Let's let the weight of a large box be L, and the weight of a small box be S.
We know that 5 large boxes and 3 small boxes is 120kg, so:
5L + 3S = 120
We also know that 7 large boxes and 9 small boxes is 234kg, so:
7L + 9S = 234
You can multiply the first equation by 3 to get:
15L + 9S = 360
See how now both equations have 9S? We can now subtract one from the other:
(15L+9S) - (7L+9S) = 360-234
8L = 126
L = 15.75
Now sub this value back into an equation:
(5x15.75) + 3S = 120
3S = 41.25
S = 13.75
Double check these values
(7x15.75) + (9x13.75)
= 110.25 + 123.75
=234, which is consistent with above.
So a large box is 15.75kg, and a small box is 13.75kg.
Hope this helped
9x squared - 6x - 6x + 4
9x squared - 12x + 4
2x2+2x-1=
4+2(-3)-1=
4+(-6)-1=
4+(-7)=
-3
Answer:
Step-by-step explanation
Hello!
Be X: SAT scores of students attending college.
The population mean is μ= 1150 and the standard deviation σ= 150
The teacher takes a sample of 25 students of his class, the resulting sample mean is 1200.
If the professor wants to test if the average SAT score is, as reported, 1150, the statistic hypotheses are:
H₀: μ = 1150
H₁: μ ≠ 1150
α: 0.05
![Z= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } } ~~N(0;1)](https://tex.z-dn.net/?f=Z%3D%20%5Cfrac%7BX%5Bbar%5D-Mu%7D%7B%5Cfrac%7BSigma%7D%7B%5Csqrt%7Bn%7D%20%7D%20%7D%20~~N%280%3B1%29)

The p-value for this test is 0.0949
Since the p-value is greater than the level of significance, the decision is to reject the null hypothesis. Then using a significance level of 5%, there is enough evidence to reject the null hypothesis, then the average SAT score of the college students is not 1150.
I hope it helps!
He only puts 1 in each glass and he has six. The answer is 6.