Sorry wrong one!!!!!!!!!!!!!!!!!!!!!!!!!!!!11

zhannawk [14.2K]

Answer:

2.5 is 12PM temp and difference is 3.9

Step-by-step explanation:

-0.5+3 because it is warmer and this is 2.5 also both B and C can be done to get the correct answer. and that is 2.5--1.4 which is the same as 2.5+1.4 which is 3.9

3 0

3 years ago

What type of decimal is this fraction: 11/20

Andre45 [30]

Answer:

I think the answer is repeating because it'd be

0.55

4 0

3 years ago

Three cylinders have bases that are the

lys-0071 [83]

SA: 2pi (r^2)+2pi(r)(h)
In order to find radius (r) of the base you must divide 10 by pi then find the square root of the result
√(10/pi)=r: 1.784cm
h: a) 8.0cm
b) 6.5cm
c) 9.4cm
2pi (3.183)+2pi(1.784)(8)=109.673cm^2
2pi (3.183)+2pi(1.784)(6.5)=92.859cm^2
2pi (3.183)+2pi(1.784)(9.4)=125.366cm^2

7 0

3 years ago

Hey! I just want to make sure if I'm right, if I'm not, whats the answer? I'm stuck on this but I will highly appreciate it if y

cluponka [151]

Answer: yes you are correct

Step-by-step explanation:

5 0

3 years ago

] It is claimed that 42% of US college graduates had a mentor in college. For a sample of college graduates in Colorado, it was

Bad White [126]

Answer:

$z=\frac{0.480 -0.42}{\sqrt{\frac{0.42(1-0.42)}{1045}}}=3.930$

The p value for this case would be given by:

$p_v =P(z>3.930)=0.0000443$

For this case the p value is lower than the significance level so then we have enough evidence to reject the null hypothesis so then we can conclude that the true proportion is significantly higher than 0.42

Step-by-step explanation:

Information given

n=1045 represent the random sample selected

X=502 represent the college graduates with a mentor

$\hat p=\frac{502}{1045}=0.480$ estimated proportion of college graduates with a mentor

$p_o=0.42$ is the value that we want to test

z would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to test if the true proportion is higher than 0.42, the system of hypothesis are.:

Null hypothesis: $p \leq 0.42$

Alternative hypothesis: $p > 0.42$

The statistic is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

Replacing the info we got:

$z=\frac{0.480 -0.42}{\sqrt{\frac{0.42(1-0.42)}{1045}}}=3.930$

The p value for this case would be given by:

$p_v =P(z>3.930)=0.0000443$

For this case the p value is lower than the significance level so then we have enough evidence to reject the null hypothesis so then we can conclude that the true proportion is significantly higher than 0.42

7 0

3 years ago