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MAVERICK [17]
2 years ago
12

Butterworth, Missouri, is trying to increase its tourism, a great source of revenue. Among other things, Butterworth prides itse

lf on its temperate climate. The tourism department would like to include in its glossy new brochure the middle range of temperatures that occur on 90% of the days there. A check with the National Weather Service finds that the average temperature in Butterworth is 76 degrees with a standard deviation of 5.7 (the temperatures were normally distributed). Help the tourism department by finding the two temperatures that cut off the middle 90% of temperatures in Butterworth.
Mathematics
1 answer:
goblinko [34]2 years ago
3 0

Answer:

Is expected that 90% of the time the temperature in Butterworth will be between 66.6 deg F and 85.37 deg F.

Step-by-step explanation:

The middle range of temperatures that occur on 90% of the days (upper limit and lower limit) can be calculated as:

UL=\mu+z*\sigma\\\\LL=\mu-z*\sigma

For a 90% interval, z is equal to 1.6449,

UL=\mu+z*\sigma=76+1.6449*5.7=76+9.376=85.376\\\\LL=\mu-z*\sigma=76-1.6449*5.7=76-9.376=66.624

Is expected that 90% of the time the temperature in Butterworth will be between 66.6 deg F and 85.37 deg F.

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Jarvis works in a garage £6 an hour. If he works on Saturday, he is paid time and a quarter. If he works on Sunday, he is paid t
ruslelena [56]

Answer: He is paid $90 last weekend.

Step-by-step explanation:

Since we have given that

Amount he earns per hour = $6

If he works on Saturday, he is paid time and a quarter .

Amount would be

6\times 1.25=\$7.5

If he works on Sunday, he is paid time and a half.

Amount would be

6\times 1.5=\$9

Number of hours he worked on Saturday = 6 hours

Number of hours he worked on Sunday = 5 hours

So, Total amount he is paid last weekend altogether is given by

6\times 7.5+5\times 9\\\\=\$90

Hence, he is paid $90 last weekend.

4 0
2 years ago
What are these fractions in simplest form?
Anton [14]
\frac{16y^{3}}{20y^{4}} =  \frac{4}{5y}
\frac{6xy}{16y} =  \frac{3x}{8}
\frac{abc}{10abc} =  \frac{1}{10}
\frac{mn^{2}}{pm^{5}n} =  \frac{n}{pm^{4}}
\frac{12h^{3}k}{16h^{2}k^{2}} =  \frac{3h}{4k}
\frac{8x}{10y} =  \frac{4x}{5y}
\frac{24n^{2}}{28n} =  \frac{6n}{7}
\frac{30hxy}{54kxy}  =  \frac{5h}{9k}
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\frac{20s^{2}t^{3}}{16st^{5}} =  \frac{5s}{4t^{2}}
8 0
2 years ago
Below is the five Number summary for a 136 hikers who recently completed the John muir trail JMT the variable is the amount of t
Helga [31]

Answer:

IQR = Q_3 -Q_1 = 28-18 = 10

Now we can find the limits in order to determine outliers like this:

Left = Q_1 -1.5 IQR = 18 -1.5*10=3

Right = Q_1 +1.5 IQR = 18 +1.5*10=33

So for this case the left boundary would be 3, if a value is lower than 3 we consider this observation as an outlier

b. 3

Step-by-step explanation:

For this case we have the following summary:

Minimum = 9 represent the minimum value

Q_1 = 18 represent the first quartile

Median =Q_2= 21 represent the median

Q_3 = 28 represent the third quartil

Maximum=56 represent the maximum

If we use the 1.5 IQR we need to find first the interquartile range defined as:

IQR = Q_3 -Q_1 = 28-18 = 10

Now we can find the limits in order to determine outliers like this:

Left = Q_1 -1.5 IQR = 18 -1.5*10=3

Right = Q_1 +1.5 IQR = 18 +1.5*10=33

So for this case the left boundary would be 3, if a value is lower than 3 we consider this observation as an outlier

b. 3

7 0
2 years ago
A baker needs 8 1/2 cups of flour for a sheet cake recipe. He has 4 2/3 cups of flour in a canister and buys 5 1/4 cups more.
Juli2301 [7.4K]

Answer: 1 5/12 cups of flour leftover

Step-by-step explanation: First, add 4 2/3 + 5 1/4. This gives you 9 11/12. 9 11/12 - 8 1/2 = 1 5/12.

6 0
2 years ago
A school cafeteria sells milk at 25 cents per carton and salads at 45 cents each. one week the total sales for these items were
denis-greek [22]

solution:

Lets start with the most amount that could have been sold.......using guess and check, we can figure out that 290 salads could have been sold, while 8 cartons of milk would have been sold.

The least amount of salads that could have been sold were none.

so,

you have  0<s<290

at least none were sold, and at most 290 were sold

but I do believe you are missing part of the question


4 0
2 years ago
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