Answer:
the answer is -1
Step-by-step explanation:
Assuming 5a3 equals 5a^3, and the same for everything else, you would plug in 4 for a, so 5(4)^3 - 2(4)^2 + 4 - 45, which equals 247.
Answer:
Bk=100
I' hope it's helpful for you
Answer:
![a)\ \ \bar x_m-\bar x_f=67.03\\\\b)\ \ E=15.7416\\\\c)\ \ CI=[51.2884, \ 82.7716]](https://tex.z-dn.net/?f=a%29%5C%20%5C%20%5Cbar%20x_m-%5Cbar%20x_f%3D67.03%5C%5C%5C%5Cb%29%5C%20%5C%20E%3D15.7416%5C%5C%5C%5Cc%29%5C%20%5C%20CI%3D%5B51.2884%2C%20%5C%2082.7716%5D)
Step-by-step explanation:
a. -Given that:

#The point estimator of the difference between the population mean expenditure for males and the population mean expenditure for females is calculated as:

Hence, the pointer is estimator 67.03
b. The standard error of the point estimator,
is calculated by the following following:

-And the margin of error, E at a 99% confidence can be calculated as:

Hence, the margin of error is 15.7416
c. The estimator confidence interval is calculated using the following formula:

#We substitute to solve for the confidence interval using the standard deviation and sample size values in a above:
![CI=\bar x_m-\bar x_f\ \pm z_{\alpha/2}\sqrt{\frac{\sigma_m^2}{n_m}+\frac{\sigma_f^2}{n_f}}\\\\=(135.67-68.64)\pm 15.7416\\\\=67.03\pm 15.7416\\\\=[51.2884, \ 82.7716]](https://tex.z-dn.net/?f=CI%3D%5Cbar%20x_m-%5Cbar%20x_f%5C%20%5Cpm%20z_%7B%5Calpha%2F2%7D%5Csqrt%7B%5Cfrac%7B%5Csigma_m%5E2%7D%7Bn_m%7D%2B%5Cfrac%7B%5Csigma_f%5E2%7D%7Bn_f%7D%7D%5C%5C%5C%5C%3D%28135.67-68.64%29%5Cpm%2015.7416%5C%5C%5C%5C%3D67.03%5Cpm%2015.7416%5C%5C%5C%5C%3D%5B51.2884%2C%20%5C%2082.7716%5D)
Hence, the 99% confidence interval is [51.2884,82.7716]
Answer:
Hey there!
5(v+3)-10v>-10
5v+15-10v>-10
-5v+15>-10
-5v>-25
v<5
Let me know if this helps :)