Answer:
In Right Δ ABC with right angle B,
∠A=(3 x -8)°, ∠B=90°, ∠C=(x-2)°
∠A+∠B+∠C=180°[∠ sum property of triangle]
3 x-8+90 + x-2=180
Adding and subtracting like terms
4 x-10+90=180
4 x+80=180
4 x=180-80
4 x=100
x=100/4
x=25°
∠A=3×25-8=75-8=67°
An engineer measures the peak current (in microamps) when a solution containing an amount of nickel (in parts per 106 ) is added
1 answer:
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Answer:
The 95% confidence interval the average maximum power is (596.0 to 644.0)
Step-by-step explanation:
Average maximum of the sample = x = 620 HP
Standard Deviation = s = 45 HP
Sample size = n = 16
We have to calculate the 95% confidence interval. The value of Population standard deviation is unknown, and value of sample standard deviation is known. Therefore, we will use one sample t-test to build the confidence interval.
Degrees of freedom = df = n - 1 = 15
Critical t-value associated with 95% confidence interval and 15 degrees of freedom, as seen from t-table = = 2.131
The formula to calculate the confidence interval is:
We have all the required values. Substituting them in the above expression, we get:
Thus, the 95% confidence interval the average maximum power is (596.0 to 644.0)
Answer:
(a) The standard deviation of the amount spent is $3229.18.
(b) The probability that a household spends between $4000 and $6000 is 0.2283.
(c) The range of spending for 3% of households with the highest daily transportation cost is $12382.86 or more.
Step-by-step explanation:
We are given that the average annual amount American households spend on daily transportation is $6312 (Money, August 2001). Assume that the amount spent is normally distributed.
(a) It is stated that 5% of American households spend less than $1000 for daily transportation.
Let X = <u><em>the amount spent on daily transportation</em></u>
The z-score probability distribution for the normal distribution is given by;
Z = ~ N(0,1)
where, = average annual amount American households spend on daily transportation = $6,312
= standard deviation
Now, 5% of American households spend less than $1000 on daily transportation means that;
P(X < $1,000) = 0.05
P( <
) = 0.05
P(Z < ) = 0.05
In the z-table, the critical value of z which represents the area of below 5% is given as -1.645, this means;
= 3229.18
So, the standard deviation of the amount spent is $3229.18.
(b) The probability that a household spends between $4000 and $6000 is given by = P($4000 < X < $6000)
P($4000 < X < $6000) = P(X < $6000) - P(X $4000)
P(X < $6000) = P( <
) = P(Z < -0.09) = 1 - P(Z
0.09)
= 1 - 0.5359 = 0.4641
P(X $4000) = P(
) = P(Z
-0.72) = 1 - P(Z < 0.72)
= 1 - 0.7642 = 0.2358
Therefore, P($4000 < X < $6000) = 0.4641 - 0.2358 = 0.2283.
(c) The range of spending for 3% of households with the highest daily transportation cost is given by;
P(X > x) = 0.03 {where x is the required range}
P( >
) = 0.03
P(Z > ) = 0.03
In the z-table, the critical value of z which represents the area of top 3% is given as 1.88, this means;
x = $6312 + 6070.86 = $12382.86
So, the range of spending for 3% of households with the highest daily transportation cost is $12382.86 or more.