Answer:
$-4
Step-by-step explanation:
Your welcome
Answer:
=53*(54 ×^9 y^12 z^15)^1/2
=53*y^6 *z^(15/2) * x^( 9/2)* (54)^1/2
Answer: 6
Step-by-step explanation:
X + y = 24....multiply by -3
3x + 5y = 100
---------------
-3x - 3y = - 72 (result of multiplying by -3)
3x + 5y = 100
--------------add
2y = 28
y = 28/2
y = 14
x + y = 24
x + 14 = 24
x = 24 - 14
x = 10
solution is (10,14)...and x represents 3 point q's and y represents 5 point q's......so this tells me there are 10 three point q's and 14 five point q's.
If inspection department wants to estimate the mean amount with 95% confidence level with standard deviation 0.05 then it needed a sample size of 97.
Given 95% confidence level, standard deviation=0.05.
We know that margin of error is the range of values below and above the sample statistic in a confidence interval.
We assume that the values follow normal distribution. Normal distribution is a probability that is symmetric about the mean showing the data near the mean are more frequent in occurence than data far from mean.
We know that margin of error for a confidence interval is given by:
Me=
α=1-0.95=0.05
α/2=0.025
z with α/2=1.96 (using normal distribution table)
Solving for n using formula of margin of error.

n=
=96.4
By rounding off we will get 97.
Hence the sample size required will be 97.
Learn more about standard deviation at brainly.com/question/475676
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The given question is incomplete and the full question is as under:
If the inspection division of a county weights and measures department wants to estimate the mean amount of soft drink fill in 2 liters bottles to within (0.01 liter with 95% confidence and also assumes that standard deviation is 0.05 liter. What is the sample size needed?