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Sonbull [250]
2 years ago
9

Find 60% of the number 60.​

Mathematics
2 answers:
faust18 [17]2 years ago
8 0
  • 60% of 60

  • 60% × 60

  • 60/100 × 60

  • 60/10 × 6

  • 6/1 × 6

  • 6 ×6

  • 36

Ans : 36

My name is Ann [436]2 years ago
8 0
The correct answer is 36.
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a triangle has side lengths of 13, 15, and 17. determine whether the triangle is right, acute, or obtuse.
Len [333]
Square of longest side = 17^2 =  289

15^2 = 225
13^2 = 169 

total 394      289 is less than this so its acute
3 0
3 years ago
Milan has 4 times as many baseball cards as Richard. Together they have 125 baseball cards. How many cards does Milan have? How
Ilia_Sergeevich [38]

Answer:

Step-by-step explanation:

500

7 0
3 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=x%20%2B%207%20%20%3C%20%20-%202" id="TexFormula1" title="x + 7 &lt; - 2" alt="x + 7 &lt; -
vladimir1956 [14]
First one because when you subtract 7 u keep the var. x alone and isolate it.
5 0
2 years ago
What is the answer for this problem?
Y_Kistochka [10]
This is a problem involving the subtraction of two functions f(x) and g(x):

<span>if f(x)=3x-1 and g(x)=x+2, find (f-g)(x).  In other words, find:

</span><span>  f(x)  = 3x-1 
-{g(x) -(x+2)
-----------------
f(x) - g(x) = 3x - 1 - x - 2 = 2x - 3  (answer)</span>
4 0
3 years ago
The overhead reach distances of adult females are normally distributed with a mean of 205 cm and a standard deviation of 7.8 cm.
devlian [24]

Answer:

Given the mean = 205 cm and standard deviation as 7.8cm

a. To calculate the probability that an individual distance is greater than 218.4 cm, we subtract the probability of the distance given (i.e 218.4 cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) from 1. Therefore, we have 1- P(Z\leq 1.72). Using the Z distribution table we have 1-0.9573. Therefore P(X >218.4)= 0.0427.

b. To calculate the probability that mean of 15 (i.e n=15) randomly selected distances is greater than 202.8, we subtract the probability of the distance given (i.e 202.8cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) divided by the square root of mean (i.e n= 15)  from 1. Therefore, we have 1- P(Z\leq -1.09). Using the Z distribution table we have 1-0.1378. Therefore P(X >202.8)= 0.8622.

c. This will also apply to a normally distributed data even if it is not up to the sample size of 30 since the sample distribution is not a skewed one.

Step-by-step explanation:

Given the mean = 205 cm and standard deviation as 7.8cm

a. To calculate the probability that an individual distance is greater than 218.4 cm, we subtract the probability of the distance given (i.e 218.4 cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) from 1. Therefore, we have 1- P(Z\leq 1.72). Using the Z distribution table we have 1-0.9573. Therefore P(X >218.4)= 0.0427.

b. To calculate the probability that mean of 15 (i.e n=15) randomly selected distances is greater than 202.8, we subtract the probability of the distance given (i.e 202.8cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) divided by the square root of mean (i.e n= 15)  from 1. Therefore, we have 1- P(Z\leq -1.09). Using the Z distribution table we have 1-0.1378. Therefore P(X >202.8)= 0.8622.

c. This will also apply to a normally distributed data even if it is not up to the sample size of 30 since the sample distribution is not a skewed one.

4 0
3 years ago
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