Answer:
I dont know sorry so much
AB = √(8^2 + 6^2)
AB = √100
AB = 10
AC = √(8^2 + 15^2)
AC = √289
AC = 17
BC = 9
P= AB + AC + BC
P = 10 + 17 + 9
P = 36 units
34)
Area of ABC = 1/2 x 8 x 9
A = 36 square units
Speed is distance traveled per block of time. As an equation, this is
Speed = distance/time
In your problem, the given distance is 300km, and the time is 2.5 hours.
Plugging into the equation,
Speed = 300km / 2.5 h
Speed = 120 km/h
Velocity is speed <em>and</em> direction.
The velocity would be 120 km/h north.
Answer:
Quarter 2
Explanation:
Quarter 1 had a 4.19505% increase in earnings, but quarter 2 had a 11.586% increase.
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!