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Kruka [31]
3 years ago
11

Which place is home of the first known written code of law?

Mathematics
2 answers:
valentina_108 [34]3 years ago
7 0

the answer to your question is b. mesopotamia


lesya692 [45]3 years ago
5 0
The answer should be B. Mesopotamia
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Solve for x. Please show work.
abruzzese [7]

Answer:

x= to 5

Step-by-step explanation:


5 0
3 years ago
The distribution of the amount of money spent by first-time gamblers at a major casino in las vegas is approximately normal in s
Lostsunrise [7]

Solution: We are given:

\mu=600, \sigma =120

We need to find the z value corresponding to probability 0.84, in order to find the how much money almost 84% of gamblers spent at casino.

Using the standard normal table, we have:

z(0.85) = 0.9945

Now we will use the z score formula to find the required amount:

z=\frac{x-\mu}{\sigma}

0.9945=\frac{x-600}{120}

0.9945 \times 120 = x - 600

119.34 = x - 600

x = 600 + 119.34        

x = 719.34

x = 720 approximately

Therefore, almost 84% of gamblers spent more than $720 amount of money at this casino.

5 0
3 years ago
drug sniffing dogs must be 95% accurate in their responses because their handlers don't want them to miss durgs and also don't w
GenaCL600 [577]

Answer:

95% Confidence interval:  (0.8449,0.9951)

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 50

Number of times the dog is right, x = 46

\hat{p} = \dfrac{x}{n} = \dfrac{46}{50} = 0.92

95% Confidence interval:

\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}

z_{critical}\text{ at}~\alpha_{0.05} = 1.96

Putting the values, we get:

0.92 \pm 1.96(\sqrt{\dfrac{0.92(1-0.92)}{50}})\\\\ = 0.92\pm 0.0751\\\\=(0.8449,0.9951)

(0.8449,0.9951) is the required 95% confidence interval for the proportion of times the dog will be correct.

7 0
3 years ago
What times 85 dives you 2790
musickatia [10]

Answer:

34.875

Step-by-step explanation:

4 0
3 years ago
Please help thank you very much
Tom [10]

Answer:

4x^{2} -6x

Step-by-step explanation:

3 0
3 years ago
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