Every cubic millimeter of blood contains about 7500 white blood cells. A count less than 1500 above this number is still conside
1 answer:
Answer:
The white cell count of 8750 is considered healthy.
Step-by-step explanation:
The question calls for any white cell count below 1500 above 7500 to be considered healthy. So, by adding the number 1500 to 7500 we find that the upper limit for being healthy is 9000 white blood cells per cubic millimeter of blood. This all can be represented mathematically with the following:
If:
<em>Then</em>
White blood cell count 'x' is healthy.
<em />
Else if:
<u>or</u>
<em>Then</em>
White blood cell count 'x' is not healthy.
Now let's test the white cell count of 8750.
<em>Does it fall between 7500 and 9000?</em>
This checks out so the answer is yes: White blood cell count '8750' is healthy.
<em />
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Answer:
The volume of the new crate is exactly 174.4% of the old crate.
Step-by-step explanation:
Volume of a rectangular prism:
The volume of a rectangular prism of dimensions x, y and z is given by:
She decided that the dimensions of the new crate needs to be increased by 40% of the original length.
Each dimension is increased by 40%, which means that each is multiplied by 1 + 0.4 = 1.4. So
Percentage:
The new volume is 2.744 the original volume. Considering the original volume is 100% = 1, it increased by 2.744 - 1 = 1.744 = 1.744*100% = 174.4%.
The volume of the new crate is exactly 174.4% of the old crate.
Lets check with every option
(a) [-4,-3]
We plug in -4 for x and -3 for x
f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.
(b) [-3,-2]
We plug in -3 for x and -2 for x
f(-2) is negative and f(-3) is negative. there is no value at x=c on the interval [-3,-2] where f(c)=0.
(c) [-2,-1]
We plug in -2 for x and -1 for x
f(-2) is negative and f(-1) is positive. there is some value at x=c on the interval [-2,-1] where f(c)=0. so there exists atleast one zero on this interval.
(d) [-1,0]
We plug in -1 for x and 0 for x
f(-1) is positive and f(0) is negative. there is some value at x=c on the interval [-1,0] where f(c)=0. so there exists atleast one zero on this interval.
(e) [0,1]
We plug in 0 for x and 1 for x
f(0) is negative and f(1) is negative. there is no value at x=c on the interval [0,1] where f(c)=0.
(f) [1,2]
We plug in 1 for x and 2 for x
f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.
so answers are (a) [-4,-3], (c) [-2,-1], (d) [-1,0], (f) [1,2]