Is it possible to have three shaded sections when graphing a single linear

A simple random sample of 32 men from a normally distributed population results in a standard deviation of 12.9 beats per minute

krok68 [10]

Answer:

Null Hypothesis, H0 = The pulse rates of men have a standard deviation equal to 10 beats per minute

Alternate Hypothesis, H1 = The pulse rates of men do not have a standard deviation equal to 10 beats per minute

Step-by-step explanation:

The null hypothesis is basically the problem statement i.e

Pulse rates of men have a standard deviation equal to 10 beats per minute

Hence, H0 = The pulse rates of men have a standard deviation equal to 10 beats per minute

The alternate hypothesis will contradict or negate the null hypothesis i.e

H1 = The pulse rates of men do not have a standard deviation equal to 10 beats per minute

4 0

2 years ago

A line passes through the point (10,3)

pishuonlain [190]

We have to find b first:

3 = -3/2(10)+b
3 = -15+b
18 = b

Now right the equation, your slope will be the same one....

Y = -3/2x + 18

5 0

3 years ago

There are 3 red and 8 green balls in a bag. If you randomly choose balls one at a time, with replacement, what is the probabilit

Kaylis [27]

So, the total number of balls is 11. We want to pick 2 red balls and 1 green ball. WLOG (since order doesnt matter here), we can say he picks red, green, red. That means on his first pick, he has a $\frac{3}{11}$ chance of picking the red ball, and he places it back in the bag. The probability of picking a green ball is $\frac{8}{11}$ , and then he places the ball back in the bag. The probability of picking the last red ball is the same as the last red ball example, and we simply multiply the probabilities together as per the multiplication rule to get:

$\frac{3}{11}\frac{3}{11}\frac{8}{11}=\frac{3*3*8}{11^3}=\frac{72}{1331}$

Now, without replacement the order does matter. He picks a red ball, a red ball then a green ball. The probability of picking the first red ball is $\frac{2}{11}$ , and the probability of picking the second red ball is $\frac{1}{10}$ and the probability of picking the green ball is $\frac{1}{9}$ . We want to multiply thm again, as per the multiplication rule like the last problem.