I need help with this!!!

Step-by-step explanation:

We will prove by mathematical induction that, for every natural [tex]n\geq 4[/tex],  

[tex]5^n\geq 2^{2n+1}+100[/tex]

We will prove our base case, when n=4, to be true.

Base case:

[tex]5^4=625\geq 612=2^{2*4+1}+100[/tex]

Inductive hypothesis:  

Given a natural [tex]n\geq 4[/tex],  

[tex]5^n\geq 2^{2n+1}+100[/tex]

Now, we will assume the induction hypothesis and then use this assumption, involving n, to prove the statement for n + 1.

Inductive step:

[tex]2^{2(n+1)+1}+100=2^{2n+1+2}+100=\\=4*2^{2n+1}+100\leq 4(2^{2n+1}+100)\leq 4*5^n<5^{n+1}[/tex]

With this we have proved our statement to be true for n+1.  

In conlusion, for every natural [tex]n\geq4[/tex].

[tex]5^n\geq 2^{2n+1}+100[/tex]

In summary, the raw data can only be retrieved from a dot plot, not a histogram, as the dot plot represents individual data points, whereas a histogram represents data in grouped intervals.

When comparing a dot plot to a histogram, there are several key points to consider:

A frequency table or a frequency polygon can also be useful tools for data representation in complement to histograms and dot plots, especially when comparing distributions.

Answer:

A. A straight line intersecting two parallel lines produces corresponding equal angles.

Step-by-step explanation:

If two straight lines are parallel, and a straight line intersects them, then we get alternate angles equal, exterior angle equal to the opposite interior angle on the same side and the interior angles on the same sides equal to two right angles.

Hence, here the rule that is not true is : A. A straight line intersecting two parallel lines produces corresponding equal angles.

Answer:

z=10

x1=0

x2=0

x3=3.33

Step-by-step explanation:

First Step convert your constraints in standard equations

so we have

x1 + 2x2 + 3x3+x4 = 10

x1 + x2 +x5= 5

x1 +x6= 1

Now we pass it all to the simplex table

Remember that we choose the column with the most negative value

Pivot Element=3

Divide all elements on Pivot Line by Pivot Element

Line x5= 0*Pivot Line +Line x5

Line x6= 0*Pivot Line+ Line X6

Line Z= 3* Pivot Line + Line Z

We finish when all the elements from the line Z are positive

Hence we have that x3=3.33 and x1=0, x2=0 and the max of z is 10

Answer:

z=10

x1=0

x2=0

x3=3.33

Step-by-step explanation:

Answer:

Length = Width = 7 inches

Height = 14 inches

Step-by-step explanation:

Let's call  

x = length and width of the bottom and top

y = heigth

The bottom and top are squares, so their total area is

[tex]A_{bt}=x^2+x^2=2x^2[/tex]

the area of one side is xy. As we have 4 sides, the total area of the sides is

[tex]A_s=4xy[/tex]

The cost for the bottom and top would be

[tex]C_{bt}=\$4.2x^2=\$8x^2[/tex]

The cost for the sides would be

[tex]C_s=\$2.4xy=\$8xy[/tex]

The total cost to made the box is

[tex]C_t=\$(8x^2+8xy)[/tex]

But the volume of the box is  

[tex]V=x^2y=686in^3[/tex]

isolating

[tex]y=\frac{686}{x^2}[/tex]

Replacing this expression in the formula of the total cost

[tex]C_t(x)=8x^2+8x(\frac{686}{x^2})=8x^2+\frac{5488}{x}[/tex]

The minimum of the cost should be attained at the point x were the derivative of the cost is zero.

Taking the derivative

[tex]C'_t(x)=16x-\frac{5488}{x^2}[/tex]

If C' = 0 then

[tex]16x=\frac{5488}{x^2}\Rightarrow x^3=\frac{5488}{16}\Rightarrow x^3=343\Rightarrow x=\sqrt[3]{343}=7[/tex]

We have to check that this is actually a minimum.

To check this out, we take the second derivative

[tex]C''_t(x)=16+\frac{10976}{x^3}[/tex]

Evaluating this expression in x=7, we get C''>0, so x it is a minimum.

Now that we have x=7, we replace it on the equation of the volume to get

[tex]y=\frac{686}{49}=14in[/tex]

The dimensions of the most economical box are

Length = Width = 7 inches

Height = 14 inches

To find the dimensions of the container of least cost, assume the length of the base is x and the width of the base is also x. The height of the container is 686/(x^2). The dimensions of the container of least cost are approximately 10.33 inches by 10.33 inches by 6.68 inches.

To find the dimensions of the container of least cost, we need to minimize the cost function. Let's assume the length of the base is x, then the width of the base will also be x since it's a square. The height of the container will be 686/(x^2) since the volume is given as 686 in³. The total cost, C, is given by C = 4(2x^2) + 2(4x(686/(x^2))). Simplifying this expression gives C = 8x^2 + 5488/x. To find the minimum cost, we can take the derivative of C with respect to x and set it equal to zero:

dC/dx = 16x - 5488/x^2 = 0

Solving this equation gives x = √(343/2) ≈ 10.33. Therefore, the dimensions of the container of least cost are approximately 10.33 inches by 10.33 inches by 6.68 inches.

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Final answer:

The sum of sequence P (YP) is greater than the sum of sequence Q (Q) by 94, with P summing to 2850 and Q summing to 2756.

Explanation:

To determine which sum is greater between P and Q, we need to first understand that P and Q represent the sum of arithmetic sequences. The first sequence, P, begins with 8 and increments by 2, ending at 106. The second sequence, Q, starts at 2 and increments by 2, ending at 104.

To find out the sums, we can use the formula for the sum of an arithmetic sequence: S = n/2 (a1 + an), where S is the sum, n is the number of terms, a1 is the first term, and an is the last term. For the sequence P, the first term a1 is 8, and the last term an is 106. To find n, the number of terms, we can use the formula n = (an - a1) / d + 1, where d is the common difference, which is 2 in this case.

Applying the formula to the sequence P, we get:

The number of terms in P: n(P) = (106 - 8) / 2 + 1 = 50

Sum of P: S(P) = 50/2 x (8 + 106) = 25 x 114 = 2850

For the sequence Q:

The number of terms in Q: n(Q) = (104 - 2) / 2 + 1 = 52

Sum of Q: S(Q) = 52/2 x (2 + 104) = 26 x 106 = 2756

Comparing these sums, we can see that the sum of sequence P, 2850, is greater than the sum of sequence Q, 2756. The difference between the sums is 2850 - 2756 = 94.

Therefore, P is greater than Q by 94.

The conversion of 1.256 grams is as follows: 1,256,000 micrograms, 1256 milligrams, and 0.001256 kilograms.

To convert 1.256 grams (g) to different units of mass, you use the following conversion factors:

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1.256 grams is equal to 1,256,000 micrograms, 1,256 milligrams, and 0.001256 kilograms. These conversions use basic multiplication and division of the metric system.

To convert 1.256 grams to different units, you should be familiar with the respective conversion factors:

Steps to Convert:

Therefore, the conversions are:

Answer:

The probability that exactly 5 are unable to complete the  race is 0.1047

Step-by-step explanation:

We are given that 25% of all who enters a race do not complete.

30 have entered.

what is the probability that exactly 5 are unable to complete the  race?

So, We will use binomial

Formula : [tex]P(X=r) =^nC_r p^r q^{n-r}[/tex]

p is the probability of success i.e. 25% = 0.25

q is the probability of failure =  1- p  = 1-0.25 = 0.75

We are supposed to find the probability that exactly 5 are unable to complete the  race

n = 30

r = 5

[tex]P(X=5) =^{30}C_5 (0.25)^5 (0.75)^{30-5}[/tex]

[tex]P(X=5) =\frac{30!}{5!(30-5)!} \times(0.25)^5 (0.75)^{30-5}[/tex]

[tex]P(X=5) =0.1047[/tex]

Hence the probability that exactly 5 are unable to complete the  race is 0.1047

Answer:

OBSERVATIONAL

Step-by-step explanation:

Let`s see the two kind of experiments mentioned:

In this case, the patient's parent only answer a questionnairewith no intervention from the doctors into the pacients at all, so it's an observational  experiment

Answer:

9.3125 hours

Step-by-step explanation:

Given:

Number of components consisting in a product = 25

Standard time for work cycle = 7.45 minutes

or

standard time for work cycle = [tex]\frac{\textup{7.45}}{\textup{60}}[/tex] hours

Number of units completed = 75

Now,

The number of standard hours

= Number of units completed × standard time for the work cycle

=  [tex]75\times\frac{\textup{7.45}}{\textup{60}}[/tex] hours

or

The number of standard hours = 9.3125 hours

Answer:

The answer is 9.3125 hours :)

Step-by-step explanation:

Answer:

0001 0110 1110 1011 0010

Step-by-step explanation:

First decimal number = 11540

second decimal number = 5972

BCD of 0 to 9 decimal number

Decimal number                               BCD

       0                                                  0000

       1                                                    0001

       2                                                   0010

       3                                                   0011

       4                                                   0100

       5                                                   0101

       6                                                   0110

       7                                                    0111

       8                                                    1000

       9                                                    1001

Hence, we can write above two numbers in BCD form as

11540 => 0001 0001 0101 0100 0000

5972 => 0101 1001 0111 0010

So, sum of 11540 and 5972 in BCD form can be given by

       0001 0001 0101 0100 0000

        +       01 01 1001 01 11  001 0

      0001  01 10 11 10 10 11  001 0

So, the sum of 11540 and 5972 in BCD form is

                  0001 0110 1110 1011 0010

Answer:

Step-by-step explanation:

First you would find the area of the lot:

50 x 36 = 1800 m^2

Then you would find the area of the house:

31 x 9 = 279 m^2

Then you subtract the are of the lot minus the area of the house:

1800 - 279 = 1521 m^2

And the final answer is:

1521 square meters is left

The area of space which is left over is [tex]1521m^{2} [/tex]

Required steps shown below,

Area of lot [tex]=length*width[/tex]

                  [tex]=50*36=1800m^{2} [/tex]

Area of house[tex]=length*width[/tex]

                       [tex]=31*9=279m^{2} [/tex]

To calculate area of space is left over, subtract area of house from area of lot.

The space left over is,

                                [tex]1800-279=1521m^{2} [/tex]

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Answer: time = 55.48 years

Explanation:

Given:

Principal amount = $10000

Interest rate = 2% p.a

Amount = $30000

We can evaluate the time taken using the following formula:

[tex]Amount=Principal(1+\frac{r}{100})^{t}[/tex]

[tex]30000 = 10000\times(1+\frac{2}{100})^{t}[/tex]

Solving the above equation, we get

[tex]1.02^{t} = 3[/tex]

Now taking log on both sides, we get;

[tex]t\ log(1.02) = log(3)[/tex]

time = 55.48 years

Answer:

a) False b) True c) False d ) False e) True f) False g) False h) False

Step-by-step explanation:

a) False

x² × x³ = [tex]x^5[/tex]

When we multiply two exponential number with same base, their powers add up.

b) True

2π + 5π = π(2+5) = 7π

The coefficients of π are added together.

c) False

x² ≥ 0. For any value of x, negative or positive x² is always positive. But for x = 0, x² = 0×0 = 0

d) False

315 - 8 = 307, which is clearly an odd number.

e)True

The sum of all integers is always even.

Let m and n be two odd integers.

Thus, they can be expressed as m = 2r + 1 and n = 2s +1, where r and s are even integers.

m + n = 2r +2s + 2, which is clearly even.

f) False

Since √2 is an irrational number. It cannot belong to z, which is collection of all integer number.

√2 ∉ z

g) False

Since -1 is a negative integer, it cannot belong to [tex]z^+[/tex], as it is collection of all positive integers.

h) False

π cannot belong to Q because Q is a collection of all rational numbers and π is not a rational number. The decimal expansion of π is non- terminating that is it does not end.

Answer:

Venom C is  0.625 times as potent as venom A.

Step-by-step explanation:

We are given that venom  A is four times as potent as venom B.

Venom C is 2.5 times as potent as Venom B.

We have to find that how many times as potent is venom C is compared to venom A.

Let Venom B =x

Then Venom A=[tex]4\times x=4x[/tex]

Venom C=[tex]2.5\times x=2.5 x[/tex]

[tex]\frac{Venom\;C}{venom\;A}=\frac{2.5x}{4x}=\frac{25}{40}=\frac{5}{8}=0.625[/tex]

Hence, venom C is  0.625 times as potent as venom A.

In relative terms, venom C is 0.625 times as potent as venom A, using venom B as a common comparison base.

To solve this, we first need to establish a common comparison base. If venom A is four times as potent as venom B, let's assign a relative potency value to venom B, let's say 1. Therefore, the potency of venom A is 4. Similarly, if venom C is 2.5 times as potent as venom B, then the potency of venom C is 2.5.

To find out how much more potent venom C is compared to venom A, we divide the potency of venom C by the potency of venom A. Mathematically, it looks like this: Potency of C ÷ Potency of A = 2.5 ÷ 4 = 0.625. Therefore, venom C is 0.625 times as potent as venom A.

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Final Answer:

No, it is not possible to find a sequence of m−1 consecutive positive integers none of which is prime for any integer m≥2.

Explanation:

In order to determine whether it is possible to find a sequence of m−1 consecutive positive integers none of which is prime, we need to consider the properties of prime numbers. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. Therefore, if we take any m−1 consecutive positive integers, at least one of them will be divisible by a prime number. This means that it is not possible to find a sequence of m−1 consecutive positive integers none of which is prime.

For example, let’s consider the sequence of 4 consecutive positive integers: 10, 11, 12, and 13. Among these numbers, 11 and 13 are prime. Even if we consider larger sequences, we will always encounter at least one prime number within the consecutive positive integers.

In conclusion, due to the nature of prime numbers and their distribution among positive integers, it is not possible to find a sequence of m−1 consecutive positive integers none of which is prime for any integer m≥2.

Answer: y = v₀tgθx - gx²/2v₀²cos²θ

a = v₀tgθ

b = -g/2v₀²cos²θ

Step-by-step explanation:

x = v₀ₓt

y = v₀y.t - g.t²/2

x = v₀.cosθt → t = x/v₀.cosθ

y = v₀y.t - g.t²/2

v₀y = v₀.senθ

y = v₀senθ.x/v₀cosθ - g/2.(x/v₀cosθ)²

y = v₀.tgθ.x - gx²/2v₀²cos²θ

a = v₀tgθ → constant because v₀ and θ do not change

b = - g/2v₀²cos²θ → constant because v₀, g and θ do not change

Final answer:

The trajectory of a projectile is shown to be parabolic by substituting time from the x-direction motion equation into the y-direction motion equation and rearranging, yielding a parabolic form y = ax + bx², with constants determined by initial velocity and gravity.

Explanation:

To prove that the trajectory of a projectile is parabolic, we start with the equations of motion in the x and y directions for a projectile that starts at the origin. For the x-direction, we have x = v_0x t, where v_0x is the initial velocity component in the x-direction and t is the time.

In the y-direction, the equation is y = v_0y t - (1/2) g t², where v_0y is the initial velocity component in the y-direction and g is the acceleration due to gravity.

To find t from the x equation: t = x / v_0x. Substituting this into the y equation yields: y = v_0y (x / v_0x) - (1/2) g (x / v_0x)^2.

Now, simplifying we get: y = (v_0y / v_0x) x - (g/2 v_0x^2) x^2, which is of the form y = ax + bx². Here, a = v_0y/v_0x and b = -g/2 v_0x²are constants depending on the initial velocity components and acceleration due to gravity. The equation y = ax + bx² represents a parabolic path, confirming that the projectile's trajectory is indeed a parabola.

Answer:

Area = 8

Step-by-step explanation:

A skecth is given in the attached file, there are two extra lines used to calculate the area with simple geometry:

We must use a double integral to obtain the area:

[tex]\int\limits^2_0 {\int\limits^b_a  \, dy } \, dx[/tex]

Where

b stands for y=2x+1

a stands for y=-x

Carring out the integrals we find the area:

[tex]\int\limits^2_0 {(2x+1 - (-x))} \, dx =  \int\limits^2_0 {3x+1} \, dx = (3x^{2}/2+x) \left \{ {{2} \atop {0}} \right.\\ A =( 3*2^{2}/2) + 2 =8[/tex]

Geometrically we can divide the area bounded by this lines as two triangles and a rectangle from the figure and the intersection of these lines we kno that the three figures have a base of 2. The heigth of the rectangle is 1 and for the triangles we have 4 for the upper triangle and 2 for the lower.

Therefore:

[tex]A = A_{upperT}+ A_{rect}+ A_{lowerT}[/tex]

and

[tex]A_{upperT}=2*4/2=4\\A_{rect}=2*1=2\\ A_{lowerT}=2*2/2=2[/tex]

Summing the four areas we have:

A=8

Greeting!

Answer:

The expended form of the provided expression is: [tex]32x^5-80x^4+80x^3-40x^2+10x-1[/tex]

Step-by-step explanation:

Consider the provided expression.

[tex](2x-1)^5[/tex]

The binomial theorem:

[tex](a+b)^{n}=\sum _{r=0}^{n}{n \choose r}a^{n-r}b^r[/tex]

Where,

[tex]{n \choose r}= ^nC_r =\frac{n!}{(n-r)!r!}[/tex]

Now by using the above formula.

[tex]\frac{5!}{0!\left(5-0\right)!}\left(2x\right)^5\left(-1\right)^0+\frac{5!}{1!\left(5-1\right)!}\left(2x\right)^4\left(-1\right)^1+\frac{5!}{2!\left(5-2\right)!}\left(2x\right)^3\left(-1\right)^2+\frac{5!}{3!\left(5-3\right)!}\left(2x\right)^2\left(-1\right)^3+\frac{5!}{4!\left(5-4\right)!}\left(2x\right)^1\left(-1\right)^4+\frac{5!}{5!\left(5-5\right)!}\left(2x\right)^0\left(-1\right)^5[/tex]

[tex]2^5\cdot \:1\cdot \:1\cdot \:x^5-1\cdot \frac{2^4\cdot \:5x^4}{1!}+1\cdot \frac{2^3\cdot \:20x^3}{2!}-1\cdot \frac{2^2\cdot \:20x^2}{2!}+1\cdot \frac{5\cdot \:2x}{1!}+1\cdot \frac{\left(-1\right)^5}{\left(5-5\right)!}[/tex]

[tex]32x^5-80x^4+80x^3-40x^2+10x-1[/tex]

Hence, the expended form of the provided expression is: [tex]32x^5-80x^4+80x^3-40x^2+10x-1[/tex]

Answer: There are 90 snack boxes.

Step-by-step explanation:

Given : The number of kinds of fruits = 5

The number of kinds of herbal teas = 3

The number of kinds of flavors of wrap sandwich = 6

Then by using the fundamental principal of counting, the number of possible snack are there will be :_

[tex]5\times3\times6= 90[/tex]

Therefore, the number of possible snacks = 90

Answer:

You must pick at least 13 pennies to be sure of getting at least five from the same year.

Step-by-step explanation:

You have:

12 1967 pennies

7 1868 pennies

11 1971 pennies.

how many must you pick to be sure of getting at least five pennies from the same year?

This value is the multiplication of the number of different pennies by the antecessor of the number of pennies you want, added by 1.

So

You have 3 differennt pennies

You want to get at least five from the same year.

[tex]3*4 + 1 = 13[/tex]

You must pick at least 13 pennies to be sure of getting at least five from the same year.

For example, if you pick 12 pennies, you can have four from each year.

Adding three values, 13 is the smallest number that you need at least one term of the addition being equal or bigger than 5.

To be sure of getting at least five pennies from the same year, you need to pick 35 pennies in total.

To ensure that you get at least five pennies from the same year, you need to consider the worst-case scenario. In this case, the worst-case scenario is where you pick pennies from each of the three different years first before getting five from the same year. So, you need to pick the maximum number of pennies from each year first before reaching the desired goal. The maximum number of pennies you need to pick is:

12 + 7 + 11 + 5 = 35 pennies

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Answer:

y = -0.9x+10.1

Step-by-step explanation:

The equation of the line is:

[tex]y=mx+b[/tex]

You have been asked to stimate m and b. To do so, first find the product between each pair of x and y and the value of x squared:

[tex]\left[\begin{array}{cccc}x&y&x*y&x^2\\1&9&9&1\\3&7&21&9\\5&7&35&25\\7&3&21&49\\9&2&18&81\end{array}\right][/tex]

Then calculate the total sum of all columns:

[tex]\left[\begin{array}{cccc}x&y&x*y&x^2\\1&9&9&1\\3&7&21&9\\5&7&35&25\\7&3&21&49\\9&2&18&81\\\bold{25}&\bold{28}&\bold{104}&\bold{165}\end{array}\right][/tex]

m can be calculated following the next equation:

[tex]m=\frac{\frac{\sum{xy}-\sum{y}}{n}}{\sum{x^2}-\frac{(\sum{x})^2}{n}}[/tex]

where n is the number of (x, y) couples (5 in our case).

Replacing the values calculated previously:

[tex]m=\frac{104-\frac{25*28}{5} }{165-\frac{25^2}{5} }=\frac{104-\frac{700}{5} }{165-\frac{625}{5} } = \frac{104-140}{165-125 } = \frac{-36}{40} = -0.9[/tex]

For b:

[tex]b=\bar{y}- m\bar{x}=\frac{\sum{y}}{n}-m\frac{\sum{x}}{n}=\frac{28}{5}-(-0.9)\frac{25}{5}= \frac{28}{5}+\frac{22.5}{5}=\frac{50.5}{5}=10.1[/tex]

In the figure attached you can see the points given and the stimated line.

Answer:

a) 0.057

b) 0.5234

c) 0.4766

Step-by-step explanation:

a)

To find the p-value if the sample average is 185, we first compute the z-score associated to this value, we use the formula

[tex]z=\frac{\bar x-\mu}{\sigma/\sqrt N}[/tex]

where

[tex]\bar x=mean\; of\;the \;sample[/tex]

[tex]\mu=mean\; established\; in\; H_0[/tex]

[tex]\sigma=standard \; deviation[/tex]

N = size of the sample.

So,

[tex]z=\frac{185-175}{20/\sqrt {10}}=1.5811[/tex]

[tex]\boxed {z=1.5811}[/tex]

As the sample suggests that the real mean could be greater than the established in the null hypothesis, then we are interested in the area under the normal curve to the right of  1.5811 and this would be your p-value.

We compute the area of the normal curve for values to the right of  1.5811 either with a table or with a computer and find that this area is equal to 0.0569 = 0.057 rounded to 3 decimals.

So the p-value is  

[tex]\boxed {p=0.057}[/tex]

b)

Since the z-score associated to an α value of 0.05 is 1.64 and the z-score of the alternative hypothesis is 1.5811 which is less than 1.64 (z critical), we cannot reject the null, so we are making a Type II error since 175 is not the true mean.

We can compute the probability of such an error following the next steps:

Step 1

Compute [tex]\bar x_{critical}[/tex]

[tex]1.64=z_{critical}=\frac{\bar x_{critical}-\mu_0}{\sigma/\sqrt{n}}[/tex]

[tex]\frac{\bar x_{critical}-\mu_0}{\sigma/\sqrt{n}}=\frac{\bar x_{critical}-175}{6.3245}=1.64\Rightarrow \bar x_{critical}=185.3721[/tex]

So we would make a Type II error if our sample mean is less than 185.3721.  

Step 2

Compute the probability that your sample mean is less than 185.3711  

[tex]P(\bar x < 185.3711)=P(z< \frac{185.3711-185}{6.3245})=P(z<0.0586)=0.5234[/tex]

So, the probability of making a Type II error is 0.5234 = 52.34%

c)

The power of a hypothesis test is 1 minus the probability of a Type II error. So, the power of the test is

1 - 0.5234 = 0.4766

The p-value is 0.014, indicating strong evidence against the null hypothesis. The probability of a type II error is 0.0907, and the power of the test is 0.9093.

To find the p-value, we need to determine the probability of observing a sample average of 185 or higher, given that the true mean foam height is 175. Since the sample size is small, we'll use the t-distribution instead of the normal distribution. With a sample size of 10, the degrees of freedom is 9. Using the t-distribution table or a calculator, we find the p-value to be 0.014 (rounded to 3 decimal places).

The probability of a type II error is the probability of failing to reject the null hypothesis when it is actually false. In this case, the null hypothesis is μ = 175, but the true mean foam height is 200. We assume α = 0.05, so the critical value is 1.645 (from the t-distribution table for a one-tailed test). Using the formula for the standard error of the sample mean, σ/√n, we can calculate the standard deviation of the sample mean to be 20/√10 = 6.32 (rounded to 2 decimal places). The difference between the critical value and the true mean is (200 - 175)/6.32 = 3.96 (rounded to 2 decimal places). Using a t-distribution table or calculator to find the area to the right of 3.96 with 9 degrees of freedom, we find the probability of a type II error to be 0.0907 (rounded to 4 decimal places).

The power of the test is 1 minus the probability of a type II error. So the power of the test is 1 - 0.0907 = 0.9093 (rounded to 4 decimal places).

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Final answer:

Explaining the impact of selling additional sandwiches on MTA's profits.

Explanation:

To prepare the schedule showing the impact on MTA's profits:

Calculate the total income from selling 800 additional sandwiches: $2.85 x 800 = $2,280.

Calculate the total cost of producing 800 sandwiches: $5 x 800 = $4,000.

Subtract the cost from the income to find the profit impact: $2,280 - $4,000 = lower profit.

Explanation (150 words): Adding 800 sandwiches at a cost of $5 each while selling at $2.85 results in a loss, impacting MTA's profits negatively. This shift reduces the overall profitability due to the increased production costs outweighing the revenue generated from the additional sandwich sales.

Answer:   100

Step-by-step explanation:

Given : The lodhl diner offers a meal combination consisting of an appetizer, a soup, a main course, and a dessert.

There are 5 appetizers, 5 soups, 4 main courses, and 5 desserts.

Also, a dessert and a appetizer are not allowed to take together.

By Fundamental counting principal ,

Number of three-course meals with dessert and without appetizer :

[tex]5\times4\times5=100[/tex]      (1)

Number of three-course meals with appetizer and without dessert :

[tex]5\times5\times4=100[/tex]      (2)

Now, the number of meals with either dessert or appetizer :-

[tex]100+100=200[/tex]           [Add (1) and (2)]

Answer:

500

Step-by-step explanation:

5x5x4x5

we do

5x5=25

25x4=100

100x5= 500

Answer:

The posterior probability that the lathe tool is properly adjusted is 94.7%

Step-by-step explanation:

This can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

[tex]P = \frac{P(B).P(A/B)}{P(A)}[/tex]

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

In your problem we have that:

-A is the probability that the part chosen is found to be acceptable.

The problem states that the probability its lathe tool is properly adjusted is 0.8. When it happens, there is a 0.9 probability that the parts produced pass inspection. There is also a 0.2 probability of  the lathe is out of adjustment, when it happens  the probability of a good part being produced is only 0.2.

So, P(A) = P1 + P2 = 0.8*0.9 + 0.2*0.2 = 0.72 + 0.04 = 0.76

Where P1 is the probability of a good part being produced when lathe tool is properly adjusted and P2 is the probability of a good part being produced when lathe tool is not properly adjusted.

- P(B) is the the probability its lathe tool is properly adjusted. The problem states that P(B) = 0.8

P(A/B) is the probability of A happening given that B has happened. We have that A is the probability that the part chosen is found to be acceptable and B is the probability its lathe tool is properly adjusted. The problem states that when the lathe is properly adjusted, there is a 0.9 probability that the parts produced pass inspection. So P(A/B) = 0.9

So, probability of B happening, knowing that A has happened, where B is the lathe tool is properly adjusted and A is that the part randomly chosen is inspected and found to be acceptable is:

[tex]P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.8*0.9}{0.76} = \frac{0.72}{0.76} = 0.947 = 94.7%[/tex]

The posterior probability that the lathe tool is properly adjusted is 94.7%

Using Bayes' Theorem, the posterior probability that the lathe tool is properly adjusted given a part passes inspection is approximately 0.947 (94.7%). We determined this by combining the prior probabilities and the conditional probabilities to find the total probability of a part passing inspection.

To find the posterior probability, we can use Bayes' Theorem. Let's define the events:

Prior Probabilities

The probability that the lathe tool is properly adjusted is 0.8. Therefore, P(A) = 0.8 and P(A') = 0.2.

Conditional Probabilities

Applying Bayes' Theorem:

Bayes' Theorem states:

First, we need to find P(B), the total probability that a part passes inspection:

Substituting values:

Now, using Bayes' Theorem, we can calculate P(A|B):

Thus, the posterior probability that the lathe tool is properly adjusted given that a part is found to be acceptable is approximately 0.947 (94.7%).

Answer:

Yes, rational numbers are closed under addition.

Step-by-step explanation:

Rational numbers are number that can be expressed in the form of fraction [tex]\frac{x}{y}[/tex], where x and y are integers and y ≠ 0.

Now, the closure property of addition of rational number states that if we add two rational number, then the sum of these two rational number will also be a rational number.

Let a and b be two rational number, then,

a+b = c, where c is the sum of and b

c is also a rational number.

Thus, rational numbers are closed under addition.

This can be explained with the help of a example.

[tex]\frac{1}{7} + \frac{2}{7}[/tex] = [tex]\frac{2}{7}[/tex]

It is clear that [tex]\frac{1}{7}, \frac{2}{7}, \frac{3}{7}[/tex] are rational numers.

Thus, rational numbers are closed under addition.

Answer:

K=6

Step-by-step explanation:

we know that

A function is: a relation from a set of inputs to a set of possible outputs where each input is related to exactly one output

so

In this problem

If the x-coordinate of the first point is equal to the x-coordinate of the second point, then the relation will not be a function

Remember that the x-coordinate is the input value and the y-coordinate is the output value

Equate the x-coordinates and solve for k

[tex]1.5k-4=-0.5k+8[/tex]

[tex]1.5k+0.5k=8+4[/tex]

[tex]2k=12[/tex]

[tex]k=6[/tex]

therefore

If the value of k is 6 the, the relation will not be a function

Final answer:

To determine when the relation A is not a function, set the first elements of the ordered pairs equal to each other. Solving for k, we find that k must equal 6 to make the relation not a function, as this value of k would result in the same input for both pairs, violating the definition of a function.

Explanation:

For a relation to be considered a function, each input value must be associated with exactly one output value. Looking at the provided relation A = {(1.5k-4, 7), (-0.5k+8, 15)}, we can see that there are two ordered pairs. For A to not be a function, the first element of both ordered pairs must be the same because this would mean that a single input is associated with two different outputs, which violates the definition of a function.

We set the first elements equal to each other to find the value of k that would make the relation not a function: 1.5k - 4 = -0.5k + 8. Solving this equation, we add 0.5k to both sides and add 4 to both sides to obtain 2k = 12, and then divide both sides by 2 to find that k = 6.

Therefore, when k equals 6, the relation A will not be a function because both ordered pairs will have the first element equal to 5, which corresponds to two different output values (7 and 15).

Answer: a. You must pay $11.19 for the item.

b.The price you would expect to pay would be $36.59

Step-by-step explanation:

Hi, for first the question you need to calculate the 30 percent of the price of the item, and then subtract that result to the original price.

So: $15.99 × 0,30 = $4.797

$15.99 - $4.797 = $11.19

You must pay $11.19 for the item.

It´s a similar resolution, first you calculate the 40 percent of the retail price, and then subtract that result to the retail price.

So:

$60.99 × 0,40= $24.396

$60.99 - $24.396 = $36.59

The price you would expect to pay would be $36.59

Answer:

The concentration of the drug is 50mg/mL.

Step-by-step explanation:

This problem can be solved by a rule of three.

In a rule of three problem, the first step is identifying the measures and how they are related, if their relationship is direct of inverse.

When the relationship between the measures is direct, as the value of one measure increases, the value of the other measure is going to increase too.

When the relationship between the measures is inverse, as the value of one measure increases, the value of the other measure will decrease.

In this problem, as the number of mg increases, so does the number of mL. It means the we have a direct rule of three.

Tha problem states that the antibody is available in a prefilled syringe containing 40 mg/0.8 mL. Calculate the concentration of drug on a mg/mL basis.

The problem wants to know how many mg are there in a mL of the drug.

40mg - 0.8mL

x mg - 1 mL

0.8x = 40

[tex]x = \frac{40}{0.8}[/tex]

x = 50mg.

The concentration of the drug is 50mg/mL.