L(t)=9-0.5t
t=time in hours,
L(t)=length of candle at time t
(A) substitute t=1,2,3,4,5,6 respectively to get
(1,8.5)
(2,8)
(3,7.5)
(4,7)
(5,6.5)
(6,6)
(B) it is a function because none of the x-values duplicate, (nor the y-values).
(C) If the rate changes to 0.45 "/hr, the relation is still linear.
All linear function with finite slope are functions.
Answer:
150 ft
Step-by-step explanation:
Answer:
The probability it will land on green every time is .
Step-by-step explanation:
We are given that a spinner is used for which it is equally probable that the pointer will land on any one of six regions. Three of the regions are colored red, two are colored green, and one is colored yellow.
The pointer is spun three times.
<u>As we know that the probability of an event is described as;</u>
Probability of an event =
Here, the favorable outcome is that the spinner will land on green every time.
So, the number of green regions = 2
Total number of regions = 3(red) + 2(green) + 1(yellow) = 6 regions
<em>Now, the probability it will land on green every time is given by;</em>
Probability =
=
=
Hence, the probability it will land on green every time is .
Answer:
Null Hypothesis: H_0: \mu_A =\mu _B or \mu_A -\mu _B=0
Alternate Hypothesis: H_1: \mu_A >\mu _B or \mu_A -\mu _B>0
Here to test Fertilizer A height is greater than Fertilizer B
Two Sample T Test:
t=\frac{X_1-X_2}{\sqrt{S_p^2(1/n_1+1/n_2)}}
Where S_p^2=\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2}
S_p^2=\frac{(14)0.25^2+(12)0.2^2}{15+13-2}= 0.0521154
t=\frac{12.92-12.63}{\sqrt{0.0521154(1/15+1/13)}}= 3.3524
P value for Test Statistic of P(3.3524,26) = 0.0012
df = n1+n2-2 = 26
Critical value of P : t_{0.025,26}=2.05553
We can conclude that Test statistic is significant. Sufficient evidence to prove that we can Reject Null hypothesis and can say Fertilizer A is greater than Fertilizer B.
Answer:
Equivalent fractions
Step-by-step explanation: