Answer:
=-3471 is the answer this may help you.
Step-by-step explanation:
=3.471×10^-3
=3.471×-1000
=-3471
Answer: I think the last option is the correct answer.
Step-by-step explanation:
Answer:
Answers below
Step-by-step explanation:
1. 77,915.06
A=2*50( this is a subscript 50)+(50+50+50)505
2. 5.47×105 or 547,000
3. 211045.04
I'm pretty confident that these are all correct but double check them first.
Answer:
14.5 units
Step-by-step explanation:
we need to find the length of the rectangle. To do that, we must find the hypotenuse of the right triangle below the rectangle.
Finding the hypotenuse of a right triangle uses the Pythagorean theorem, or a² + b² = c²
let a be 9 and b be 7.
9² + 7² = c²
81 + 49 = c²
130 = c²
√130 = c
c = approx. 11.4
Now, we know the length of the rectangle.
Once a rectangle is cut diagonally, it forms 2 equal right triangles. We will use the same process to find the hypotenuse of the second right triangle.
9² + 11.4² = c²
81 + 130 = c²
81 + 130 = approx. 211
c² = 211
√211 = c
c = approx. 14.5
Given that the mean of 15 bowlers that have been selected at random is distributed normally with mean 157 and std dev of 12
The probability that a random sample of 15 bowlers would have an average score greater than 165 will be:
mean=157
std dev,σ =12
std error=σ/√n=12/√15=3.0984
standardizing xbar to z=(xbar-μ)/(σ/√n)
P(xbar>165)=P(165-157)/3.0984
=P(z>2.582)
using normal probability tables we get:
P(z>2.582)=0.0049
Next we calculate the probability that a random sample of 150 bowlers will have an average score greater than 165.
μ=157
σ=12
std error=12/√150=0.9797=0/98
standardizing the xbar we get:
z=(165-157)/0.98
=P(z>8.165)
from normal table this will give us:
P(z>8.165)=0.00