4b^2 +8b + 7 = 4show your work I already have the answer!!
2 answers:
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<h3>Answer: choice A) 55.3%</h3>
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Explanation:
Use a Z table found in the back of your book to find that
P(Z < -0.41) = 0.3409
P(Z < 1.25) = 0.8944
So,
P(-0.41 < Z < 1.25) = P(Z < 1.25) - P(Z < -0.41)
P(-0.41 < Z < 1.25) = 0.8944 - 0.3409
P(-0.41 < Z < 1.25) = 0.5535
Now convert this to a percentage by multiplying by 100, which is the same as moving the decimal point over 2 places
0.5535 ---> 55.35%
Round this to the nearest tenth of a percent. You could argue that 55.35% rounds to either 55.3% or 55.4% since that last digit is a 5. I'm going with 55.3% since 55.4% isn't listed as an answer choice. The table I used only lists approximate values, so there is likely some rounding error somewhere. When I used my TI83 (see image below) I got roughly 0.5534 which is fairly close to 0.5535. If you want to use your TI83 or TI84 calculator, then the normalcdf function can be found by pressing the yellow "2ND" button (top left corner) and then pressing the VARS key (3rd row from the top, just to the left of the CLEAR key).
Answer: Choice D
(0,-4) (4,0) and (-1,-5)
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Explanation:
Plug in x = -4
y = x-4
y = -4-4
y = -8
The point (-4,-8) is on the line instead of (-4,0). This rules out choice A and choice B.
Now plug in x = 1
y = x-4
y = 1-4
y = -3
We get (1,-3) instead of (1,5). So we can rule out choice C.
The answer must be choice D.
The graph is below.
Answer:
Step-by-step explanation:
We are looking for a trigonometric function which contains the point (0, 3), and has an amplitude of 3.
We know that for a sine function ,
; therefore the function we a looking for cannot be a sine function because it is zero at
.
However, the cosine function gives non-zero value at
therefore, a cosine function can be our function.
Now, cosine function with amplitude has the form
this is because the cosine function is maximum at and therefore, has the property that
in other words it contains the point .
The function we are looking for contains the point ; therefore, its amplitude must be 3, or
we see that this function satisfies our conditions: has amplitude of 3, and it passes through the point (0, 3) because