h(t)=(t+3) 2 +5 h, left parenthesis, t, right parenthesis, equals, left parenthesis, t, plus, 3, right parenthesis, squared, plu
lesya692 [45]
Answer:
1
Step-by-step explanation:
If I understand the question right, G(t) = -((t-1)^2) + 5 and we want to solve for the average rate of change over the interval −4 ≤ t ≤ 5.
A function for the rate of change of G(t) is given by G'(t).
G'(t) = d/dt(-((t-1)^2) + 5). We solve this by using the chain rule.
d/dt(-((t-1)^2) + 5) = d/dt(-((t-1)^2)) + d/dt(5) = -2(t-1)*d/dt(t-`1) + 0 = (-2t + 2)*1 = -2t + 2
G'(t) = -2t + 2
This is a linear equation, and the average value of a linear equation f(x) over a range can be found by (f(min) + f(max))/2.
So the average value of G'(t) over −4 ≤ t ≤ 5 is given by ((-2(-4) + 2) + (-2(5) + 2))/2 = ((8 + 2) + (-10 + 2))/2 = (10 - 8)/2 = 2/2 = 1
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Answer:
A=z
B=-2
C=0
D=30
Step-by-step explanation:
First lets simply the equation
21 - (2x +3)2 + 4(1-1) + z^3
21 - (2x^2 + 9) + 4(0) + z ^3
21 - 2x^2 + 9 + z^3
- 2x^2 + z ^3 + 30
This means that:
A=z
B=-2
C=0
D=30
<u>Explanation:</u>
a) First, note that the Type I error refers to a situation where the null hypothesis is rejected when it is actually true. Hence, her null hypothesis would be H0: mean daily demand of her clothes in this region should be greater than or equal to 100.
The implication of Type I error in this case is that Mary <u>rejects</u> that the mean daily demand of her clothes in this region is greater than or equal to 100 when it is actually true.
b) While, the Type II error, in this case, is a situation where Mary accepts the null hypothesis when it is actually false. That is, Mary <u>accepts</u> that the mean daily demand of her clothes in this region is greater than or equal to 100 when it is actually false.
c) The Type I error would be important to Mary because it shows that she'll be having a greater demand (which = more sales) for her products despite erroneously thinking otherwise.
I think this is the answer
A)Cathy had $6 at first
B)Selena has $24 at first
