MATHEMATICS COLLEGE

Stuck on letter b, please help

Answers

Answer 1

Answer:

Answer: f(3)=10

Step-by-step explanation:

In order to calculate f(3) you need to substitute 3 for x in function

f(x)=3x +1 :

f(3)=( 3x3) + 1

f(3)= 9+1

f(3)=10

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1. Consider the following hypotheses:H1 : ∃x (p(x) ∧ q(x)) H2 : ∀x (q(x) → r(x))
Use rules of inference to prove that the following conclusion follows from these hypotheses:
C : ∃x (p(x) ∧ r(x))
Clearly label the inference rules used at every step of your proof.

2. Consider the following hypotheses:
H1 : ∀x (¬C(x) → ¬A(x)) H2 : ∀x (A(x) → ∀y B(y)) H3 : ∃x A(x)
Use rules of inference to prove that the following conclusion follows from these hypotheses:
C : ∃x (B(x) ∧ C(x))
Clearly label the inference rules used at every step of your proof.

3. Consider the following predicate quantified formula:
∃x ∀y (P (x, y) ↔ ¬P (y, y))
Prove the unsatisfiability of this formula using rules of inference.

Answers

Answer:

See deductions below

Step-by-step explanation:

a) p(y)∧q(y) for some y (Existencial instantiation to H1)

b) q(y) for some y (Simplification of a))

c) q(y) → r(y) for all y (Universal instatiation to H2)

d) r(y) for some y (Modus Ponens using b and c)

e) p(y) for some y (Simplification of a)

f) p(y)∧r(y) for some y (Conjunction of d) and e))

g) ∃x (p(x) ∧ r(x)) (Existencial generalization of f)

a) ¬C(x) → ¬A(x) for all x (Universal instatiation of H1)

b) A(x) for some x (Existencial instatiation of H3)

c) ¬(¬C(x)) for some x (Modus Tollens using a and b)

d) C(x) for some x (Double negation of c)

e) A(x) → ∀y B(y) for all x (Universal instantiation of H2)

f) ∀y B(y) (Modus ponens using b and e)

g) B(y) for all y (Universal instantiation of f)

h) B(x)∧C(x) for some x (Conjunction of g and d, selecting y=x on g)

i) ∃x (B(x) ∧ C(x)) (Existencial generalization of h)

3) We will prove that this formula leads to a contradiction.

a) ∀y (P (x, y) ↔ ¬P (y, y)) for some x (Existencial instatiation of hypothesis)

b) P (x, y) ↔ ¬P (y, y) for some x, and for all y (Universal instantiation of a)

c) P (x, x) ↔ ¬P (x, x) (Take y=x in b)

But c) is a contradiction (for example, using truth tables). Hence the formula is not satisfiable.

O A. (-2, 1)
O B. (-2,-1)
O c. (-1,-2)
O D. (1.-2)

Answers

Answer:

the answer is C -1-2 to the equation

-17 + 25 + -6 - 18 can you please show work too, thank youu

Answers

Answer:

-16

Step-by-step explanation:

-17+25+-6-18

you just basically combined -17,-6 and -18

so -41+25

= -16

Stephanie claims that 26540is greater than 50. Do you agree or disagree? Explain.

Answers

Agree. 50 is a smaller

A study published in 1993 found that babies born at different times of the year may develop the ability to crawl at different ages! The author of the study suggested that these differences may be related to the temperature at the time the infant is 6 months old. (Benson and Janette, Infant Behavior and Development [1993]. The study found that 32 babies born in January crawled at an average age of 29.84 weeks, with a standard deviation of 7.08 weeks. Among 21 July babies, crawling ages averaged 33.64 weeks, with a standard deviation of 6.91 weeks. Is this difference significant?

Answers

Answer:

$t=\frac{(29.84-33.64)-0}{\sqrt{(7.08^2)/(32)+(6.91^2)/(21)}}}=-1.939$

$p_v =2*P(t_(51)<-1.939)=0.0580$

Comparing the p value with the significance assumed $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the difference between the two groups is NOT significantly different at 1% of significance.

Step-by-step explanation:

Data given

$\bar X_(1)=29.84$ represent the mean for sample January

$\bar X_(2)=33.64$ represent the mean for sample July

$s_(1)=7.08$ represent the sample standard deviation for 1

$s_(2)=6.91$ represent the sample standard deviation for 2

$n_(1)=32$ sample size for the group 2

$n_(2)=21$ sample size for the group 2

$\alpha=0.01$ Significance level provided

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the difference in the population means, the system of hypothesis would be:

Null hypothesis: $\mu_(1)-\mu_(2)=0$

Alternative hypothesis: $\mu_(1) - \mu_(2)\neq 0$

We don't have the population standard deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{(\bar X_(1)-\bar X_(2))-\Delta}{\sqrt{(\sigma^2_(1))/(n_(1))+(\sigma^2_(2))/(n_(2))}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=32+21-2=51$

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

With the info given we can replace in formula (1) like this:

$t=\frac{(29.84-33.64)-0}{\sqrt{(7.08^2)/(32)+(6.91^2)/(21)}}}=-1.939$

P value

Since is a bilateral test the p value would be:

$p_v =2*P(t_(51)<-1.939)=0.0580$

Suppose that the weather forecast indicates a 10% chance that cold weather will reduce the citrus grower’s profit from $100,000 to $85,000 and a 10% chance that cold weather will reduce the profit to $75,000. Should the grower spend $5000 to protect the citrus fruit against the possible bad weather?

Answers

Answer:

We can think in this situation as:

There is 10% (or a probability equal to 0.1) where the citrus grower's profit will be reduced by $15,000

There is 10% (or a probability equal to 0.1) where the citrus grower's profit will be reduced by $25,000

And in the remaining 80% (or a probability equal to 0.8), the profit does not change.

Then the expected value for the total profit can be written as:

EV = $100,000 + ( 0.1*(-$15,000) + 0.1*(-$25,000) + 0.8*$0)

= $96,000

In the case where the citrus grower spends $5000 to protect the fruits against possible bad weather, there is a 100% that is profit will not change, but he must pay $5,000

Then his profit will be:

P = $100,000 - $5,000 = $95,000

So in this case, the profit is $1000 less than the expected profit in the prior case. So the scenario where he does not buy the protection has a larger expected profit, which may mean that is better to not buy it (in a straight mathematical point of view)

One also could think that the values are really close together, so buying the protection does not mean a big change, and increases the security