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kobusy [5.1K]
4 years ago
5

State the domain and range for the given relations. {(1,2), (2,2), (1,2)}

Mathematics
1 answer:
marshall27 [118]4 years ago
5 0

Answer:

Domain: 1, 2

Range: 2

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The square root of a number is 2k. what is half of the number?​
shtirl [24]

Answer:

2 milion

Step-by-step explanation:

2,000 x 2,000= 4,000,000

4,000,000÷ 2= 2,000,000

5 0
4 years ago
What are the leading coefficient, constant term and degree, if any, of the algebraic expression
nasty-shy [4]
Coefficients are the numbers in from with the variable, the constant is just a number and the degree is the exponent.
6 0
3 years ago
The shear strength of each of ten test spot welds is determined, yielding the following data (psi).
Mumz [18]

Answer:

a) Mean = 382.3 psi

Standard deviation = 20.8 psi

b) 392.7 psi

c) P(X<400)=0.8026

Step-by-step explanation:

a) The population mean can be estimated as equal to the mean of the sample, and the population standard deviation can be estimated from the sample standard deviation:

M=\frac{1}{n}\sum x_i=\frac{1}{10}(389+405+409+367+358+415+376+375+367+362)\\\\M=\frac{1}{10}(3823)=382.3\\\\\mu\approx M=382.3

s=\sqrt{\frac{1}{n-1}\sum (x_i-M)^2}=\sqrt{\frac{1}{10-1}(389-382.3)^2+...+(362-383.2)^2}\\\\\\s=\sqrt{\frac{1}{9}(3906.1)}=\sqrt{434}=20.8\\\\\\\sigma\approx s=20.8

b) We start by searching for the z-value for the 95th percentile. This value is

z=1.645:

P(z

Then, the strength value below which 95% of all welds will have their strengths is:

X=\mu+z\cdot \sigma/\sqrt{n}=382.3+1.645*20.8/\sqrt{10}\\\\X=382.3+32.9/3.2=382.3+10.4\\\\X=392.7

c) We calculate the probability of X being equal or less than 400 as:

z=(X-\mu)/\sigma=(400-382.3)/20.8=17.7/20.8=0.851\\\\\\P(X\leq400)=P(z

6 0
3 years ago
Question 16 please...
Juliette [100K]
Brita have 323 in her account before the withdrawal
5 0
4 years ago
How many times can you put 40÷746 in
prisoha [69]
Well 746÷40=18.65 so would that be 18?
7 0
3 years ago
Read 2 more answers
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