Answer:
(A) We failed to reject the null hypothesis
Step-by-step explanation:
Given that a high school teacher hypothesized that in her class, boys were taller than girls. If the probability value of her null hypothesis is 0.34, it means:
Here p value is 0.34
This is greater than 0.05 or 0.1 i.e. both for 5% significance and 10% significance.
Whenever p > alpha, we fail to reject null hypothesis
A) We failed to reject the null hypothesis
This would be the correct option
i.e. final conclusion is that there is no significant difference between mean height of girls and mean height of boys at 5% or 10% significance level.
Answer:
Hi there!
Your answer is:
2 & 1/4 <u>OR</u> 2.25 cups fills the container!
Step-by-step explanation:
If 1 & 1/2 (aka 3/2) fills 2/3 of the container, then:
we know that <em>half</em> of 3/2s fills 1/3 of the container
To find half of 3/2, we multiply 3/2 by the <em><u>inverse</u></em> of 2, which is 1/2.
3/2÷2
3/2×1/2 = 3/4 ths
Now that we know 3/4ths fills 1/3 of the container, we add that to 1& 1/2.
3/4+ 1&1/2 = 2 & 1/4
Check your work!
3/4 × 3 should equal 2&1/4 if we are correct
3/4 × 3/1 = 9/4.
Simplify!
9/4 = 2&1/4
That tells us that <u>we</u><u> </u><u>are</u><u> </u><u>correct</u><u>!</u>
Hope this helps!
8, 3, -2, -7
basically
you have to figure how to get from one term to the other. so in this case, to get from 8 to 3, you have to minus 5. and same again to get from 3 to -2, you again have to -5.
so the first past of the answer is -5n.
then after that, you have to figure out how to get from -5, to the first term of your sequence. so you have to add 13, in this case.
your answer would then be
<h2>
<em><u>
-5n + 13</u></em></h2>
Best Answer: 2 LiCl = 2 Li + Cl2
mass Li = 56.8 mL x 0.534 g/mL=30.3 g
moles Li = 30.3 g / 6.941 g/mol=4.37
the ratio between Li and LiCl is 2 : 2 ( or 1 : 1)
moles LiCl required = 4.37
mass LiCl = 4.37 mol x 42.394 g/mol=185.3 g
Cu + 2 AgNO3 = Cu/NO3)2 + 2 Ag
the ratio between Cu and AgNO3 is 1 : 2
moles AgNO3 required = 4.2 x 2 = 8.4 : but we have only 6.3 moles of AgNO3 so AgNO3 is the limiting reactant
moles Cu reacted = 6.3 / 2 = 3.15
moles Cu in excess = 4.2 - 3.15 =1.05
N2 + 3 H2 = 2 NH3
moles N2 = 42.5 g / 28.0134 g/mol=1.52
the ratio between N2 and H2 is 1 : 3
moles H2 required = 1.52 x 3 =4.56
actual moles H2 = 10.1 g / 2.016 g/mol= 5.00 so H2 is in excess and N2 is the limiting reactant
moles NH3 = 1.52 x 2 = 3.04
mass NH3 = 3.04 x 17.0337 g/mol=51.8 g
moles H2 in excess = 5.00 - 4.56 =0.44
mass H2 in excess = 0.44 mol x 2.016 g/mol=0.887 g