If EF bisects angle CEB, angle CEF=7x + 21 and angle FEB = 10x - 3, find the measure of DEB.
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Answer: Rectangular container is preferred to minimize the costs and there are 142 rectangular container is needed to pack 6400 candles.
Explanation:
Since we have given that
Number of candles in a rectangular container = 45
Number of candles in a cylindrical container = 41
Cost of each rectangular container = $15
Cost of each cylinder container = $16
That means ,
In case of rectangular container,
Cost of 45 candles = $15
Cost of 1 candle is given by
Similarly,
In case of cylindrical container,
Cost of 41 candles = $16
Cost of 1 candle is given by
So, if the company needs to package 6,400 candles , it should use <u>rectangular containers </u>to minimize costs.
Number of candle he need to pack = 6400
Number of rectangular container for 6400 candles is given by
Hence, there are 142 rectangular container is needed to pack 6400 candles.
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.