What is the constant of proportionality?

The mean annual tuition and fees for a sample of 15 private colleges was with a standard deviation of . A dotplot shows that it

Fudgin [204]

Answer:

Step-by-step explanation:

The question is incomplete. The complete question is:

The mean annual tuition and fees for a sample of 15 private colleges was $35,500 with a standard deviation of $6500. A dotplot shows that it is reasonable to assume that the population is approximately normal. You wish to test whether the mean tuition and fees for private colleges is different from $32,500. State the null and alternate hypotheses. A) H0: 4 = 32,500, H:4=35,500 C) H: 4 = 35,500, H7:35,500 B) H: 4 = 32,500, H : 4 # 32,500 D) H0:41 # 32,500, H : 4 = 32,500

Solution

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

H0: µ = 32500

For the alternative hypothesis,

Ha: µ ≠ 32500

This is a two tailed test.

Since the number of samples is small and the population standard deviation is not given, the distribution is a student's t.

Since n = 15,

Degrees of freedom, df = n - 1 = 15 - 1 = 14

t = (x - µ)/(s/√n)

Where

x = sample mean = 35500

µ = population mean = 32500

s = samples standard deviation = 6500

t = (35500 - 32500)/(6500/√15) = 1.79

We would determine the p value using the t test calculator. It becomes

p = 0.095

Assuming alpha = 0.05

Since alpha, 0.05 < than the p value, 0.095, then we would fail to reject the null hypothesis.

7 0

3 years ago

A lot of 80 bulbs contains 45 defective and 32 non warranty bulbs. One bulb is drawn at random from the lot. What is the probabi

kenny6666 [7]

Step-by-step explanation:

Total bulbs = 80

Probability of defective bulbs = 45/80 = 9/16

Probability of non warranty bulbs = 32/80 = 4/5

5 0

3 years ago

If 7x-2y=z then -14x+4y+3z

creativ13 [48]

I'm confused what are you suppose to do

8 0

3 years ago

Help. I need help with these questions ( see image).<br> Please show workings.

Andrew [12]

9514 1404 393

Answer:

4) 6x

5) 2x +3

Step-by-step explanation:

We can work both these problems at once by finding an applicable rule.

$\text{For $f(x)=ax^n$}\\\\\lim\limits_{h\to 0}\dfrac{f(x+h)-f(x)}{h}=\lim\limits_{h\to 0}\dfrac{a(x+h)^n-ax^n}{h}\\\\=\lim\limits_{h\to 0}\dfrac{ax^n+anx^{n-1}h+O(h^2)-ax^n}{h}=\boxed{anx^{n-1}}$

where O(h²) is the series of terms involving h² and higher powers. When divided by h, each term has h as a multiplier, so the series sums to zero when h approaches zero. Of course, if n < 2, there are no O(h²) terms in the expansion, so that can be ignored.

This can be referred to as the <em>power rule</em>.

Note that for the quadratic f(x) = ax^2 +bx +c, the limit of the sum is the sum of the limits, so this applies to the terms individually:

lim[h→0](f(x+h)-f(x))/h = 2ax +b

__

4. The gradient of 3x^2 is 3(2)x^(2-1) = 6x.

5. The gradient of x^2 +3x +1 is 2x +3.

__

If you need to "show work" for these problems individually, use the appropriate values for 'a' and 'n' in the above derivation of the power rule.

3 0

3 years ago

I'm stuck on graphing this quadratic equation:<br><img src="https://tex.z-dn.net/?f=y%20%5Cleqslant%20%20%7Bx%7D%5E%7B2%7D%20%20

rewona [7]

Your welcome I hope it is right

7 0

3 years ago