f(x) = log2 x
f(40) = log2 40
40 = 2^y
2^5 = 32 and 2^6 = 64
so f(40) lies between integers 5 and 6.
Answer:
£13,831.72
Step-by-step explanation:
Answer:
The bolts with diameter less than 5.57 millimeters and with diameter greater than 5.85 millimeters should be rejected.
Step-by-step explanation:
We have been given that the diameters of bolts produced in a machine shop are normally distributed with a mean of 5.71 millimeters and a standard deviation of 0.08 millimeters.
Let us find the sample score that corresponds to z-score of bottom 4%.
From normal distribution table we got z-score corresponding to bottom 4% is -1.75 and z-score corresponding to top 4% or data above 96% is 1.75.
Upon substituting these values in z-score formula we will get our sample scores (x) as:


Therefore, the bolts with diameters less than 5.57 millimeters should be rejected.
Now let us find sample score corresponding to z-score of 1.75 as upper limit.


Therefore, the bolts with diameters greater than 5.85 millimeters should be rejected.
Recall the double angle identity for cosine:

It follows that

Since 0° < 22° < 90°, we know that sin(22°) must be positive, so csc(22°) is also positive. Let x = 22°; then the closest answer would be C,

but the problem is that none of these claims are true; cot(32°) ≠ 4/3, cos(44°) ≠ 5/13, and csc(22°) ≠ √13/2...
X × x = x^2
x × -3 = -3x
3 × x = 3x
3 × -3 = -9
x^2 + 3x - 3x - 9
So I would agree with C because the -3x and the 3x cancel each other out. I hope this helps!