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Natalija [7]
3 years ago
12

PLEASE HELP ME FAST!!!!!!!!11

Mathematics
1 answer:
elena-14-01-66 [18.8K]3 years ago
7 0
I think it’s the second one please correct me if I’m wrong not perfect at this
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Tanzania [10]
Aas i belive . If not consult the other answer
6 0
3 years ago
C2d3 has a solubility product constant of 9.14×10−9. what is the molar solubility of c2d3?
UkoKoshka [18]
The correct answer for the question that is being presented above is this one:

We need to express the ksp expression of C2D3
<span>C2D3
= (2x)2(3x)3
= 108x5 </span>

<span>Then set that equation equal to your solubility constant </span>

<span>9.14x10-9 = 108x5 </span>
<span>x = 9.67x10-3 
</span>
<span>So the molar solubility is 9.67x10-3</span>
4 0
3 years ago
Who knows how to solve these? I need help!!
Elina [12.6K]
1= 0.25
2=1.12
3= -075
4 0
3 years ago
A sample of 12 radon detectors of a certain type was selected, and each was exposed to 100 pCI/L of radon. The resulting reading
valkas [14]

Answer:

Null hypothesis:\mu = 100  

Alternative hypothesis:\mu \neq 100  

t=\frac{98.375-100}{\frac{6.109}{\sqrt{12}}}=-0.921  

p_v =2*P(t_{11}  

Step-by-step explanation:

1) Data given and notation  

Data: 105.6, 90.9, 91.2, 96.9, 96.5, 91.3, 100.1, 105.0, 99.6, 107.7, 103.3, 92.4

We can calculate the sample mean and deviation for this data with the following formulas:

\bar X =\frac{\sum_{i=1}^n X_i}{n}

s=\sqrt{\frac{\sum_{i=1}^n (X_i- \bar X)^2}{n-1}}

The results obtained are:

\bar X=98.375 represent the sample mean  

s=0.6.109 represent the sample standard deviation  

n=12 sample size  

\mu_o =100 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

2) State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is equal to 100pCL/L, the system of hypothesis are :  

Null hypothesis:\mu = 100  

Alternative hypothesis:\mu \neq 100  

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

3) Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{98.375-100}{\frac{6.109}{\sqrt{12}}}=-0.921  

4) P-value  

First we need to find the degrees of freedom for the statistic given by:

df=n-1=12-1=11

Since is a two sided test the p value would given by:  

p_v =2*P(t_{11}  

5) Conclusion  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significant different from 100 at 5% of significance.  

3 0
3 years ago
10,000 more than 395,040is?
Marizza181 [45]
405,040 hope that helps.
7 0
3 years ago
Read 2 more answers
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