........Help Please.......

Answer:

  A) The annual multiplier was 1.0339; the annual increase was 0.0339 of the value.

  B) 3.39% per year

  C) $182,000

Step-by-step explanation:

A) Let's let t represent years since 1987. Then we can fill in the numbers and solve for r.

  165000 = 100000(1 +r)^15

  1.65^(1/15) = 1 +r . . . . . divide by 100,000; take the 15th root

  1.03394855265 -1 = r ≈ 0.0339

The value was multiplied by about 1.0339 each year.

__

B) The value increased by about 3.39% per year.

__

C) S = $100,000(1.03394855265)^18 ≈ $182,000

Answer:

In mixed number format: 6 1/2

Step-by-step explanation:

To solve the following equation: 7(x - 2) = 3(x + 4), first we need to apply the distributive property:

7(x - 2) = 3(x + 4) → 7x -14 = 3x + 12

Solving for 'x' → 4x = 26 → x = 6.5

→ In mixed number format: 6 1/2

For this case we must solve the following equation:

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

Applying distributive property to the terms within the parenthesis we have:

[tex]7x-14 = 3x + 12[/tex]

We subtract 3x on both sides of the equation:

[tex]7x-3x-14 = 12\\4x-14 = 12[/tex]

Adding 14 to both sides of the equation:

[tex]4x = 12 + 14\\4x = 26[/tex]

Dividing between 4 on both sides of the equation:

[tex]x = \frac {26} {4} = \frac {13} {2}[/tex]

ANswer:

[tex]x = \frac {13} {2}\\x = 6 \frac {1} {2}[/tex]

Answer:

The value of k is 6

Step-by-step explanation:

we need to find the value of k

Given : - [tex]y=e^{2x}[/tex] is the solution [tex]y''-5y'+ky=0[/tex]

[tex]y=e^{2x}[/tex]                               ........(1)                  

differentiate  [tex]y=e^{2x}[/tex] with respect to 'x'

[tex]\frac{dy}{dx}=\frac{d}{dx}e^{2x}[/tex]

Since, [tex]\frac{d}{dx}e^{x} =e^{x}\frac{d}{dx}(x)[/tex]

[tex]\frac{dy}{dx}=e^{2x}\frac{d}{dx}(2x)[/tex]

[tex]\frac{dy}{dx}=e^{2x}\times 2[/tex]

[tex]\frac{dy}{dx}=2e^{2x}[/tex]

so, [tex]y'=2e^{2x}[/tex]                     ..........(2)

Again differentiation above with respect to 'x'

[tex]\frac{d}{dx}\frac{dy}{dx}=\frac{d}{dx}2e^{2x}[/tex]

[tex]\frac{d^{2}y}{dx^{2}}=2e^{2x}\frac{d}{dx}(2x)[/tex]

[tex]\frac{d^{2}y}{dx^{2}}=2e^{2x}\times 2[/tex]

[tex]\frac{d^{2}y}{dx^{2}}=4e^{2x}[/tex]

so, [tex]y''=4e^{2x}[/tex]                         ........(3)

Now, put the value of [tex]y\ ,y' \ \text{and} \ y''[/tex] in [tex]y''-5y'+ky=0[/tex]

[tex]4e^{2x}-5(2e^{2x})+(e^{2x})k=0[/tex]

[tex]4e^{2x}-10e^{2x}+e^{2x}k=0[/tex]

[tex]-6e^{2x}+e^{2x}k=0[/tex]

add both the sides by [tex]6e^{2x}[/tex]

[tex]e^{2x}k=6e^{2x}[/tex]

Cancel out the same terms from left and right sides

[tex]k=6[/tex]

Hence, the value of k is 6

Answer: Hence, a) No, they are not independent

b) 0.193

c) 0.60

Step-by-step explanation:

Since we have given that

Probability that a child's gestational age is less than 37 weeks say P(A)= 0.142

Probability that his or her birth weight is less than 2500 grams say P(B) = 0.051

P(A∩B) = 0.031

We need to check whether it is independent or not.

Since ,

[tex]P(A).P(B)=0.142\times 0.051=0.0072[/tex]

and

[tex]P(A\cap B)=0.051[/tex]

So, we can see that

[tex]P(A).P(B)\neq P(A\cap B)[/tex]

So, it is not independent.

a) Hence, A and B are not independent.

b) P(A∪B) is given by

[tex]P(A\ or B\ or\ both)=P(A)+P(B)\\\\P(A\ or\ B\ or\ both)=0.142+0.051=0.193[/tex]

c) P(A|B) is given by

[tex]P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{0.031}{0.051}=0.60[/tex]

Hence, a) No,

b) 0.193

c) 0.60

Answer:

624

Step-by-step explanation:

The sequence is 49, 47, 45,...., 7, 5, 3.  This is an arithmetic sequence, because the difference between terms is the same.

The sum of the first n terms of an arithmetic sequence is:

S = n/2 (a₁ + an)

where a₁ is the first term and an is the nth term.

Here, we know that a₁ = 49 and an = 3.  But we need to find what n is.  To do that, we use definition of an arithmetic sequence:

an = a₁ + (n-1) d

where d is the common difference (in this case, -2)

3 = 49 + (n-1) (-2)

2(n-1) = 46

n - 1 = 23

n = 24

So there are 24 terms in the sequence.

The sum is:

S = 24/2 (49 + 3)

S = 12 (52)

S = 624

There are 624 bricks in the wall.

The total number of bricks in the wall is 624. This is a math problem that involves arithmetic sequence, where each term is obtained from the previous one by subtracting a fixed number (2, in this case), and concepts from algebra (equations).

The problem describes a scenario where each row of a brick wall has two fewer bricks than the row below it, which characterizes a sequence in mathematics. More specifically, this is an arithmetic sequence, which is characterized by a common difference between terms, in this case, the difference is -2.

To solve the problem, we need to find the sum of an arithmetic sequence. The formula of the sum is given by:

S = n/2 * (a1 + an)

Where S is the sum, n the number of terms, a1 the first term, and an the last term. Here, a1 is 49 and an is 3. To find n, we use the formula n = (a1 - an) / d + 1, with d being the common difference which is -2. Solving the equation we find that n = 24.

We now plug these values into the sum formula and find that the sum S, which represents the total number of bricks in the wall is

S = 24/2 * (49 + 3) = 12 * 52 = 624.

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⇒ The sum of the slopes of line b and line c is 0.

⇒   False ⇒ NOT true

Answer:

The sum of the slopes of line b and line c is 0.

Step-by-step explanation:

Remember that the product of the slopes of two parallel lines is -1, so in order to be -1 you have to multiply M*-1/m=-1 so since to add them up you would do it like this m+(-1/m) taht wouldn´t get as result 0, so that would be the option that is not correct, remember that parallel lines have the same slope, so that also eliminates all of the other options.

Answer:

Step-by-step explanation:

This sort of problem is best worked by a tool such as a graphing calculator or spreadsheet.

Answer:

$8000 is invested for 7% interest and $2000 is invested for 9% interest

Step-by-step explanation:

Points to remember

Simple interest formula

I = PNR/100

P - Principle amount

N - Number of years

R - Rate of interest

To find the amount of investment

It is given that total amount = 10,000 and total interest = $740

Let 'x' be the amount invested at the rate of 7%

10,000 - x be the amount invested at the rate of 9%

I = PNR/100

740 = (x*1*7)/100 + (10000 - x)*1*9/100

740 = 7x/100 + 90000/100 - 9x/100

740 = 7x/100 + 900 - 9x/100

740-900 = -2x/100

-160 = -2x/100

x = 16000/2 = 8000

10000-8000 = 2000

Therefore $8000 is invested for 7% interest and $2000 is invested for 9% interest

Answer:

$8000 is invested for 7% interest and $2000 is invested for 9% interest

Step-by-step explanation:

Points to remember

Simple interest formula

I = PNR/100

P - Principle amount

N - Number of years

R - Rate of interest

To find the amount of investment

It is given that total amount = 10,000 and total interest = $740

Let 'x' be the amount invested at the rate of 7%

10,000 - x be the amount invested at the rate of 9%

I = PNR/100

740 = (x*1*7)/100 + (10000 - x)*1*9/100

740 = 7x/100 + 90000/100 - 9x/100

740 = 7x/100 + 900 - 9x/100

740-900 = -2x/100

-160 = -2x/100

x = 16000/2 = 8000

10000-8000 = 2000

Therefore $8000 is invested for 7% interest and $2000 is invested for 9% interest

I really hope it helped idiot noob die

Answer: Simple random sampling

Step-by-step explanation:

Given: A market researcher obtains a list of all streets in a town. She randomly samples 10 street names from the list, and then administers survey questions to every family living on those 10 streets.

Since she randomly samples street names , therefore the type of sampling is simple random sampling.

Answer:

Width of box= 2inches

Length of box= 4inches

Height of box= 6inches.

Step-by-step explanation:

Let width of box=x inches

Length of box = twice of width=[tex]2\times x[/tex]=[tex]2x[/tex]

Height of box= 4 inches greater than width= [tex]x+4[/tex]

Volume of box= 48 cubic inches

We know that the formula of volume of cuboid= [tex] length\times breadth\times height[/tex]

Apply the formula

Volume of box= [tex]x\times 2x\times (x+4)[/tex]

Volume of cube = [tex]2x^2(x+4)[/tex]

[tex]2x^2(x+4)=48[/tex]

[tex]x^2(x+4)=24[/tex]

[tex]x^3+x^2-24[/tex]

Apply inspection method to solve the equation

Put [tex]x=0[/tex]

Then we get [tex]-24\neq0 [/tex]

Hence, x=0 is not the solution of x

Put x=1 in the equation then we get

[tex]-22\neq 0[/tex]

Hence x=1 is not the solution of equation.

Put x=2 then we get

[tex](2)^3+(4)^2-24[/tex]

8+16-24=0

Hence, x=2 is the solution of equation .

[tex] (x-2)(x^2+6x+12)[/tex]=0

Now substitute equation [tex]x^2+6x+12[/tex]=0

Sum roots =6

Product of roots=12

When sum of roots  is greater than zero and product of roots is greater than zero then value of roots of equation is imaginary.

Hence, the roots of equation [tex]x^2+6x+12=0[/tex] are imaginary.

Lenght , widht and height are dimensions of box therefore, imaginary value are not possible.

Hence,[tex] x=2 [/tex] is the only real values of root of equation .Therefore, it is possible and other two imaginary value of roots are not possible .

Widht of box=2 inches

Length of box = [tex]2\times2[/tex]=4inches

Height of box=[tex]x+4[/tex]=2+4=6 inches

The dimensions of the box are: length = 4 inches, width = 2 inches, and height = 6 inches.

Let's use the given information to solve for the dimensions of the box:

Let the width of the box be represented by x inches.

The length of the box is twice as long as the width, so the length is 2x inches.

The height is 4 inches greater than the width, so the height is (x + 4) inches.

The volume of a box can be calculated by multiplying the length, width, and height. Since the volume of the box is given as 48 cubic inches, we can set up the equation: 2x * x * (x + 4) = 48.

Simplifying the equation, we get 2x^3 + 8x^2 - 48 = 0.

Factoring the equation, we find that (x - 2)(x + 4)(x + 6) = 0.

The possible solutions are x = 2, x = -4, or x = -6.

Since we are dealing with dimensions, the width cannot be negative, so we can disregard the negative solutions. The width, therefore, is 2 inches.

The length is twice as long as the width, so the length is 2 * 2 = 4 inches.

The height is 4 inches greater than the width, so the height is 2 + 4 = 6 inches.

Therefore, the dimensions of the box are: length = 4 inches, width = 2 inches, and height = 6 inches.

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Answer:

Speed of boat x = 84 km/hr

Speed of current = 12 km/hr

Step-by-step explanation:

Let 'x' be the speed of boat  and 'y' be the speed of still water

Upstream speed = x - y  and

Downstream speed = x + y

It is given that, A motorboat travels 180 km in 3 hours going upstream and 504 in 6 hours going downstream

Upstream speed = x - y = 180/3 = 60 km/hr

Downstream speed = x + y =  504/6 = 84 km/hr

To find the value of x and y

x + y = 84  ----(1)

x - y = 60   ----(2)

(1) + (2) ⇒

x + y = 84  ----(1)

x - y = 60  ----(2)

2x  + 0  = 144

x = 144/2 = 72

x + y = 84

y = 84 - 72 = 12

Therefore speed of boat x = 84 km/hr

Speed of current = 12 km/hr

Answer:

p = 0.718 and n = 20

Step-by-step explanation:

p is the probability of success and n is the number of trials.

Here, p = 0.718 and n = 20.

Answer:

There is a 29.50% probability of winning more than 15 times.

Step-by-step explanation:

For each time you play the arcade game, there are only two possible outcomes. Either you win, or you lose. This means that we can solve this problem using the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinatios of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

In this problem we have that:

The probability of winning a game is 0.718. So [tex]p = 0.718[/tex].

The game is going to be played 20 times, so [tex]n = 20[/tex].

If you play the arcade game 20 times, we want to know the probability of winning more than 15 times.

This is

[tex]P(X > 15) = P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.2950[/tex].

There is a 29.50% probability of winning more than 15 times.

1.052 is equal to 105.2 percent.

Given that a decimal number 1.052, we need to write 1.052 as a percent,

To express a decimal number as a percent, you need to multiply it by 100.

Let's calculate 1.052 as a percent:

1.052 x 100 = 105.2

To understand this, let's break it down:

The number 1.052 represents 105.2% because it is greater than 1 (100%). By multiplying it by 100, we shift the decimal point two places to the right, resulting in 105.2.

In percentage terms, 105.2% means that 1.052 is 105.2 parts out of 100. This can also be interpreted as 105.2 per hundred or simply 105.2 out of every 100 units.

Therefore, 1.052 can be written as 105.2%.

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Answer:

A vertical line has an infinite or undefined slope since the denominator is zero.

Step-by-step explanation:

Parallel lines by definition refers to lines that never intersect or meet since they have identical slopes. The slope of line is defined as;

(change in y)/(change in x)

For a vertical line, the y values are changing while the x values remain constant. The slope of this line will thus have a zero value in the denominator implying that its slope will not defined or will be infinity.

Answer:

A vertical line has an infinite or undefined slope since the denominator is zero.

Step-by-step explanation:

checking the vertex of this upside-down parabola, it has a vertex at (1000, 2000000), so that's the U-turn, when as the price "p" increases the revenue goes down.

[tex]\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ y=\stackrel{\stackrel{a}{\downarrow }}{a}x^2\stackrel{\stackrel{b}{\downarrow }}{+b}x\stackrel{\stackrel{c}{\downarrow }}{+c} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)[/tex]

now, if we solve the quadratic using the value of 500000

[tex]\bf \stackrel{R(p)}{500000}=-2p^2+4000p\implies 250000=-p^2+2000p \\\\\\ p^2-2000p+250000=0[/tex]

and we run the quadratic formula on it, we get the values of x = 133.97 and x = 1866.03, one value is obviously when going upwards, the first one, and the other is when going downwards.

so we know that the R(p) is 500,000 at x = 133.97, and it keeps on going up, up to the vertex above at x = 1000, so we can say from x = [134, 1000] R(p) > 500000.

Answer:

The correct answer is second option

4a²b²c²∛b)

Step-by-step explanation:

It is given an expression, ∛(64a⁶b⁷c⁹)

Points to remember

Identities

ⁿ√x = x¹/ⁿ

To find the equivalent expression

We have,  ∛(64a⁶b⁷c⁹)

∛(64a⁶b⁷c⁹) =  (64a⁶b⁷c⁹)1/3

 = (4³/³ a⁶/³ b⁷/³ c⁹/³)          [Since 64 = 4³]

 = 4a² b² b¹/³ c³

 = 4a²b²c³(b¹/³)

 = 4a²b²c³ (∛b)

Therefore the correct answer is second option

4a²b²c³(∛b)

Answer: 0.8664

Step-by-step explanation:

Given : Mean : [tex]\mu =  60,000\text{ miles}[/tex]

Standard deviation : [tex]\sigma = 4,000\text{ miles}[/tex]

Sample size : [tex]n=40[/tex]

The formula to calculate the z-score :-

[tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]

For x=  59,050

[tex]z=\dfrac{59050-60000}{\dfrac{4000}{\sqrt{40}}}\approx-1.50[/tex]

For x= 60,950

[tex]z=\dfrac{60950-60000}{\dfrac{4000}{\sqrt{40}}}\approx1.50[/tex]

The P-value : [tex]P(-1.5<z<1.5)=P(z<1.5)-P(z<-1.5)[/tex]

[tex]=0.9331927-0.0668072=0.8663855\approx0.8664[/tex]

Hence, the likelihood of finding that the sample mean is between 59,050 and 60,950=0.8664

The likelihood of finding that the sample mean is between 59,050 and 60,950 miles, according to the given normal distribution, is approximately 86.64%.

To solve this problem, we consider that the population mean is 60,000 and the standard deviation is 4,000. If we choose a sample of 40 tires, the standard deviation of the sample mean (standard error) is the standard deviation divided by the square root of the sample size (σ/√n).

This gives us 4,000/√40 = 633. The z-scores for the lower and upper bounds of our interval (59,050 and 60,950) are calculated by subtracting the population mean from these values, and dividing by the standard error. For 59,050: (59,050 - 60,000)/633 = -1.5 and for 60,950: (60,950 - 60,000)/633 = 1.5.

Using standard normal distribution tables, we know that the probability associated with a z-value of 1.5 is 0.9332. Since the normal distribution is symmetric, the probability associated with -1.5 is also 0.9332. Therefore, the probability that the sample mean lies between 59,050 and 60,950 is 0.9332 - (1 - 0.9332) = 0.8664 or approximately 86.64%.

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Since there is a right angle, you can use Pythagoras' Theorem:

So x = √(24² + 7²) = 25

---------------------------------------------------------

Answer:

25

The future value of ​$510 per year for 8 years compounded annually at 9 ​percent is $1,016.21.

The investment's future value refers to the compounded value of the present cash flows in the future, using an interest rate.

The future value can be determined using the future value table or formula.

We can also determine the future value using an online finance calculator as below.

N (# of periods) = 8 years

I/Y (Interest per year) = 9%

PV (Present Value) = $510

PMT (Periodic Payment) = $0

Results:

FV = $1,016.21 ($510 + $506.21)

Total Interest = $506.21

Thus, the future value of ​$510 per year for 8 years compounded annually at 9 ​percent is $1,016.21.

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Answer:

The simplest form of [tex]\sqrt[3]{27a^{3}b^{7}}[/tex] is

3ab²(∛b)

Step-by-step explanation:

The given term is:

[tex]\sqrt[3]{27a^{3}b^{7}}[/tex]

To convert it into its simplest form, we will apply simple mathematical rules to simplify the power of individual terms.

[tex]\sqrt[3]{27a^{3}b^{7}}\\= \sqrt[3]{3^{3} a^{3}b^{7}}\\= \sqrt[3]{3^{3}a^{3}b^{6}b}\\= 3^{3/3} a^{3/3}b^{6/3}b^{1/3}}\\= 3ab^{2}(\sqrt[3]{b})[/tex]

While simplifying the term, we basically took the cube root of individual terms. The powers cancelled out cube root for some terms. In the end, we were left with the simplest form of the expression.

Answer:

[tex]1000±100\sqrt{55}[/tex]

Step-by-step explanation:

To simplify that expression, first we need to find the largest common of the expression inside the radical, in this case: 2.200.000.

We know that 2.200.000 = 2 · 2 · 2 · 2 · 2 · 2 · 5 · 5 · 5 · 5 · 5 · 11 = [tex]2^{6}[/tex] ×[tex]5^{5}[/tex]× [tex]11[/tex]

Now, [tex]\sqrt{2^{6}5^{5}11} = 200\sqrt{55}[/tex].

Now we have: [tex]\frac{2000±200\sqrt{55}}{2}[/tex]

Dividing by 2: [tex]1000±100\sqrt{55}[/tex]

So the simplified expression is: [tex]1000±100\sqrt{55}[/tex]

The derivative of the function f(x) is:

                 [tex]f'(x)=-2x[/tex]

We are given a function f(x) as:

[tex]f(x)=5-x^2[/tex]

We have:

[tex]f(x+h)=5-(x+h)^2\\\\i.e.\\\\f(x+h)=5-(x^2+h^2+2xh)[/tex]

( Since,

[tex](a+b)^2=a^2+b^2+2ab[/tex] )

Hence, we get:

[tex]f(x+h)=5-x^2-h^2-2xh[/tex]

Also, by using the definition of f'(x) i.e.

[tex]f'(x)= \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}[/tex]

Hence, on putting the value in the formula:

[tex]f'(x)= \lim_{h \to 0} \dfrac{5-x^2-h^2-2xh-(5-x^2)}{h}\\\\\\f'(x)=\lim_{h \to 0} \dfrac{5-x^2-h^2-2xh-5+x^2}{h}\\\\i.e.\\\\f'(x)=\lim_{h \to 0} \dfrac{-h^2-2xh}{h}\\\\f'(x)=\lim_{h \to 0} \dfrac{-h^2}{h}+\dfrac{-2xh}{h}\\\\f'(x)=\lim_{h \to 0} -h-2x\\\\i.e.\ on\ putting\ the\ limit\ we\ obtain:\\\\f'(x)=-2x[/tex]

      Hence, the derivative of the function f(x) is:

          [tex]f'(x)=-2x[/tex]

Answer:

The derivative of given function is -2x.

Step-by-step explanation:

The first principle of differentiation is

[tex]f'(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}[/tex]

The given function is

[tex]f(x)=5-x^2[/tex]

[tex]f'(x)=lim_{h\rightarrow 0}\frac{5-(x+h)^2-(5-h^2}{h}[/tex]

[tex]f'(x)=lim_{h\rightarrow 0}\frac{5-(x^2+2xh+h^2)-5+h^2}{h}[/tex]

[tex]f'(x)=lim_{h\rightarrow 0}\frac{5-x^2-2xh-h^2-5+h^2}{h}[/tex]

[tex]f'(x)=lim_{h\rightarrow 0}\frac{-x^2-2xh}{h}[/tex]

[tex]f'(x)=lim_{h\rightarrow 0}\frac{-x^2}{h}-\frac{2xh}{h}[/tex]

[tex]f'(x)=lim_{h\rightarrow 0}\frac{-x^2}{h}-2x[/tex]

Apply limit.

[tex]f'(x)=\frac{-x^2}{0}-2x[/tex]

[tex]f'(x)=0-2x[/tex]

[tex]f'(x)=-2x[/tex]

Therefore, the derivative of given function is -2x.

Answer: July

Step-by-step explanation:

Formula of z score :

[tex]z=\dfrac{X-\mu}{\sigma}[/tex]

Given: The mean high temperature in January = [tex]\mu_1=36^{\circ} F[/tex]

Standard deviation : [tex]\sigma_1=8^{\circ}F[/tex]

For X = [tex]57^{\circ}F[/tex]

[tex]z=\dfrac{57-36}{8}=2.625[/tex]

The mean high temperature in July = [tex]\mu_1=72^{\circ} F[/tex]

Standard deviation : [tex]\sigma_1=9^{\circ}F[/tex]

[tex]z=\dfrac{57-72}{8}=-1.875[/tex]

⇒ 57° F is about 2.6 standard deviations above the mean of January high temperatures, and  57° F is about 1.9 standard deviations below the mean of July’s high temperatures.

A general rule says that z-scores lower than -1.96 or higher than 1.96 are considered unusual .

Hence, the 57˚F is  more unusual in January.

Final answer:

A high temperature of 57 degrees is more unusual in January than in July, as it is 2.625 standard deviations above the January mean, compared to 1.667 standard deviations below the July mean.

Explanation:

To determine in which month it is more unusual to have a high temperature of 57 degrees Fahrenheit, we can calculate the z-score for each month. The z-score tells us how many standard deviations away from the mean a particular value is.

For January, the z-score is calculated as follows:

Z = (57 - 36) / 8 = 21 / 8 = 2.625

This means that a temperature of 57 degrees in January is 2.625 standard deviations above the January mean.

For July, the z-score is calculated as follows:

Z = (57 - 72) / 9 = -15 / 9 = -1.667

This means that a temperature of 57 degrees in July is 1.667 standard deviations below the July mean.

Since the absolute value of the January z-score (2.625) is higher than the absolute value of the July z-score (-1.667), a high temperature of 57 degrees is more unusual in January than in July.

Answer:

Answer is contained in the explanation

Step-by-step explanation:

[tex]g(x)=5-x\\g'(x)=\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\\g'(x)=\lim_{h \rightarrow 0} \frac{[5-(x+h)]-[5-x]}{h}\\g'(x)=\lim_{h \rightarrow 0} \frac{5-x-h-5+x}{h}\\g'(x)=\lim_{h \rightarrow 0} \frac{-h}{h}\\g'(x)=\lim_{h \rightarrow 0} -1\\g'(x)=-1[/tex]

g(x)=5-x has domain all real numbers (you can plug an a number and always get a number back)

So in interval notation this is [tex](-\infty, \infty)[/tex]

g'(x)=-1 has domain all real numbers (the original function had domain issues... and no matter the number you plug in you do get a number, that number being -1)

So in interval notation this is [tex](-\infty, \infty)[/tex]

The derivative of given function g(x) is

g'(x)=-1

Domain of function g(x) is (-∞,∞)

Domain of derivative is  (-∞,∞)

Given :

[tex]g(x) = 5 - x[/tex]

Lets find derivative using definition of derivative

[tex]\lim_{h \to 0} \frac{g(x+h)-g(x)}{h} \\g(x)=5-x\\g(x+h)=5-(x+h)\\g(x+h)=5-x-h\\\lim_{h \to 0} \frac{5-x-h-(5-x)}{h} \\\\\lim_{h \to 0} \frac{5-x-h-5+x}{h} \\\\\lim_{h \to 0} \frac{-h}{h} \\\\-1[/tex]

Derivative g'(x)=-1

g(x) is a linear function . for all linear function the domain is set of all real numbers

Domain of function g(x) is (-∞,∞)

Derivative function g'(x) =-1. For all values of x  the value of y is -1

So domain is set of all real numbers

Domain of derivative is  (-∞,∞)

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Answer:

In the account that paid 6% Susan invest [tex]\$6,000[/tex]

In the account that paid 5% Susan invest [tex]\$4,000[/tex]

Step-by-step explanation:

we know that

The simple interest formula is equal to

[tex]I=P(rt)[/tex]

where

I is the Final Interest Value

P is the Principal amount of money to be invested

r is the rate of interest  

t is Number of Time Periods

Part a) account that paid 6% simple interest per year

in this problem we have

[tex]t=1\ years\\ P=\$x\\r=0.06[/tex]

substitute in the formula above

[tex]I1=x(0.06*1)[/tex]

[tex]I1=0.06x[/tex]

Part b) account that paid 5% simple interest per year

in this problem we have

[tex]t=1\ years\\ P=\$10,000-\$x\\r=0.05[/tex]

substitute in the formula above

[tex]I2=(10,000-x)(0.05*1)[/tex]

[tex]I2=500-0.05x[/tex]

we know that

[tex]I1+I2=\$560[/tex]

substitute and solve for x

[tex]0.06x+500-0.05x=560[/tex]

[tex]0.01x=560-500[/tex]

[tex]0.01x=60[/tex]

[tex]x=\$6.000[/tex]

therefore

In the account that paid 6% Susan invest [tex]\$6,000[/tex]

In the account that paid 5% Susan invest [tex]\$4,000[/tex]

Susan invested $6,000 at 6% and the remainder, $4,000, at 5% interest.

Susan invested $10,000 in two different accounts, one with a 6% simple interest and the other with a 5% simple interest. After one year, she received a total of $560 in interest. We need to find out how much she invested in each account.

Let's denote x as the amount invested at 6% and (10,000 - x) as the amount invested at 5%. Using the formula for simple interest, interest = principal × rate × time, we can set up two equations based on the given information:

The interest from the account with 6% interest: 0.06 × x

The interest from the account with 5% interest: 0.05 × (10,000 - x)

The sum of these interests is $560, so the equation is:

0.06x + 0.05(10,000 - x) = 560

Now we solve for x:

0.06x + 500 - 0.05x = 560

0.01x = 60

x = 60 / 0.01

x = $6,000

Therefore, Susan invested $6,000 at 6% and the remainder, $4,000, at 5% interest.

Answer:

The solutions of the system of equations are (-1,8) and (5,2)

Step-by-step explanation:

[tex]y=-x+7[/tex] -------> equation A (equation of a line)

[tex]y=0.5(x-3)^{2}[/tex] ----> equation B (vertical parabola open upward)

Solve the system of equations by graphing

Remember that the solution is the intersection points both graphs

using a graphing tool

The intersection points are (-1,8) and (5,2)

see the attached figure

therefore

The solutions of the system of equations are (-1,8) and (5,2)

Answer: (-1,8) and (5,2)

Step-by-step explanation: The person above me is correct. Give him five stars and a thanks!

Answer:  Probability that he will receive a grade of B or better in both courses is 0.30.

Step-by-step explanation:

Since we have given that

Probability that he will get a grade of B or better in introduction to Finance say P(A) = 0.6

Probability that he will get a grade of B or better in introduction to Accounting say P(B) = 0.5

Since A and B are independent events.

We need to find the probability that he will receive a grade of B or better in both the courses.

So, it becomes,

[tex]P(A\cap B)=P(A).P(B)\\\\P(A\cap B)=0.6\times 0.5\\\\P(A\cap B)=0.30[/tex]

Hence, Probability that he will receive a grade of B or better in both courses is 0.30.

The probability that you will receive a grade of B or better in both courses is 0.3.

The probability that you will receive a grade of B or better in both Introduction to Finance and Introduction to Accounting, given that the events are independent, is calculated by multiplying the individual probabilities of each event.

[tex]\( P(F) = 0.6 \) \( P(A) = 0.5 \)[/tex]

Since the events are independent, the probability of both events occurring is given by the product of their individual probabilities:

[tex]\( P(F \text{ and } A) = P(F) \times P(A) \)[/tex]

Substituting the given probabilities:

[tex]\( P(F \text{ and } A) = 0.6 \times 0.5 \) \( P(F \text{ and } A) = 0.3 \)[/tex]

The final answer is [tex]\(\boxed{0.3}\).[/tex]

Answer:

Step-by-step explanation:

I would formulate the problem like this. Let f, a, c represent the numbers of packages bought from Fred Motors, Admiral Motors, and Chrysalis, respectively. Then the function to minimize (in thousands) is …

  objective = 500f +400a +300c

The constraints on the numbers of cars purchased are …

  5f +5a +10c >= 700

  5f +10a +5c >= 600

  10f +5a +5c >= 700

Along with the usual f >=0, a>=0, c>=0. Of course, we want all these variables to be integers.

Any number of solvers are available in the Internet for systems like this. Shown in the attachments are the input and output of one of them.

The optimal purchase appears to be …

The total cost of these is $40 million.

This is a linear programming problem that requires to minimize a cost function subject to several constraints about the total number of small, medium, and large cars in the fleet. It can be set up using the system of inequalities and then solved using methods like the Simplex one.

This problem can be solved through linear programming, which involves creating a system of inequalities to represent the constraints of the problem, and then optimizing a linear function. To start, let's define the variables: x is the number of packages bought from Fred Motors, y is the number from Admiral Motors, and z is the number from Chrysalis.

The fleet requirements translate to the following constraints: 5x + 5y + 10z ≥ 700 (small cars), 5x + 10y + 5z ≥ 600 (medium cars), and 10x + 5y + 5z ≥ 700 (large cars).

Then, the cost to minimize is: $500,000x + $400,000y + $300,000z.

This is a linear programming problem and can be solved using various methods, such as the Simplex method or graphically. Exact solutions would require a more detailed analysis.

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[tex]\bf \stackrel{height}{h(t)}=-16t^2+32t+48\implies \stackrel{48~ft}{~~\begin{matrix} 48 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}=-16t^2+32t~~\begin{matrix} +48 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix} \\\\\\ 0=-16t^2+32t\implies 16t^2-32t=0\implies 16t(t-2)=0\implies t= \begin{cases} 0\\ 2 \end{cases}[/tex]

t = 0 seconds, when the ball first took off, and t = 2, 2 seconds later.

Answer: [tex]t_1=0\\t_2=2[/tex]

Step-by-step explanation:

We know that the function [tex]h(t)=-16t^2+32t+48[/tex] models the height  "h" of the ball above the ground as a function of time "t".

Then, to find the times in which the ball will be 48 feet above the ground, we need to substitute [tex]h=48[/tex] into the function and solve fot "t":

[tex]48=-16t^2+32t+48\\0=-16t^2+32t+48-48\\0=-16t^2+32t[/tex]

Factorizing, we get:

[tex]0=-16t(t-2)\\t_1=0\\t_2=2[/tex]

For this case we have that by definition, the GCF or (Greatest Common Factor) is given by the greatest common factor that divides both terms without leaving a residue.

15: 1,3,5,15

3: 1.3

Then we have the GCF of the expression is:

[tex]3t ^ 2s (t ^ 3-5s ^ 2)[/tex]

ANswer:

Option C