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Andre45 [30]
3 years ago
5

9s - 5s + 2s - s Please explain how you did this question.

Mathematics
2 answers:
Maslowich3 years ago
4 0
Just add them as if it were normal then put the s behind it
masha68 [24]3 years ago
4 0
Just take out the s' to make it easier
So 9-5+2-1
9-5= 4
4+2=6
And then +1 is 7
And all you have to do is put the s back in so the answer is:
7s
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Steve paid 10% tax on a purchase of $40. Select the dollar amount of the tax and the total dollar amount Steve paid on the numbe
lesya [120]
The answer is $36

hope this helped :)
alisa202
5 0
3 years ago
Read 2 more answers
Suppose we have a binomial experiment in which success is defined to be a particular quality or attribute that interests us.
mestny [16]

For a binomial experiment in which success is defined to be a particular quality or attribute that interests us, with n=36 and p as 0.23,  we can approximate p hat by a normal distribution.

Since n=36 , p=0.23 , thus q= 1-p = 1-0.23=0.77

therefore,

n*p= 36*0.23 =8.28>5

n*q = 36*0.77=27.22>5

and therefore, p hat can be approximated by a normal random variable, because n*p>5 and n*q>5.

The question is incomplete, a possible complete question is:

Suppose we have a binomial experiment in which success is defined to be a particular quality or attribute that interests us.

Suppose n = 36 and p = 0.23. Can we approximate p hat by a normal distribution? Why? (Use 2 decimal places.)

n*p = ?

n*q = ?

Learn to know more about binomial experiments at

brainly.com/question/1580153

#SPJ4

3 0
2 years ago
Describe how the graph of y = x2 can be transformed to the graph of the given equation. y = (x+4)2
NNADVOKAT [17]
Moving 4 units to the left would change from y=x^2 to y=(x+4)^2.
6 0
3 years ago
Fractions closer to 0 than 1
vagabundo [1.1K]

Answer:

Any legal fraction (denominator not equal to zero) with a numerator equal to zero has an overall value of zero. all have a fraction value of zero because the numerators are equal to zero

Step-by-step explanation:

5 0
3 years ago
1. Consider the following hypotheses:
Andrej [43]

Answer:

See deductions below

Step-by-step explanation:

1)

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)

2)

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.

7 0
3 years ago
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