

4(-2)³ - 6(-2)² + (-2) - 1

= -32 - 24 - 2 - 1

= - 59

Answer:

please provide the diagram

Step-by-step explanation:

for the solution figure is necessary so provides us to solve this problem

The range of y will be all real numbers such that 0≤y≤40

The complete question is:

Raj’s bathtub is clogged and is draining at a rate of 1.5 gallons of water per minute. The table shows that the amount of water remaining in the bathtub, y, is a function of the time in minutes, x, that it has been draining. What is the range of this function? all real numbers such that y ≤ 40 all real numbers such that y ≥ 0 all real numbers such that 0 ≤ y ≤ 40 all real numbers such that 37.75 ≤ y ≤ 40

The relation between two expressions that are not equal, employing a sign such as ≠ ‘not equal to’, > ‘greater than, or < ‘less than.

Raj’s bathtub is clogged and is draining at a rate of 1.5 gallons of water per minute.

The amount of water remaining in the bathtub = y

The function of  time in minutes, that it has been draining = x

At 0 minutes the amount of water is 40 gallons.

The highest volume of water is 40 which is decreasing at the rate of 1.5 gallons per minute.

The given function is a linear function  

y = 0

However, the volume of water can be 0 but cannot ever be negative.

Therefore the range of y will be all real numbers such that 0≤y≤40

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[tex]{\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}[/tex]

★ Curved surface area of right circular cylinder is 4.4 m².

★ Radius of base of the cylinder is 0.7 m.

★ [tex]\tt \pi = \dfrac{22}{7}[/tex]

[tex]{\large{\textsf{\textbf{\underline{\underline{To \: Find :}}}}}}[/tex]

★ The height of cylinder.

[tex] {\large{\textsf{\textbf{\underline{\underline{Using \: Formula :}}}}}}[/tex]

[tex] \star \: \tt C.S.A \: of \: cylinder = \boxed{ \tt \pink{{ 2πrh}}}[/tex]

[tex] {\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}[/tex]

Let,

❍ The height of circular cylinder be [tex]h[/tex]

❍ Radius [r] of base of cylinder be 0.7 m

We know,

[tex] \star \: \tt C.S.A \: of \: cylinder = 2πrh[/tex]

Putting,

☆ [tex]\tt \pi = \dfrac{22}{7}[/tex]

☆ r as 0.7

[tex] \longrightarrow \tt 4.4 {m}^{2} = 2πrh[/tex]

[tex] \longrightarrow \tt 4.4 {m}^{2} = \bigg( 2 \times \dfrac{22}{7} \times 0.07 \times h \bigg)m[/tex]

[tex] \longrightarrow \tt \dfrac{44}{10} {m}^{2} = \bigg( {2 }\times \dfrac{22}{ \cancel{7}} \times \dfrac{ \cancel{7}}{10} \times h \bigg)m[/tex]

[tex] \longrightarrow \tt \dfrac{44}{10} {m}^{2} = \bigg( 44 \times \dfrac{ {1}}{10} \times h \bigg)m[/tex]

[tex] \longrightarrow \tt \dfrac{44}{ \cancel{10}} \times \cancel{10} \: m = \bigg( 44 \times 1 \times h \bigg)[/tex]

[tex] \longrightarrow \tt 44 m = 44h[/tex]

[tex]\longrightarrow \tt \dfrac{44}{44} m = h[/tex]

[tex]\longrightarrow \tt \red{ 1 m } = h[/tex]

Therefore, the height of cylinder = 1 m.

[tex]\begin{gathered} {\underline{\rule{290pt}{2pt}}} \end{gathered}[/tex]

An equivalent ratio to the tape diagram shown is 1 to 2

From the tape diagram, we have:

Red = 2

Blue = 4

Express as a ratio

Red : Blue = 2 : 4

Divide by 2

Red : Blue = 1 : 2

Hence, an equivalent ratio to the tape diagram shown is 1 to 2

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College students (ages 18-26) tend to make decisions that are tentative (more short-range) and support a desire for autonomy. This depicts more permanent choices.

It should be noted that values are a compass that helps us make decisions and choices.

Choices characterize the stage of life we are in. For example 18-26 ages tend to make tentative choices, later 27-31 ages tend to make permanent choices, and ages 32-42 people make stable choices.

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The recursive formula would be: 100 - XN = B.

When a function calls itself and uses its own previous terms to define its subsequent terms, it is called a recursive function. It is the technical recursive function’s definition, i.e., a recursive function builds on itself.

X: represents how many weeks

N: represents books per week

B: represents books she has at anytime

So the recursive formula would be: 100 - XN = B

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Answer:

7%

Step-by-step explanation:

[tex] \frac{14766 - 13800}{13800} \times 100 = 7 [/tex]

The percent increase in the property's value is 7% if the parcel of land is valued at $13,800. Today, that same parcel of land has a value of $14,766.

It's the ratio of two integers stated as a fraction of a hundred parts. It is a metric for comparing two sets of data, and it is expressed as a percentage using the percent symbol.

We have:

A few years ago, Sarah acquired a parcel of land valued at $13,800. Today, that same parcel of land has a value of $14,766

Percentage increase:

[tex]= \rm \dfrac{14766-13800}{13800}\times 100[/tex]

= 7%

Thus, the percent increase in the property's value is 7% if the parcel of land is valued at $13,800. Today, that same parcel of land has a value of $14,766.

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Check the forward differences of the sequence.

• first-order differences

5 - 1 = 4

12 - 5 = 7

22 - 12 = 10

35 - 22 = 13

• second-order differences (i.e. differences of the first differences)

7 - 4 = 3

10 - 7 = 3

13 - 10 = 3

The second differences are all 3 (as far as we know), so the sequence of first differences is arithmetic/linear, which means the original sequence is quadratic. Let the [tex]n[/tex]-th term be

[tex]x_n = an^2 + bn + c[/tex]

Given that [tex]x_1=1[/tex], [tex]x_2=5[/tex], and [tex]x_3=12[/tex], we have

[tex]\begin{cases} a + b + c = 1 \\ 4a + 2b + c = 5 \\ 9a + 3b + c = 12 \end{cases} \implies a=\dfrac32, b=-\dfrac12, c=0[/tex]

and so the [tex]n[/tex]-th term of the sequence is generated by the rule

[tex]x_n = \dfrac{3n^2 - n}2[/tex]

which most closely resembles the last option,

Square the term number, multiply the result by 3, subtract the term number, and divide the result by 2.

..................................

The probability you will draw a blue marble on your first draw and then draw a blue marble on your second draw is 0.063.

It is defined as the ratio of the number of favorable outcomes to the total number of outcomes, in other words, the probability is the number that shows the happening of the event.

Total marbles = 20

Number of blue marbles = 5

P(blue) = 5/20

P(blue∩blue) = P(blue)×P(blue)

= (5/20)(5/20)

= 25/400

= 1/16

= 0.0625 ≈ 0.063

Thus, the probability you will draw a blue marble on your first draw and then draw a blue marble on your second draw is 0.063.

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The range, standard deviation, and variance of the numbers of medals won by these countries are 28, 8.845, and 78.2353, respectively.

A range is given to a parameter to allow maximum leverage to the parameter. for example, if a vendor wants a rod of diameter 20 cm, then he may give a range of ±1 cm., which means he will accept the rod of 19(20-1) cm to 21(20+1) cm.

The range  of the numbers of medals won by these countries is,

Range =  Max - Min = 29 - 1 = 28

To find the standard deviation we need to know the following details,

Now, the standard deviation of medals won by these countries is,

[tex]\sigma = \sqrt{\dfrac{\sum x^2 - \frac1n (\sum x)^2}{n-1}}\\\\\sigma = \sqrt{\dfrac{\4372 - \frac{234^2}{18}}{18-1}}\\\\\sigma = 8.845[/tex]

The variance of the numbers of medals won by these countries is,

v = σ²

v = 78.2353

Hence, the range, standard deviation, and variance of the numbers of medals won by these countries are 28, 8.845, and 78.2353, respectively.

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Using the probability concept, we have that:

a) It would not be unusual to observe one component​ fail, since the probability that one component​ fails is greater than 0.05.

b) It would be unusual to observe two components​ fail, since the probability that two components​ fail is less than 0.05.

A probability is given by the number of desired outcomes divided by the number of total outcomes. If a probability is less than 0.05, the event is considered unusual.

In this problem, the probabilities are given as follows:

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The approximations of the mean and the standard deviation are 233.3 and 229.82, respectively

The table of values is given as:

Savings Lower Limit Upper Limit     Frequency

0-199         0         199                       345

200-399   200   399                         97

400-599   400   599                         52

600-799   600   799                         21

800-999   800   999                         9

1000-1199 1000  1199                         8

1200-1399 1200  1399                      3

Rewrite the table to include the class midpoint and the frequency

x               f

99.5         345

299.5        97

499.5        52

699.5        21

899.5        9

1099.5        8

1299.5        3

The mean is calculated as:

[tex]\bar x = \frac{\sum fx}{\sum f}[/tex]

So, we have:

[tex]\bar x = \frac{99.5* 345 + 299.5* 97 + 499.5* 52 + 699.5 * 21 + 899.5 * 9 + 1099.5 * 8 + 1299.5 * 3}{345 + 97 + 52 + 21 + 9 + 8 +3}[/tex]

Evaluate

[tex]\bar x = 233.331775701[/tex]

Approximate

[tex]\bar x = 233.3[/tex]

Hence, the approximation of the mean is 233.3

The standard deviation is calculated as:

[tex]\sigma = \sqrt{\frac{\sum f(x - \bar x)^2}{\sum f}}[/tex]

So, we have:

[tex]\sigma= \sqrt{\frac{(99.5-233.3)^2* 345 + (299.5-233.3)^2* 97 +...... + (1299.5 -233.3)^2* 3}{345 + 97 + 52 + 21 + 9 + 8 +3}[/tex]

Evaluate

[tex]\sigma = 229.82[/tex]

Hence, the approximation of the standard deviation is 229.82

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Yes it is

Step-by-step explanation:

it is a linear function as it is written in the standard form Y=MX+C

The solution set for the given inequality is 3<x<5.

The given inequality is 6<x+3<8.

In mathematics, a solution set is the set of values that satisfy a given set of equations or inequalities.

Now, solve the given inequality:

6<x+3<8⇒6<x+3 and x+3<8

6-3<x ⇒3<x

x+3<8⇒x<5

Thus, 3<x<5.

Therefore, the solution set is 3<x<5.

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Inequality and graph the solution set and a number line, express the Solution set in interval notation is 3 < x < 5

6<x+3<8

[tex]6 < x+3\quad \mathrm{and}\quad \:x+3 < 8[/tex]

[tex]\mathrm{If}\:a < u < b\:\mathrm{then}\:a < u\quad \mathrm{and}\quad \:u < b[/tex]

[tex]\mathrm{Combine\:the\:intervals}[/tex]

[tex]x > 3\quad \mathrm{and}\quad \:x < 5[/tex]

[tex]3 < x < 5[/tex]

Therefore the Inequality and graph of the solution set and a number line, express the Solution set in interval notation as 3 < x < 5.

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Answer: 19683/64

Step-by-step explanation:

[tex]a_{n}=8\left(\frac{3}{2} \right)^{n-1}\\\\\implies a_{10}=8 \left(\frac{3}{2} \right)^{10-1}=\boxed{\frac{19683}{64}}[/tex]

Answer:

Day 5 ................

the end behavior is:

If we have a polynomial of even degree, then the end behavior in both ends is the same one.

Particularly, in these cases (even degree) we only need to look at the leading coefficient. If it is positive, then as x tends to infinity and negative infinity, the function tends to infinity.

In this case, the polynomial is:

y = 9x⁴ + 8x - 2

Notice that the degree is 4, even, and the leading coefficient is 9 (positive).

Then the end behavior is:

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The weight of each pile yesterday will be 6 and 8 pounds.

The unitary method is a method for solving a problem by the first value of a single unit and then finding the value by multiplying the single value.

It is Given that Yesterday the ratio of seeds in these piles was 3:4.

Let 3x and 4x represent the seeds.

If the 2 pounds of seeds are added in the bigger pile, then

4x + 2

He also ate 1/4 pound from the smaller pile,

3x - 1/4

The quantities of seeds in those piles is in the ratio of 5:16.

So, 4x + 2 = 5x

3x - 1/4 = 16x

Solve;

So, 4x + 2 = 5x

2 = 5x - 4x

x = 2

The weight of each pie will be

3x = 6 pound

4x = 8 pound

Hence, The weight of each pile yesterday would be 6 and 8 pounds.

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An inequality relating 100n and n³ is 100n ≥ n³ for n ≤ 10 and 100n ≤ n³ for n ≥ 10.

An inequality is comparison of  two values, showing if one is less than, greater than, or simply not equal to another value.

Since 100n and n³ for n = 1, 2, 3, . . . 9, 10, 11 are 100, 200, 300, . . . 900, 1000, 1100 and 1, 8, 27, . . . 729, 1000, 1331 respectively.

Therefore, an inequality relating 100n and n³ will be 100n ≥ n³ for n ≤ 10 and 100n ≤ n³ for n ≥ 10.

Induction hypothesis:

Suppose 100n ≤ k³ for some positive integer k ≥ 10.

We need to show that 100( k + 1 ) ≤ ( k + 1 )³ = k³ + 3k² +3k + 1.

Note 100( k + 1 ) = 100k + 100 ≤ k³ + 100

                                                 ≤ k³ + 3k²    (∵ k ≥ 10 )

                                                 ≤ k³ + 3k² + 3k

                                                 ≤ k³ + 3k²+3k + 1

So 100( k + 1 ) ≤ ( k + 1 )³, which is true.

Hence by the principle of mathematical induction, 100n ≤ k³ for every integer  k ≥ 10.

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Answer:

t = 2

Step-by-step explanation:

[tex]2t + 4 = 14 - 3t[/tex]

[tex]5t = 10[/tex]

[tex]t = 2[/tex]

Checking:

[tex] log(2(2) + 4) = log(8) [/tex]

[tex] log(14 - 3(2)) = log(8) [/tex]

Using the Factor Theorem, the polynomials are given as follows:

1. [tex]P(x) = x^5 + 2x^4 - 7x^3 + x^2[/tex]

2. [tex]P(x) = 0.8(x^4 - 4x^3 - 16x^2 + 64x)[/tex]

3. P(x) = -0.1(x³ - 4x² - 3x + 18)

The Factor Theorem states that a polynomial function with roots [tex]x_1, x_2, \codts, x_n[/tex] is given by:

[tex]f(x) = a(x - x_1)(x - x_2) \cdots (x - x_n)[/tex]

In which a is the leading coefficient.

Item a:

The parameters are:

[tex]a = 1, x_1 = x_2 = 1, x_3 = x_4 = 0, x_5 = -4[/tex]

Hence the equation is:

P(x) = (x - 1)²x²(x + 4)

P(x) = (x² - 2x + 1)(x + 4)x²

P(x) = (x³ + 2x² - 7x + 1)x²

[tex]P(x) = x^5 + 2x^4 - 7x^3 + x^2[/tex]

Item b:

The roots are:

[tex]x_1 = x_2 = 4, x_3 = 0, x_4 = -4[/tex]

Hence:

P(x) = a(x - 4)²x(x + 4)

P(x) = a(x² - 16)x(x - 4)

P(x) = a(x³ - 16x)(x - 4)

[tex]P(x) = a(x^4 - 4x^3 - 16x^2 + 64x)[/tex]

It passes through the point x = 5, P(x) = 36, hence:

45a = 36.

a = 4/5

a = 0.8

Hence:

[tex]P(x) = 0.8(x^4 - 4x^3 - 16x^2 + 64x)[/tex]

Item 3:

The roots are:

[tex]x_1 = x_2 = 3, x_3 = -2[/tex]

Hence:

P(x) = a(x - 3)²(x + 2)

P(x) = a(x² - 6x + 9)(x + 2)

P(x) = a(x³ - 4x² - 3x + 18)

For the y-intercept, x = 0, y = -1.8, hence:

18a = -1.8 -> a = -0.1

Thus the function is:

P(x) = -0.1(x³ - 4x² - 3x + 18)

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Answer:

$38.50

Step-by-step explanation:

(100-30)% of $55

= 70% of $55

= $38.50

Hope this helps :)