5280•5= (5000•5) + (200•5) + (80•5) + (0•5) = 26400.
Answer:
0.2755
Step-by-step explanation:
We intend to make use of the normal approximation to the binomial distribution.
First we'll check to see if that approximation is applicable.
For p=10% and sample size n = 500, we have ...
pn = 0.10(500) = 50
This value is greater than 5, so the approximation is valid.
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The mean of the distribution we'll use as a model is ...
µ = p·n = 0.10(500)
µ = 50
The standard deviation for our model is ...
σ = √((1-p)µ) = √(0.9·50) = √45
σ ≈ 6.708204
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A continuity correction can be applied to better approximate the binomial distribution. We want p(t ≤ 9.1%) = p(t ≤ 45.5). For our lookup, we will add 0.5 to this limit, and find p(t ≤ 46).
The attached calculator shows the probability of fewer than 45.5 t's in the sample is about 0.2755.
Answer:
The image of the point is (1, 2)
Step-by-step explanation:
- If the point (x, y) rotated about the origin by angle 90° counterclockwise, then its image is (-y, x)
- If the point (x, y) rotated about the origin by angle 180° counterclockwise, then its image is (-x, -y)
- If the point (x, y) rotated about the origin by angle 270° counterclockwise, then its image is (y, -x)
Let us use the rules above to solve the question
∵ The point (-1, -2) is rotated 180° counterclockwise about the origin
→ By using the 2nd rule above change the signs of x and y coordinates
∴ Its image is (1, 2)
∴ The image of the point is (1, 2)
Okay, so here is a small lesson on rounding. When you round you MUST know the place value of each number. Ones, tens, hundreds, thousands, millions.
So, when we look at this number what do we have in the ones place --there is a 0 -- in the tens place we have a 9 and in the the thousands place we have 5. to round this number look at the tens place. if the number is greater than 5 you round 73590 to 73600, if it is 4 and lower it stays the same and the numbers behind change to zero--> like so 73500. But for this problem it would round up to 73600.
9^2*4^4 is the correct answer I believe.
