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katrin2010 [14]
2 years ago
12

Kari is flying a kite. She releases 50 feet of string. What is the approximate difference in the height of the kite when the str

ing makes a 25o angle with the ground and when the string makes a 45o angle with the ground? Round to the nearest tenth.
Mathematics
2 answers:
nirvana33 [79]2 years ago
5 0
Sinα=h/L  where h=height, L=string length...

h=Lsinα  so

h(25°)=50sin25≈21.1ft

h(45°)=50sin45≈35.4ft
Vika [28.1K]2 years ago
4 0

Answer: 14.2 feet.

<3 hope you do well

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7, 16, 25, 34, 43, ..... nth term is
vladimir2022 [97]

Answer:

79 is the ninth term.

Step-by-step explanation:

You add nine to each one.

7, 16, 25, 34, 43, 52, 61, 70, 79

79 is the ninth term.

6 0
2 years ago
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Coffee with sugar and milk added to it can contain up to calories. To consume a maximum of 600 mg if caffeine per day, how many
Marina CMI [18]
We know there is 330 mg of caffeine per 16-ounce serving. So if we want 8-ounce servings:
330 mg : 2 = 165 mg
Also we have to consume a maximum of 600 mg.
165 mg * 3 = 495 mg < 600 mg
165 mg * 4 = 660 mg > 600 mg
Answer: Someone could drink 3 full 8-ounce servings in a day and still stay below the limit.
8 0
2 years ago
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Mrs Atkins is going to choose two students from her class to take part in a competition.
Schach [20]

Given:

Total number of girls in her class = 16

Total number of boys in her class = 14

To find:

The number of different ways of choosing one girl and one boy.

Solution:

We have,

Total number of girls = 16

Total number of boys = 14

So,

Total number of ways to select one girl from 16 girls = 16

Total number of ways to select one boy from 14 boys = 14

Now, number of different ways of choosing one girl and one boy is

16\times 14=224

Therefore, the required number of different ways is 224.

6 0
2 years ago
According to a Los Angeles Times study of more than 1 million medical dispatches from 2007 to 2012, the 911 response time for me
Reil [10]

Answer:

a) \bar X=10.65

Median =\frac{10.7+10.7}{2}=10.7

Mode= 10.7

b) Range = 11.8-8.3=3.5

s= 0.948

c)IQR = Q_3 -Q_1 = 11.05-10.55=0.5

And we can find the usual limits with:

Lower = Q_1 -1.5 IQR = 10.55 -1.5*0.5=9.8

Upperer = Q_3 +1.5 IQR = 10.55 +1.5*0.5=11.3

And since 8.3 <9.8 we can consider this value too low or as an outlier for this case.

d) The mean for this case was 10.65 and the usual values are between 9.8 and 11.3, so as we can see all are above the specified value of 6 minutes, and we can conclude that the times are not satisfying the quality standards for this case.

And they should be considered apply some strategies to reduce the response time, adding more stations around points selected at the city could be useful in order to reduce the response time.

Step-by-step explanation:

We have the following data:

11.8 10.3 10.7 10.6 11.5 8.3 10.5 10.9 10.7 11.2

Part a

We can calculate the sample mean with the following formula:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

And if we replace we got: \bar X=10.65

For the median we need to sort the values on increasing order and we have:

8.3 10.3 10.5 10.6 10.7 10.7 10.9 11.2 11.5 11.8

Since n =10 we can calculate the median as the average between the 5th and 6th position of the dataset ordered and we got:

Median =\frac{10.7+10.7}{2}=10.7

The mode would be the most repeated value on this case:

Mode= 10.7

Part b

The range is defined as Range =Max-Min and if we replace we got:

Range = 11.8-8.3=3.5

We can calculate the standard deviation with the following formula:

s= sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

And if we replace we got:

s= 0.948

Part c

For this case we can use the IQR method in order to determine if 8.3 is an outlier or not.

We can calculate the first quartile with these values: 8.3 10.3 10.5 10.6 10.7 10.7 and Q_1= \frac{10.6+10.7}{2}=10.55

And for the Q3 we can use: 10.7 10.7 10.9 11.2 11.5 11.8 and we got Q_3 = \frac{10.9+11.2}{2}=11.05

Then we can find the IQR like this:

IQR = Q_3 -Q_1 = 11.05-10.55=0.5

And we can find the usual limits with:

Lower = Q_1 -1.5 IQR = 10.55 -1.5*0.5=9.8

Upperer = Q_3 +1.5 IQR = 10.55 +1.5*0.5=11.3

And since 8.3 <9.8 we can consider this value too low or as an outlier for this case.

Part d

The mean for this case was 10.65 and the usual values are between 9.8 and 11.3, so as we can see all are above the specified value of 6 minutes, and we can conclude that the times are not satisfying the quality standards for this case.

And they should be considered apply some strategies to reduce the response time, adding more stations around points selected at the city could be useful in order to reduce the response time.

6 0
2 years ago
What does the digit 7 represent in 170,280?
Sidana [21]

Answer:

that is the ten thousands place

Step-by-step explanation:

3 0
2 years ago
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