Answer: 0.8461
Step-by-step explanation:
Given : 
Let x be the random variable that represents the cost for the hospital emergency room visit.
We assume that cost for the hospital emergency room visit is normally distributed .
z-score for x=1000 ,
![z=\dfrac{1000-1400}{392}\approx-1.02\ \ \ [\because z=\dfrac{x-\mu}{\sigma}]](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7B1000-1400%7D%7B392%7D%5Capprox-1.02%5C%20%5C%20%5C%20%5B%5Cbecause%20z%3D%5Cdfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%5D)
Using z-value table , we have
P-value =P(x>1000)=P(z>-1.02)=1-P(z≤ -1.02)=1-0.1538642
=0.8461358≈0.8461 [Rounded nearest 4 decimal places]
Hence, the probability that the cost will be more than $1000 = 0.8461
Answer:
56
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<u>Answer-</u>
<em>The amount will be </em><em>$8944.62</em><em> after 5 years.</em>
<u>Solution-</u>
We know that,
![\text{FV of annuity}=P[\dfrac{(1+r)^n-1}{r}]](https://tex.z-dn.net/?f=%5Ctext%7BFV%20of%20annuity%7D%3DP%5B%5Cdfrac%7B%281%2Br%29%5En-1%7D%7Br%7D%5D)
Where,
P = Payment = $50 monthly
r = rate of interest compounded monthly= 
n = number of period = 5 years = 5×12 = 60 months
Putting the values in the formula,
![\text{FV of annuity}=50[\dfrac{(1+0.0325)^{60}-1}{0.0325}]](https://tex.z-dn.net/?f=%5Ctext%7BFV%20of%20annuity%7D%3D50%5B%5Cdfrac%7B%281%2B0.0325%29%5E%7B60%7D-1%7D%7B0.0325%7D%5D)
![=50[\dfrac{(1.0325)^{60}-1}{0.0325}]](https://tex.z-dn.net/?f=%3D50%5B%5Cdfrac%7B%281.0325%29%5E%7B60%7D-1%7D%7B0.0325%7D%5D)
![=50[\dfrac{6.8140-1}{0.0325}]](https://tex.z-dn.net/?f=%3D50%5B%5Cdfrac%7B6.8140-1%7D%7B0.0325%7D%5D)
![=50[\dfrac{5.8140}{0.0325}]](https://tex.z-dn.net/?f=%3D50%5B%5Cdfrac%7B5.8140%7D%7B0.0325%7D%5D)
Therefore, the amount will be $8944.62 after 5 years.
Answer:
x = .06418
Step-by-step explanation:
2 + (9.2)−8x = 2.32
(9.2)−8x = 2.32 − 2
log (9.2)−8x = log .32
−8x log 9.2 = log .32
−8x log 9.2
−8 log 9.2
=
log .32
−8 log 9.2
x = .06418
