To solve this problem, we can already start looking at which equation we can make, since we have all the data. To find the time it takes, we know that we will have to have the distance on the top and the speed on the bottom. That eliminates the bottom two options. But, since we know that the speedboat is moving against the current of the lake, that means we subtract x from the total speed, 70. That means that the answer to this problem is f(x) = 
In first grade, Rebecca's school lunch cost $2.50. By seventh grade the cost of lunch increased By 75%
1 answer:
You might be interested in
Answer:
(a) Probability that a family of a returning college student spend less than $150 on back-to-college electronics is 0.0537.
(b) Probability that a family of a returning college student spend more than $390 on back-to-college electronics is 0.0023.
(c) Probability that a family of a returning college student spend between $120 and $175 on back-to-college electronics is 0.1101.
Step-by-step explanation:
We are given that according to an NRF survey conducted by BIG research, the average family spends about $237 on electronics in back-to-college spending per student.
Suppose back-to-college family spending on electronics is normally distributed with a standard deviation of $54.
Let X = <u><em>back-to-college family spending on electronics</em></u>
SO, X ~ Normal()
The z score probability distribution for normal distribution is given by;
Z = ~ N(0,1)
where, = population mean family spending = $237
= standard deviation = $54
(a) Probability that a family of a returning college student spend less than $150 on back-to-college electronics is = P(X < $150)
P(X < $150) = P( <
) = P(Z < -1.61) = 1 - P(Z
1.61)
= 1 - 0.9463 = <u>0.0537</u>
The above probability is calculated by looking at the value of x = 1.61 in the z table which has an area of 0.9463.
(b) Probability that a family of a returning college student spend more than $390 on back-to-college electronics is = P(X > $390)
P(X > $390) = P( >
) = P(Z > 2.83) = 1 - P(Z
2.83)
= 1 - 0.9977 = <u>0.0023</u>
The above probability is calculated by looking at the value of x = 2.83 in the z table which has an area of 0.9977.
(c) Probability that a family of a returning college student spend between $120 and $175 on back-to-college electronics is given by = P($120 < X < $175)
P($120 < X < $175) = P(X < $175) - P(X $120)
P(X < $175) = P( <
) = P(Z < -1.15) = 1 - P(Z
1.15)
= 1 - 0.8749 = 0.1251
P(X < $120) = P( <
) = P(Z < -2.17) = 1 - P(Z
2.17)
= 1 - 0.9850 = 0.015
The above probability is calculated by looking at the value of x = 1.15 and x = 2.17 in the z table which has an area of 0.8749 and 0.9850 respectively.
Therefore, P($120 < X < $175) = 0.1251 - 0.015 = <u>0.1101</u>