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Sholpan [36]
3 years ago
7

At a concert, 20% of the people are wearing black dresses or suits, 1/4 are wearing navy, 0.35 are wearing brown, and the rest a

re wearing a variety of colors. (other). What percent are wearing “other”? Type in your answer using the decimal equivalent of this amount.
Mathematics
1 answer:
Elan Coil [88]3 years ago
5 0

Answer: 20%

Step-by-step explanation:

20% black dresses or suits

.35 = 35% wearing brown

1/4 = 25% wearing navy

100% -20% - 35% - 25% = 20% left

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3(4a+b) What matches this equation
vivado [14]

Answer:

12a + 3b.

Step-by-step explanation:

3(4a + b)

= 3 * 4a + 3 * b

= 12a + 3b.

Hope this helps!

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3 years ago
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A researcher has two percentages and wants to know if the percentages are statistically different. The researcher calculates the
love history [14]

Answer:

p_v =2*P(Z>4.21) =2.55x10^{-5}

And we can use the following excel code to find it:"=2*(1-NORM.DIST(4.21,0,1,TRUE)) "

With the p value obtained and using the significance level assumed for example\alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the percentage 1 is significantly different from the percentage 2.

D) Are statistically different.

Step-by-step explanation:

The system of hypothesis on this case are:  

Null hypothesis: \mu_1 = \mu_2  

Alternative hypothesis: \mu_1 \neq \mu_2  

Or equivalently:  

Null hypothesis: \mu_1 - \mu_2 = 0  

Alternative hypothesis: \mu_1 -\mu_2\neq 0  

Where \mu_1 and \mu_2 represent the percentages that we want to test on this case.

The statistic calculated is on this case was Z=4.21. Since we are conducting a two tailed test the p value can be founded on this way.

p_v =2*P(Z>4.21) =2.55x10^{-5}

And we can use the following excel code to find it:"=2*(1-NORM.DIST(4.21,0,1,TRUE)) "

With the p value obtained and using the significance level assumed for example\alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the percentage 1 is significantly different from the percentage 2.

And the best option on this case would be:

 D) Are statistically different.

4 0
4 years ago
In a study of the accuracy of fast food​ drive-through orders, Restaurant A had 229229 accurate orders and 7070 that were not ac
tiny-mole [99]

Answer:

a) 90% Confidence interval for true proportion of orders that are not accurate

= (0.194, 0.274)

In percentage terms, (19.4%, 27.4%)

b) Since the two confidence intervals overlap, neither restaurant appears to have a significantly different percentage of orders that are not accurate.

Step-by-step explanation:

a) Confidence Interval for the population proportion of orders that are not accurate is basically an interval of range of values where the true population proportion of orders that are not accurate can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample proportion) ± (Margin of error)

Sample proportion of orders that were not accurate = 70 ÷ (70 + 229) = 0.2341

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error)

Critical value at 90% confidence interval for sample size of (70+229=299) is obtained from the z-tables because although, the population standard deviation is not known, the sample size is large enough.

Critical value = 1.645 (from the z-tables)

Standard error of the mean = σₓ = √[p(1-p)/N]

p = sample proportion of orders that are not accurate = 0.2341

n = sample size = 299

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95% Confidence Interval = (Sample proportion) ± [(Critical value) × (standard Error)]

CI = 0.2341 ± (1.645 × 0.02449)

CI = 0.2341 ± 0.04028

90% CI = (0.19382, 0.27438)

90% Confidence interval = (0.194, 0.274)

b) Confidence Interval for true proportion of orders that are not accurate

For Restaurant A = (0.194, 0.274)

For Restaurant B = (0.216, 0.297)

There is overlap. For example, p = 0.235 is possible for both intervals, which means that both restaurants could have a 23.5% inaccuracy rate. If there was no overlap of the intervals, then it would be impossible for p to be the same value for both restaurants. So again, we could conclude that the restaurants could have the same inaccurate order rate due to the overlap. There just isn't enough evidence to suggest that the true percentage of orders that are not accurate for the 2 restaurants are very different.

Hope this Helps!!!

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creativ13 [48]

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