Answer: 214 - 5d
Which of the following real-world scenarios could be represented by the algebraic expression 7.65 - 2p? Select all that apply.
Answers:
1) You buy 2 equally priced party favors, spending from your $7.65 gift card.
4) you give Tim twice the amount of face paint he had from your supply of 7.65 ounces.
You decide to order pizzas for your party. A large pizza is cut into 12 slices. If each pizza is shared evenly among G guests, which expression represents the number of slices that each guest will eat?
Answer: 12/g
(these are the rest of the answers for the rest of the 7 slides of "Writing expression" assignment on e2020
4th: $25 dollars is the constant team so the cost for h hours is 15h and the correct expression is 15h+25 and 15(h+25) would be interpreted as adding 25 hours to the number of hours rented, and then multiplying by 15.
5th: m + 2m - s
6th: No it's not true. Each table began with x/8 balloons. If two pop then you will not be able to distribute it equally among 8 tables. There are x/8 - 2 balloons left. Not all tables will be equal.
7th: n - (1/4)n - 15
8th: 12n/r
9th: (x+3)/5
10th: (3r)/2 is what the second DJ had. 3 times as many songs as 3r and half as many is translated as division to 2 so half as many is translated as multiplication by 1/2. So therefore, the second DJ had (3r)/2 songs.
Step-by-step explanation:
Answer:
Step-by-step explanation:
Given that:
The sample mean
The standard deviation = 9
Population mean = 20
Null hypothesis:
Alternative hypothesis:
(a)
When Sample size = 10
t = 1.0541
Degree of freedom df:
df = n -1
df = 10 - 1
df = 9
P(value) for t = 1.0541 at df = 9:
P(value) = P(Z > 1.0541)
P(value) = 1 - P(< 1.0541)
P(value) = 1 - 0.8403
P(value) = 0.1597
There is no enough evidence to infer at the 5% significance since p-value is greater than the level of significance.
(b) When sample size = 30
t = 1.8257
Degree of freedom df:
df = n -1
df = 30 - 1
df = 29
P(value) for t = 1.8257 at df = 29:
P(value) = P(Z > 0.9609)
P(value) = 1 - P(< 0.9609)
P(value) = 1 - 0.9609
P(value) = 0.0391
There is enough evidence to infer that the mean is greater than 20 at the 5% significance level as the p-value is less than the significance level.
(c) When sample size = 50
t = 2.3570
Degree of freedom df:
df = n -1
df = 50 - 1
df = 49
P(value) for t = 2.3570 at df = 49:
P(value) = P(Z > 0.9888)
P(value) = 1 - P(< 0.9888)
P(value) = 1 - 0.9888
P(value) = 0.0112
There is enough evidence to infer that the mean is greater than 20 at the 5% significance level as the p-value is less than the significance level.
Answer is b because angle BPD is greater than 90 degree
Answer:
a) All dogs have fleas.
X the set of all animals
D(y)=y is a dog
F(y)=y has fleas
∀X ( D(x) → F(x) )
¬∃X ( D(x) ^ F(x) )
b) There exists a horse that can add.
X the set of all animals
H(y)=y is a horse
A(y)=y can add
∃X ( H(x) ^ A(x) )
∀X ( H(x) → ¬A(x) )
c) Every koala can climb.
X the set of all animals
K(y)=y is a koala
C(y)=y can climb
∀X ( K(x) → C(x) )
¬∃X ( K(x) ^ C(x) )
d) No monkey can speak French.
X the set of all animals
M(y)=y is a Monkey
F(y)=y can speak French
∀X ( M(x) → ¬F(x) )
∃X ( M(x) ^ F(x) )
e) There exists a pig that can swim and catch fish.
X the set of all animals
P(y)=y is a pig
S(y)=y can swim
C(y)=y can catch fish
∃X ( P(x) ^ ( S(x) ^ C(x) )
∀X ( P(x) → ( ¬S(x) v ¬C(x) ) )
Answer:
(-5, 11.3333)
Step-by-step explanation: