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3 years ago

The length of a rectangular portrait is 1.5 times the width. The perimeter is 10 feet.

1 answer:

8 0

In order to solve the problem above, represent the width by x. With this, the length of the rectangular portrait is 1.5x. The perimeter of the rectangle is two times the sum of the length and width. This translates to,

2 (1.5x + x) = 10 ft, x = 2 ft.

Thus, the length of the rectangular portrait is 3 ft.

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I =125 r=6% t=1 what is p

vfiekz [6]

There is no P in here, you can't solve for P without a P, Or I just don't get the question.

4 0

3 years ago

A computer manufacturing company has sent a mail survey to 2,800 of its randomly selected customers that have purchased a new la

vodka [1.7K]

Answer:

The 96% confidence interval for the population proportion of customers satisfied with their new computer is (0.77, 0.83).

Step-by-step explanation:

We have to calculate a 96% confidence interval for the proportion.

We consider the sample size to be the customers that responded the survey (n=800), as we can not assume the answer for the ones that did not answer.

The sample proportion is p=0.8.

$p=X/n=640/800=0.8$

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.8*0.2}{800}}\\\\\\ \sigma_p=\sqrt{0.0002}=0.014$

The critical z-value for a 96% confidence interval is z=2.054.

The margin of error (MOE) can be calculated as:

$MOE=z\cdot \sigma_p=2.054 \cdot 0.014=0.03$

Then, the lower and upper bounds of the confidence interval are:

$LL=p-z \cdot \sigma_p = 0.8-0.03=0.77\\\\UL=p+z \cdot \sigma_p = 0.8+0.03=0.83$

The 96% confidence interval for the population proportion is (0.77, 0.83).

4 0

2 years ago

Factor completely −5x2 + 10x − 25.

garri49 [273]

5(2x-7)

Hope this helps !

Have a great day!

3 0

2 years ago

A group of 11 people are going to the movies one Saturday. Adult tickets cost $12.50 each and child tickets cost $7.75 each. If

Lostsunrise [7]

The answer is B. 5 !!

3 0

2 years ago

Assume that when adults with smartphones are randomly selected, 54% use them in meetings or classes (based on data from an lg sm

GuDViN [60]

Answer: 0.951%

Explanation:

Note that in the problem, the scenario is either the adult is using or not using smartphones. So, we have a yes or no scenario involved with the random variable, which is the number of adults using smartphones. Thus, the number of adults using smartphones follows the binomial distribution.

Let x be the number of adults using smartphones and n be the number of randomly selected adults. In Binomial distribution, the probability that there are k adults using smartphones is given by

$P(x = k) = \frac{n!}{k!(n-k)!}p^k (1-p)^{n-k}$

Where p = probability that an adult is using smartphones = 54% (since 54% of adults are using smartphones).

Since n = 12 and k = 3, the probability that fewer than 3 are using smartphones is given by

$P(x \ \textless \ 3) = P(x = 0) + P(x = 1) + P(x = 2)
\\ \indent = \frac{12!}{0!(12-0)!}(0.54)^0 (1-0.54)^{12-0} + \frac{12!}{1!(12-1)!}(0.54)^1 (1-0.54)^{12-1} + \\ \indent \frac{12!}{2!(12-2)!}(0.54)^2 (1-0.54)^{12-2}
\\
\\ \indent = \frac{12!}{(1)(12!)}(0.46)^{12} + \frac{12(11!)}{(1)(11!)}(0.54)(0.46)^{11}+ \frac{12(11)(10!)}{(2)(10!)}(0.54)^2(0.46)^{10}
\\
\\ \indent = (1)(0.46)^{12} + (12)(0.54)(0.46)^{11}+ (66)(0.54)^2(0.46)^{10}
\\ \indent \boxed{P(x \ \textless \ 3) \approx 0.00951836732 }
$

Therefore, the probability that there are fewer than 3 adults are using smartphone is 0.00951 or 0.951%.

5 0

3 years ago