12.5 cups of flour are required to make 2.5 batches
Answer:
y = 3x - 1
Step-by-step explanation:
The equation of a line in slope- intercept form is
y = mx + c ( m is the slope and c the y- intercept )
Given
3x - y + 12 = 0 ( subtract 3x + 12 from both sides )
- y = - 3x - 12 ( multiply through by - 1 )
y = 3x + 12 ← in slope- intercept form
with slope m = 3
Parallel lines have equal slopes , thus
y = 3x + c ← is the partial equation
To find c substitute (2, 5) into the partial equation
5 = 6 + c ⇒ c = 5 - 6 = - 1
y = 3x - 1 ← equation of parallel line
Answer:
4650 add 1500+1800+3400 u get 6700
then you get the tuition cost then subtract it by 6700
your final answer is 4650 dollar
Answer:
a. CI=[128.79,146.41]
b. CI=[122.81,152.39]
c. As the confidence level increases, the interval becomes wider.
Step-by-step explanation:
a. -Given the sample mean is 137.6 and the standard deviation is 20.60.
-The confidence intervals can be constructed using the formula;

where:
is the sample standard deviation
is the s value of the desired confidence interval
we then calculate our confidence interval as:
![\bar X\pm z\frac{s}{\sqrt{n}}\\\\=137.60\pm z_{0.05/2}\times\frac{20.60}{\sqrt{21}}\\\\=137.60\pm1.960\times \frac{20.60}{\sqrt{21}}\\\\=137.60\pm8.8108\\\\\\=[128.789,146.411]](https://tex.z-dn.net/?f=%5Cbar%20X%5Cpm%20z%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D%5C%5C%5C%5C%3D137.60%5Cpm%20z_%7B0.05%2F2%7D%5Ctimes%5Cfrac%7B20.60%7D%7B%5Csqrt%7B21%7D%7D%5C%5C%5C%5C%3D137.60%5Cpm1.960%5Ctimes%20%5Cfrac%7B20.60%7D%7B%5Csqrt%7B21%7D%7D%5C%5C%5C%5C%3D137.60%5Cpm8.8108%5C%5C%5C%5C%5C%5C%3D%5B128.789%2C146.411%5D)
Hence, the 95% confidence interval is between 128.79 and 146.41
b. -Given the sample mean is 137.6 and the standard deviation is 20.60.
-The confidence intervals can be constructed using the formula in a above;
![\bar X\pm z\frac{s}{\sqrt{n}}\\\\=137.60\pm z_{0.01/2}\times\frac{20.60}{\sqrt{21}}\\\\=137.60\pm3.291\times \frac{20.60}{\sqrt{21}}\\\\=137.60\pm 14.7940\\\\\\=[122.806,152.394]](https://tex.z-dn.net/?f=%5Cbar%20X%5Cpm%20z%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D%5C%5C%5C%5C%3D137.60%5Cpm%20z_%7B0.01%2F2%7D%5Ctimes%5Cfrac%7B20.60%7D%7B%5Csqrt%7B21%7D%7D%5C%5C%5C%5C%3D137.60%5Cpm3.291%5Ctimes%20%5Cfrac%7B20.60%7D%7B%5Csqrt%7B21%7D%7D%5C%5C%5C%5C%3D137.60%5Cpm%2014.7940%5C%5C%5C%5C%5C%5C%3D%5B122.806%2C152.394%5D)
Hence, the variable's 99% confidence interval is between 122.81 and 152.39
c. -Increasing the confidence has an increasing effect on the margin of error.
-Since, the sample size is particularly small, a wider confidence interval is necessary to increase the margin of error.
-The 99% Confidence interval is the most appropriate to use in such a case.