4/5 is your answer
Hope this helps
Answer:
138
Step-by-step explanation:
70 + 32 = 102
102 + 36 = 138
138 + 40 = 178
178 + 44 = 222
<em><u>Question:</u></em>
<em>Mrs. Magdalino kept records on how much she spent on gasoline and the maintenance of her car. She found that it cost $485 to drive 500 mi in a month.</em>
<em>a) Find the cost per mile. Write an equation that relates the cost c for gasoline and maintenance of a car to the number of miles m the car is driven.</em>
<em>b) Use the equation to find the cost for driving 1200 miles.</em>
<em>c) About how many miles are driven for a cost of $820?</em>
<em><u>Answer:</u></em>
a)
$485/500mi = <u><em>$0.97/mi</em></u>
c = cost
m = miles driven
cost per miles driven = cost / mile driven = <u><em>c/m</em></u>
b)
(1200 mi)*($0.97/mi) = <u><em>$1164</em></u>
c)
(x miles)*($0.97/mi) = $820
*divide both sides by $0.97/mi
x miles = $820/($0.97/mi)
<u><em>x = 845.36 miles</em></u>
The ratio would be 3:1
( the 6 triangles are blue, which I’m pretty sure they are.)
Since there are 18 yellows and 6 blues, and the question calls for the ratio of the yellows to the blues, the ratio is 18:6. But you always have to simplify. The GCF is 6, so you would divide both of them by 6. The ratio is then 3:1.
Answer: A: 0.0031
Step-by-step explanation:
Given : In a study of wait times at an amusement park, the most popular roller coaster has a mean wait time of 17.4 minutes with a standard deviation of 5.2 minutes.
i.e. 
and 
We assume that the wait times are normally distributed.
samples size : n= 30
Let x denotes the sample mean wait time.
Then, the probability that the mean wait time is greater than 20 minutes will be :
![P(x>20)=1-P(x\leq20)\\\\=1-P(\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}\leq\dfrac{20-17.4}{\dfrac{5.2}{\sqrt{30}}})\\\\=1-P(z\leq2.74)\ \ [\because\ z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}]\\\\=1-0.9969\ \ [\text{ By z table}]\\\\=0.0031](https://tex.z-dn.net/?f=P%28x%3E20%29%3D1-P%28x%5Cleq20%29%5C%5C%5C%5C%3D1-P%28%5Cdfrac%7Bx-%5Cmu%7D%7B%5Cdfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D%5Cleq%5Cdfrac%7B20-17.4%7D%7B%5Cdfrac%7B5.2%7D%7B%5Csqrt%7B30%7D%7D%7D%29%5C%5C%5C%5C%3D1-P%28z%5Cleq2.74%29%5C%20%5C%20%5B%5Cbecause%5C%20z%3D%5Cdfrac%7Bx-%5Cmu%7D%7B%5Cdfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D%5D%5C%5C%5C%5C%3D1-0.9969%5C%20%5C%20%5B%5Ctext%7B%20By%20z%20table%7D%5D%5C%5C%5C%5C%3D0.0031)
Hence, the probability that the mean wait time is greater than 20 minutes.= 0.0031
Thus , the correct answer is A: 0.0031 .
