Who would be the most likely audience for Jack London's story?
2 answers:
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Answer:
Now we are supposed to find probabilities that the response time is between 5 and 10 minutes i.e P(5<x<10)
Formula :
at x = 5
at x = 10
P(-2<z<0.6315)=P(z<0.6315)-P(z<-2)
Refer the z table
P(-2<z<10)=0.7357-0.0228=0.7129
So, the probability that response time is between 5 and 10 minutes is 0.7129
b)the response time is less than 5 minutes
at x = 5
P(x<5)=P(z<-2)=0.0228
So, the probability that the response time is less than 5 minutes is 0.0228
c)the response time is more than 10 minutes
at x = 10
P(x>10) = 1-P(x<10) = 1-P(z<0.63) = 1-0.7357 = 0.2643
So, The probability that the response time is more than 10 minutes is 0.2643
Answer:
The length of the ribbons in all were 25.3 meters.
Step-by-step explanation:
Given:
Lengths of the three pieces of ribbon were 10 meters and 9 decimeters, 3 meters and 10 decimeters, and finally 10 meters and 4 decimeters.
Now, to find the length of the ribbons in all.
So, by adding the length of all three pieces we get the length of all the ribbons:
.
Now, converting the decimeters into meters = 23 meters + 2.3 meters = 25.3 meters.
(1 decimeter = 0.10 meter)
(23 decimeter = 2.3 meter.)
Therefore, the length of the ribbons in all were 25.3 metres.
Answer:
f(-3) = -20
Step-by-step explanation:
We observe that the given x-values are 3 units apart, and that the x-value we're concerned with is also 3 units from the first of those given. So, a simple way to work this is to consider the sequence for x = 6, 3, 0, -3. The corresponding sequence of f(x) values is ...
34, 10, -8, ?
The first differences of these numbers are ...
10 -34 = -24
-8 -10 = -18
And the second difference is ...
-18 -(-24) = 6
For a quadratic function, second differences are constant. This means the next first-difference will be ...
? -(-8) = -18 +6
? = -12 -8 = -20
The value of the function at x=-3 is -20.
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The attachment shows using a graphing calculator to do a quadratic regression on the given points. The graph can then be used to find the point of interest. There are algebraic ways to do this, too, but they are somewhat more complicated than the 5 addition/subtraction operations we needed to find the solution. (Had the required x-value been different, we might have chosen a different approach.)
