The upper quartile for the data set given below is. 14, 8, 23, 9, 11, 27, 22, 3, 17, 12, 29
LekaFEV [45]
Ok, first we need to organize.
<span>14, 8, 23, 9, 11, 27, 22, 3, 17, 12, 29
</span>3, 8, 9, 11, 12, 14, 17, 22, 23, 27, 29.
First, we need to find the median, the or the middle, which is 14.
Now that we know the median is fourteen, we take away all numbers to the left of 14, including 14: 17, 22, 23, 27, 29. Now just find the median in this new set of numbers. This will give us our upper quartile, which is 23. Hope this helped, and don't forget to drop a like.
The sweater cost more originally. It costs $2.50 more from jacket.
Step-by-step explanation:
Given,
Discount rate of sweater = 20%
Sale price of sweater = (100-20)% = 80%
Discount rate of jacket = 15%
Sale price of jacket = (100-15)% = 85%
Sale price of each item = $34.00
Let,
x be the original price of sweater.
y be the original price of jacket.
80% of x = 34

Dividing both sides by 0.80

The original price of sweater is $42.50
85% of y = 34

Dividing both sides by 0.85

The original cost of jacket is $40
Difference = 42.50 - 40 = $2.50
The sweater cost more originally. It costs $2.50 more from jacket.
Keywords: subtraction, division
Learn more about subtraction at:
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B) p=9t
he gets paid $9/hr, and his pay depends how many hours he works.
Answer:
Tuesday's Z-score is 7.56
Step-by-step explanation:
We are given that a department store, on average, has daily sales of 28,651.79 and the standard deviation of sales is $1000.
Also, it is given that on Tuesday, the store sold $36,211.08 worth of goods.
Let X = Daily sales of goods
So, X ~ N(
)
The z-score probability distribution is given by;
Z =
~ standard normal N(0,1)
Now, Tuesday,s Z-score is given by;
Z =
= 7.56
Yes, Tuesday was an unusually good day as on this day more worth of sales takes place as compared to the average daily sales of $28,651.79 .
Let’s say it takes time t hours for the interception.
In that time the carrier travels 28t nm west and the helicopter 130t nm.
Now we use the sine rule to find the angle x which lies between the distance 175 and 130t.
sin(x)/28t=sin35/130t. So sin(x)=28sin35/130=14sin35/65=0.1235 and x=7.0964 degrees.
Therefore the helicopter should use a bearing of 35+7.0964=42.0964 degrees north of east or 47.9 degrees east of north approx.