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zlopas [31]
3 years ago
13

According to records from the last 50 years, there is a 2/3 chance of June temperatures being above 80 F. Azul needs to find the

probability that it will be above 80' F for 20 or more days this June. There are 30 days in June.
Mathematics
2 answers:
MA_775_DIABLO [31]3 years ago
8 0

Answer:

The answer is 0.59

Step-by-step explanation:

We can find the probability the event by thinking the event as binomial distribution. Probability of a day temperature higher than 80F is p=0.67 and the probability of a day temperature equal or lower that 80F is q=0.33. Therefore, the probability of more that 20 days are hotter than 80F in the June will be:

P(X\geq 20)=combination(30,20)*0.67^{20}*0.33^{10}+combination(30,21)*0.67^{21}*0.33^{9}+....+combination(30,30)*0.67^{30}*0.33^{0}=0.59

umka21 [38]3 years ago
6 0
I think the answer is 52% because I got it correct on i-Ready. I really hope this helps because I hate that when you fail you have to redo a lesson. 

Have a great day and let me know if this is the right answer. 
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Chris wants to order blu-rays over the internet. Each Blu-ray cost $29.99 and shipping for
astra-53 [7]

Answer:

4 blu-rays

Step-by-step explanation:

*Note: I'm rounding up to full dollars to make this easier. You will get the same answer, whether you round up or not.

First, get rid of the shipping cost from the budget.

$150 - $10 = <u>$140 left for blu-rays</u>

Then divide $140 by $30, which is the price for a single blu-ray to get the number of blu-rays Chris can get.

$140 ÷ $30 ≈ 4.6 repeating. You can't have a fraction of a movie, so Chris can get 4 blu-rays

4 0
3 years ago
-5x + 10y =40<br> Solve intercept form
Tema [17]
I thought I knew the answer but I don’t Srry I need points
5 0
2 years ago
Fine the exact distance between the line 6x-y=-3 and the point (6,2). Show your work. Explain your answer.
AVprozaik [17]

Answer:

\sqrt{37} units

Step-by-step explanation:

If we draw a perpendicular from point (6,2) on the line 6x - y = - 3, then we have to find the length of the perpendicular.  

We know the formula of length of perpendicular from a point (x_{1}, y_{1} ) to the straight line ax + by + c = 0 is given by  

\frac{|ax_{1} + by_{1} +c |}{\sqrt{a^{2}+b^{2}}}

Therefore, in our case the perpendicular distance is  

\frac{|6(6)-2+3|}{\sqrt{6^{2} +(-1)^{2} } } = \frac{37}{\sqrt{37} } = \sqrt{37}  units. (Answer)

6 0
3 years ago
Lcm of 3x^(2) and 10
Ahat [919]

Answer:

the least common multiple is 18

Step-by-step explanation:

7 0
3 years ago
After once again losing a football game to the college'ss arch rival, the alumni association conducted a survey to see if alumni
antoniya [11.8K]

Answer:

a) The 90% confidence interval would be given (0.561;0.719).

b) p_v =P(z>2.8)=1-P(z

c) Using the significance level assumed \alpha=0.01 we see that p_v so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion is higher than 0.5 or 50%.    

Step-by-step explanation:

1) Data given and notation  

n=100 represent the random sample taken    

X=64 represent were in favor of firing the coach

\hat p=\frac{64}{100}=0.64 estimated proportion for were in favor of firing the coach

p_o=0.5 is the value that we want to test since the problem says majority    

\alpha represent the significance level (no given, but is assumed)    

z would represent the statistic (variable of interest)    

p_v represent the p value (variable of interest)    

p= population proportion of Americans for were in favor of firing the coach

Part a

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 90% confidence interval the value of \alpha=1-0.9=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=1.64

And replacing into the confidence interval formula we got:

0.64 - 1.64 \sqrt{\frac{0.64(1-0.64)}{100}}=0.561

0.64 + 1.64 \sqrt{\frac{0.64(1-0.64)}{100}}=0.719

And the 90% confidence interval would be given (0.561;0.719).

Part b

We need to conduct a hypothesis in order to test the claim that the proportion exceeds 50%(Majority). :    

Null Hypothesis: p \leq 0.5  

Alternative Hypothesis: p >0.5  

We assume that the proportion follows a normal distribution.    

This is a one tail upper test for the proportion of  union membership.  

The One-Sample Proportion Test is "used to assess whether a population proportion \hat p is significantly (different,higher or less) from a hypothesized value p_o".  

Check for the assumptions that he sample must satisfy in order to apply the test  

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.  

b) The sample needs to be large enough  

np_o =100*0.64=64>10  

n(1-p_o)=100*(1-0.64)=36>10  

Calculate the statistic    

The statistic is calculated with the following formula:  

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

On this case the value of p_o=0.5 is the value that we are testing and n = 100.  

z=\frac{0.64 -0.5}{\sqrt{\frac{0.5(1-0.5)}{100}}}=2.8

The p value for the test would be:  

p_v =P(z>2.8)=1-P(z

Part c

Using the significance level assumed \alpha=0.01 we see that p_v so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion is higher than 0.5 or 50%.    

6 0
3 years ago
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