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storchak [24]
1 year ago
14

3 A buoy is 30 feet from the shore. You swim 3/5 the way to the buoy. How much farther do you have to swim to reach the buoy?​

Mathematics
1 answer:
disa [49]1 year ago
8 0

Answer:

12 feet

Step-by-step explanation:

30 divided by 5= 6

6 times 3= how far youve already swam(18 feet)

6 times 2= how far you have left(12 feet)

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A control chart is developed to monitor the analysis of iron levels in human blood. The lines on the control chart were obtained
sergejj [24]

Answer:

Step-by-step explanation:

Hello!

The variable is X: iron level on human blood.

It has a mean of μ= 51.50 mg/dl and a standard deviation of σ= 3.50 mg/dl.

According to the control chart, the process should be shut down for troubleshooting when the analysis shows values X[bar]≥ 53.42 mg/dl and X[bar]≤ 49.58 mg/dl

Warnings are received at levels X[bar]≥52.78 mg/dl and X[bar]≤ 50.22 mg/dl

The system works between levels 49.58<X[bar]<53.42 and works without warnings between 50.22<X[bar]<52.78.

Using these parameters you have to analyze if the lists of sample means to see which ones are within the working values are wich ones are outside this interval.

<u>Sample 1</u>

Min= 50.15

Max= 51.99

Mean= 51.61

Without warning 50.22<X[bar]<52.78

Without Action interval 49.58<X[bar]<53.42

<em>This sample's min value is below the lower limit of the warning interval but not low enough to reach action levels, the max value is within the working range.</em>

<em><u /></em>

<u>Sample 2 </u>

Min= 50.32

Max= 52.56

Mean= 51.16

Without warning 50.22<X[bar]<52.78

Without Action interval 49.58<X[bar]<53.42

<em>Both the max and min values of the sample are within the working range without warning.</em>

<em />

<u>Sample 3</u>

Min= 50.25

Max= 53.12

Mean= 51.83

Without warning 50.22<X[bar]<52.78

Without Action interval 49.58<X[bar]<53.42

<em>The max value of this sample is above the upper limit of the warning interval but does not surpass the upper bond of the troubleshoot interval. Min value is within working values.</em>

<em />

<u>Sample 4</u>

Min= 50.05

Max= 53.01

Mean= 51.70

Without warning 50.22<X[bar]<52.78

Without Action interval 49.58<X[bar]<53.42

<em>Both min and max values surpass the warning interval but do not reach action levels</em>.

<u>Sample 5</u>

Min= 50.35

Max= 52.71

Mean= 51.37

Without warning 50.22<X[bar]<52.78

Without Action interval 49.58<X[bar]<53.42

<em>Both min and max values are within working levels without warnings.</em>

<em />

Considering that samples reaching warning levels should be shut down as a precaution, they are classified as:

Shutdown: Sample 1, 3 and 4

Do not shutdown: Sample 2 and 5.

I hope it helps!

8 0
2 years ago
Solve 10m -28 = 6 -7m​
Finger [1]

Answer:

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
A rectangular parking lot has a perimeter of 232 feet. The length of the parking lot is 36 feet less than with. Find the length
Ainat [17]

Answer:

Length: 80 feet

Width: 36 feet

Step-by-step explanation:

232/2 = 116

116-36 = 80

(i.e. 80+80+36+36=232)

7 0
2 years ago
Customer: "I buy the same items every week and my bill is normally $90.00. Why am I now being charged $108.00 for the exact same
scZoUnD [109]
18
108.00- 90.00= 18.00
7 0
2 years ago
An unloaded truck and trailer, with the driver aboard, weighs 30{,}00030,00030, comma, 000 pounds. When fully loaded, the truck
kari74 [83]

Answer:

1131 pounds.

Step-by-step explanation:

We have been given that an unloaded truck and trailer, with the driver aboard, weighs 30,000 pounds. When fully loaded, the truck holds 26 pallets of cargo, and each of the 18 tires of the fully loaded semi-truck bears approximately 3,300 pounds.

First of all, we will find weight of 18 tires by multiplying 18 by 3,300 as:

\text{Weight of tires of the fully loaded semi-truck}=18\times 3,300

\text{Weight of tires of the fully loaded semi-truck}=59,400

The weight of 26 pallets would be weight of 18 tires minus weight of unloaded truck.

\text{Weight of 26 pallets of cargo}=59,400-30,000

\text{Weight of 26 pallets of cargo}=29,400

Now, we will divide 29,400 by 26 to find average weight of one pallet of cargo.

\text{Average weight of one pallet of cargo}=\frac{29,400}{26}

\text{Average weight of one pallet of cargo}=1130.769230769

\text{Average weight of one pallet of cargo}\approx 1131

Therefore, the average weight of one pallet of cargo is approximately 1131 pounds.

3 0
1 year ago
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