Can someone help me I need it to be broken down

Answer:

tomatoes = 1.28800 kg

Step-by-step explanation:

12kg = 7kg 670g + 3kg 42gm + x

                   add together

12kg = 10kg 712gm + x

                subtract

1.28800 kg = x

The function F(x)'s valid domain is therefore (-, -3), (-3, -2), and (-2, ).

We enter each value of x into the function and simplify as follows in order to evaluate the function F(x) at the values of the supplied domain:

\(F(-3) = (2(-3) + 6)/((-3)^2 + 5(-3) + 6) = 0/0\) (undefined) (undefined)

\(F(-2) = (2(-2) + 6)/((-2)^2 + 5(-2) + 6) = 0/0\)(undefined) (undefined)

\(F(0) = (2(0) + 6)/(0^2 + 5(0) + 6) = 1\)

\(F(2) = (2(2) + 6)/(2^2 + 5(2) + 6) = 2/3\)

\(F(3) = (2(3) + 6)/(3^2 + 5(3) + 6) = 2/4 = 1/2\)

The function is undefined at x=-3 and x=-2 because the fraction's

denominator is zero at these points, and division by zero is undefined, as can be seen from the aforementioned evaluations. All real numbers, with the exception of -3 and -2, fall inside the given function's valid domain.

The function F(x)'s valid domain is therefore (-, -3), (-3, -2), and (-2, ).

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Answer:

x= 25+20y

Step-by-step explanation:

De acuerdo a la situación planteada, la expresión algebraica debe indicar que 25€ que es el valor por la visita más 20€ multiplicado por el número de horas trabajadas es igual al dinero total que se debe pagar. La expresión es:

x= 25+20y, donde

x= valor total a pagar

y= número de horas trabajadas

The probability that i reach into the jar and randomly grab a dime and then, without replacement a quarter is equals to the 0.0807.

We have a jar which contains the following:

Number of pennies = 14

Number of dimes = 16

Number of nickels = 17

Number of quarters = 28

So, total number of coins in jar= 75 ( 14+16+17+28)

Now, randomly, a coin is grabbed from jar. So, probability of grabbing coin is dim = favourable outcomes/total possible outcomes, P₁

= 16/75 = 0.2133

After it without replacement one another coin will grab from jar, so total possible outcomes or coins in jar = 75 - 1 = 74

Thus, the probability that this time grabbed coin is quarter, P₂ = 28/74 = 0.3784

As we see events in P₁ and P₂ both are independent. So, the total probability that first randomly grab a dime and then, without replacement, a quarter = P₁ × P₂

= 0.3784 × 0.2133

= 0.08071

Hence, the required probability is 0.0807.

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Answer:

$4.7

Step-by-step explanation:

Answer:

uumm

Step-by-step explanation:

On solving the provided question, we can say that the relationship between the domain and range of f(x) = \(y = 9x + 1 = 10 + (x-1)(9)\)

Domain of a function is the set of possible values that it can accept. X-values of a function like f are represented by these integers (x). A function's domain is the set of possible values on which it can be used. Set the value that the function returns after the insertion of the x value. Y = f is the definition of a function with x as the independent variable and y as the dependent variable (x). A value of x is said to be in a function's domain if it can be successfully utilised to produce a single value of y by using the value of x.

\(f(x) = y = 9x + 1 = 10 + (x-1)(9)\\f(1) = 9+1 = 10\\f(2) = 9(2)+1 = 19\\f(3) = 9(3)+1 = 28\\f(4) = 9(4)+1 = 37\)

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Answer:

16 and 12

Step-by-step explanation:

A.1 IN=2FT

8IN =2(8)FT

= 16

B.1 IN=2FT

6IN=2(6)FT

=12 FT

Answer:

n=-0.769

Step-by-step explanation:

Since there is no number after the equal sign in the equation given, we can put a zero in that place.

117n+90=0

Subtract 90 from both sides.

117n=-90

Divide 117 and -90

n=-0.769

Hope this helps!

If not, I am sorry.

True statements are (a), (b) and (e)

a. True. The rules in a recursively defined set ensure that each new element is unique and different from any previously generated element.

b. True. In a structural induction proof, you need to show that the statement holds for the initial population (base case) and that it is passed on through the recurrence relations (inductive step) that create new elements from old elements.

c. False. To prove a statement P(n) for all natural numbers n, you need to use mathematical induction, which consists of proving the base case (usually P(1)) and the inductive step (assuming P(k) is true, then showing P(k+1) is true).

d. False. In a structural induction proof, you need to show P(n) for all n in the initial population (base case) and that whenever P(n) is true for some n, P(n+1) is also true (inductive step).

e. True. Induction is a special case of structural induction where the set being considered is the set of natural numbers. Structural induction is a more general concept that applies to recursively defined sets in general.

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Answer:

49.50

Step-by-step explanation:

All you do is turn the fraction into a decimal so:

1/2 = .50

so the answer would be 49.50

Ace Carlos

Answer:

3x + 7 = -20

3x = -20-7

3x = -27

3x ÷ 3 = x

- 27 ÷ 3 = -9

x= -9

a. The probability that a flaw is detected by the end of the second fixation is given by the formula: P(flaw is detected by the end of the second fixation) = 1 - P(flaw is not detected in first fixation) * P(flaw is not detected in second fixation).

b. Similarly, the probability that a flaw will be detected by the end of the nth fixation is given by the formula: P(flaw is detected by the nth fixation) = 1 - P(flaw is not detected in first fixation) * P(flaw is not detected in second fixation) * ... * P(flaw is not detected in n-th fixation).

c. To calculate the probability that a flawed item will pass inspection, we can use the formula: P(B'|A), where A is the event that an item has a flaw and B is the event that the item passes inspection. Thus, P(B'|A) is the probability that the item passes inspection given that it has a flaw. Since the item is passed if a flaw is not detected in the first three fixations, and the probability that a flaw is not detected in any one fixation is 1 - p, we have P(B'|A) = P(flaw is not detected in first fixation) * P(flaw is not detected in second fixation) * P(flaw is not detected in third fixation) = (1 - p)³.

d. To find the probability that an item is chosen at random and passes inspection, we can use the formula: P(C) = P(item is not flawed and passes inspection) + P(item is flawed and passes inspection). We can calculate this as (1 - 0.1) * 1 + 0.1 * P(B|A'), where A' is the complement of A. Since P(B|A') = P(flaw is not detected in first fixation) * P(flaw is not detected in second fixation) * P(flaw is not detected in third fixation) = (1 - p)³, we have P(C) = 0.91 + 0.1 * (1 - p)³.

e. It's important to note that all of these formulas assume certain conditions about the inspection process, such as the number of fixations and the probability of detecting a flaw in each fixation. These assumptions may not hold in all situations, so the results obtained from these formulas should be interpreted with caution.

The given problem deals with calculating the probability that an item is flawed given that it has passed inspection. Let us define the events, where D denotes the event that an item has passed inspection, and E denotes the event that the item is flawed.

Using Bayes’ theorem, we can calculate the probability that an item is flawed given that it has passed inspection. That is, P(E|D) = P(D|E) * P(E) / P(D). Here, P(D|E) is the probability that an item has passed inspection given that it is flawed. P(E) is the probability that an item is flawed. And, P(D) is the probability that an item has passed inspection.

Since the item is passed if a flaw is not detected in the first three fixations, we can find P(D|E) = (1 - p)³. Also, given that 10% of all items contain a flaw, we have P(E) = 0.1.

Now, to find P(D), we can use the law of total probability. P(D) = P(item is not flawed and passes inspection) + P(item is flawed and passes inspection). This is further simplified to (1 - 0.1) * 1 + 0.1 * (1 - p)³.

Finally, we have P(E|D) = (1 - p)³ * 0.1 / [(1 - 0.1) * 1 + 0.1 * (1 - p)³], where p = 0.5. Therefore, we can use this formula to calculate the probability that an item is flawed given that it has passed inspection.

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To compare the values - 1.4 and - 1.1 using a number line, the correct expression for the comparison will be the less than sign, Hence, expressed as - 1.4 < -1.1

On a number line, negative values decreases as we move to the left of the number line ; Hence, values which are situated to the leftmost part are lesser than those to their right.

The value - 1.4 tends more to the left than the value - 1.1 ; therefore, - 1.4 is less than - 1.1

Therefore, the correct expression which shows the relation is - 1.4 < - 1.1

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Answer:

3x + 3 = -2x + 13 doesn't

Step-by-step explanation:

Answer:

(x-3) (x-2)

Step-by-step explanation:

\(x^{2}\) - 5x + 6

How to break down the equation and factorise it:

-3 x -2 = 6

-3 + -2 = -5

Final Answer:

(x-3) (x-2)

it can be concluded that the population standard deviation is within or less than 12.

To construct a hypothesis test to determine whether the population standard deviation of σ≦12 should be rejected for Costco, we can use a chi-square test for variance.

Step 1: State the null and alternative hypotheses:
- Null hypothesis (H₀): σ ≤ 12
- Alternative hypothesis (H₁): σ > 12

Step 2: Determine the level of significance (α = 0.05) and degrees of freedom (df = n - 1 = 11 - 1 = 10).

Step 3: Calculate the test statistic:
- χ² = (n - 1) * (s² / σ²) = 10 * (11.3² / 12²) = 10 * 0.94 = 9.4

Step 4: Determine the critical value:
- The critical value at α = 0.05 with df = 10 is χ²ₐ = 18.307

Step 5: Compare the test statistic with the critical value:
- Since χ² = 9.4 < χ²ₐ = 18.307, we fail to reject the null hypothesis.

Step 6: Conclusion:
- Based on the given sample data, there is not enough evidence to reject the hypothesis that the population standard deviation of σ≤12 for Costco.

Therefore, it can be concluded that the population standard deviation is within or less than 12.

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Answer:

b

Step-by-step explanation:

12x12

11x11

10x10

9x9

8x8

7x7

6x6

5x5

4x4

3x3

2x2

1x1

add em together equal 650

If this is a sample of 6 scores, then  SS using the definitional formula would equal 17.5.

To find the SS (sum of squares) using the definitional formula, you need to first calculate the mean of the scores. Here's

1. Calculate the mean (µ) using ∑x and the number of scores (n):
Mean (µ) = (∑x) / n
µ = 39 / 6
µ = 6.5

2. Use the computational formula for SS:
SS = ∑x² - ( (∑x)² / n )
SS = 271 - (39² / 6)
SS = 271 - (1521 / 6)
SS = 271 - 253.5

3. Calculate sample score SS:
SS = 17.5

So, the answer is 17.5.

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The population of an island in 1950 was 2 million. The population grew exponentially for 63 years and reached 6.5 million in 2013. The carrying capacity of the island is estimated to be 10 million.

If the population growth rate in the future is similar to the last 50 years, what will the population be in 2050

The population is given to be increasing exponentially, which means it will follow the equation:

\($P(t) = P_0 e^{rt}$\)Here,\($P(t)$\) is the population after a period of time \($t$, $P_0$\) is the initial population, $r$ is the annual growth rate (which we are given is the same as the growth rate of the last 50 years), and \($t$\) is the time.

We can find the annual growth rate $r$ using the formula:\($$r = \frac{\ln{\frac{P(t)}{P_0}}}{t}$$\)
We know\($P_0 = 2$ million, $P(t) = 6.5$ million, and $t = 63$\) years. Substituting these values, we get:

\($r = \frac{\ln{\frac{6.5}{2}}}{63} = 0.032$\) (rounded to 3 decimal places)

Since the carrying capacity of the island is 10 million, we know that the population will not exceed this limit.

Therefore, we can use the logistic model to find the population growth over time. The logistic growth model is:

\($$\frac{dP}{dt} = r P \left(1 - \frac{P}{K}\right)$$\)

where $K$ is the carrying capacity of the environment. This can be solved to give:\($P(t) = \frac{K}{1 + A e^{-rt}}$\)

where \($A = \frac{K-P_0}{P_0}$. We know $K = 10$ million, $P_0 = 2$ million, and $r = 0.032$\). Substituting these values, we get:\($A = \frac{10-2}{2} = 4$\)

Therefore, the equation for the population of the island is:\($P(t) = \frac{10}{1 + 4 e^{-0.032t}}$\)

To find the population in 2050, we substitute\($t = 100$\) (since 63 years have already passed and we want to find the population in 2050, which is 100 years after 1950):

\($P(100) = \frac{10}{1 + 4 e^{-0.032 \times 100}} \approx \boxed{8.76}$ million\)
Therefore, the estimated population of the island in 2050, assuming constant carrying capacity and consumption per capita, is approximately 8.76 million.

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Answer:

B.No. The input is not possible.

Step-by-step explanation:

According to the given scenario, as we know that f(x) indicates that the baskets number and x the number of players at the practice.

Therefore, if we are having the point (12, 36) we can conclude that during practice there were 36 baskets and 12 players.

Hence, it does not make any sense which results that input is not possible

Answer:

1/3( 3x + 12)

Step-by-step explanation:

0.7994 portion of persons with myopia will differ from the population portion by less than 3%.

Here we have to implement the central limit theorem,

therefore,

the formula is Z = x- a /s where, a = mean , d= standard deviation

let us consider that 42% has myopia

then p = 0.42

size of random sample given is 442

therefore, n = 442

then, a = p = 0.42

standard deviation is

s = \(\sqrt{p(1-p)/n}\)

=\(\sqrt{0.42* 0.58/442}\)

=0.0235

portion between 0.42 +0.03 = 0.45 and 0.42 - 0.03 = 0.39

there for there are two values of X = 0.45 , X = 0.39

Z = X - a/s

Z = 0.45 - 0.42 / 0.0235

Z = 1.28 have a p-value of 0.8997

Z = X - a /s

Z = 0.39 - 0.42 / 0.0235

Z = -1.28 have a p-value of 0.1003

0.8997 - 0.1003 = 0.7994

0.7994 portion of persons with myopia will differ from the population portion by less than 3%.

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Answer:

9-18-141

Step-by-step explanation:

9-18= -9

-9-141= (-150)

Answer:

k=-3

Explanation:

Plug in x and y values

-40=k(14)+2

-2 both sidea

-42=k(14)

divide by 14 both sides

k=-3

k = -3

Step-by-step explanation:

y = kx + 2 through the point a(14, -40)

Subtitute…

y = kx + 2

-40 = k(14) + 2

-40 = 14k + 2

-40 - 2 = 14k

-42 = 14k

(-42)/14 = k

-3 = k

A type II error is a statistical term used to describe the failure to reject a false null hypothesis. It occurs when the null hypothesis is actually false but is not rejected because the statistical test failed to find significant evidence against it.

In other words, it happens when an alternative hypothesis is true, but we fail to reject the null hypothesis.

Types of errors in hypothesis testing

There are two types of errors in hypothesis testing:

Type I error: Rejecting a true null hypothesis

Type II error: Failing to reject a false null hypothesis.Type I error occurs when a true null hypothesis is rejected while Type II error occurs when a false null hypothesis is not rejected. These types of errors are inversely related, meaning that a decrease in one type of error causes an increase in the other.

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Part a)Given, utility function of the consumer as:U(x1, x2) = log(x1) + log(x2)X1 ≤ 0.5Let p2 = 1 and m = 1, and p1 is unknown. The consumer has a budget constraint as: p1x1 + p2x2 = m = 1Now we have to find the minimal p1 such that the consumer consumes x1 strictly less than 0.5.

We need to find the value of p1 such that the consumer spends the entire budget (m = 1) on the two goods, but purchases only less than 0.5 units of the first good. In other words, the consumer spends all his money on the two goods, but still cannot afford more than 0.5 units of good 1.

Mathematically we can represent this as:

p1x1 + p2x2 = 1......(1)Where, x1 < 0.5, p2 = 1 and m = 1

Substituting the given value of p2 in (1), we get:

p1x1 + x2 = 1x1 = (1 - x2) / p1Given, x1 < 0.5 => (1 - x2) / p1 < 0.5 => 1 - x2 < 0.5p1 => p1 > (1 - x2) / 0.5

Now we know, 0 < x2 < 1.So, we will maximize the expression (1 - x2) / 0.5 for x2 ∈ (0,1) which gives the minimum value of p1 such that x1 < 0.5.On differentiating the expression w.r.t x2, we get:d/dx2 [(1-x2)/0.5] = -1/0.5 = -2

Therefore, (1-x2) / 0.5 is maximum at x2 = 0.

Now, substituting the value of x2 = 0 in the above equation, we get:p1 > 1/0.5 = 2So, the minimal value of p1 is 2.Part b)Now, we have to show mathematically that whether the threshold on p1 found in Part a increases/decreases/stays the same when p2 increases.

That is, if p2 increases then the minimum value of p1 will increase/decrease/stay the same.Since p2 = 1, the consumer’s budget constraint is given by:

p1x1 + x2 = m = 1Suppose that p2 increases to p2′.

The consumer’s new budget constraint is:

p1x1 + p2′x2 = m = 1.

Now we will find the minimal p1 denoted by pi, such that the consumer purchases less than 0.5 units of good 1. This can be expressed as:

p1x1 + p2′x2 = 1Where, x1 < 0.5

The budget constraint is the same as that in Part a, except that p2 has been replaced by p2′. Now, using the same argument as in Part a, the minimum value of p1 is given by:

p1 > (1 - x2) / 0.5.

We need to maximize (1 - x2) / 0.5 w.r.t x2.

As discussed in Part a, this occurs when x2 = 0.Therefore, minimal value of p1 is:

pi > 1/0.5 = 2

This value of pi is independent of the value of p2′.

Hence, the threshold on p1 found in Part a stays the same when p2 increases.

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