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3 years ago

If X = 2 centimeters, Y = 4 centimeters, and Z = 6 centimeters, what is the area of the object?

Mathematics

2 answers:

Tatiana [17]3 years ago

5 0

Answer:

Step-by-step explanation:

the shape is a trpizoid with height y

A=(a+b)/2 *h (a=2x=4cm, h=y=4cm, b=2z=12 cm

A=(4+12)/2 * 4

Area=16/2 *4=8*4= 32 cm²

kirill [66]3 years ago

4 0

Answer:

24 Centimeters

Step-by-step explanation:

what you want to do is split it up, make two triangles (the sides), and one big square. The square has the dimensions 4 by 4 which has the area of 16. The two triangles have the dimensions 2 by 4, which would be 4 and because there are two, the triangles equal 8. 16+8=24

Hope this helps. :)

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Write the quadratic equation in standard form going through the points (-1,-5) (0,-4) (2,10)

Alexxandr [17]

Dismal high dash app nah hell shall facho call hi Chuck coco go scanners cop

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2 years ago

You recently sent out a survey to determine if the percentage of adults who use social media has changed from 66%, which was the

il63 [147K]

Answer:

The 98% confidence interval estimate of the proportion of adults who use social media is (0.56, 0.6034).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

Of the 2809 people who responded to survey, 1634 stated that they currently use social media.

This means that $n = 2809, \pi = \frac{1634}{2809} = 0.5817$

98% confidence level

So $\alpha = 0.02$ , z is the value of Z that has a pvalue of $1 - \frac{0.02}{2} = 0.99$ , so $Z = 2.327$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5817 - 2.327\sqrt{\frac{0.5817*4183}{2809}} = 0.56$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5817 + 2.327\sqrt{\frac{0.5817*4183}{2809}} = 0.6034$

The 98% confidence interval estimate of the proportion of adults who use social media is (0.56, 0.6034).

6 0

2 years ago

The average amount of time that students use computers at a university computer center is 36 minutes with a standard deviation o

frosja888 [35]

Answer:

2119 students use the computer for more than 40 minutes. This number is higher than the threshold estabilished of 2000, so yes, the computer center should purchase the new computers.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 36, \sigma = 5$

The first step to solve this question is finding the proportion of students which use the computer more than 40 minutes, which is 1 subtracted by the pvalue of Z when X = 40. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{40 - 36}{5}$

$Z = 0.8$

$Z = 0.8$ has a pvalue of 0.7881.

1 - 0.7881 = 0.2119

So 21.19% of the students use the computer for longer than 40 minutes.

Out of 10000

0.2119*10000 = 2119

2119 students use the computer for more than 40 minutes. This number is higher than the threshold estabilished of 2000, so yes, the computer center should purchase the new computers.

8 0

3 years ago

Suppose f varies inversely with g and that f = 28 when g = 2. What is the value of f when g = 7?

Lesechka [4]

Make up the proportion

$28*2=x*7\\7x=56\\x=8$

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3 0

3 years ago

Read 2 more answers

Simplify

OlgaM077 [116]

Answer:

see below

Step-by-step explanation:

$\frac{x-1}{x^2-3x+2}+ \frac{x-2}{x^2-5x+6} +\frac{x-5}{x^2-8x+15}$

we need to simplify that

$x^2-3x+2=(x-1)(x-2)\\\\x^2-5x+6=(x-2)(x-3)\\\\x^2-8x+15=(x-3)(x-5)$

so we can continue

$\frac{x-1}{(x-1)(x-2)}=\frac{1}{x-2}\\\\\frac{x-2}{(x-2)(x-3)} =\frac{1}{x-3}\\\\\frac{x-5}{(x-3)(x-5)} =\frac{1}{x-3}$

and we can put all together

$\frac{1}{x-2}+ \frac{1}{x-3}+ \frac{1}{x-3}\\\\\frac{1}{x-2} +\frac{2}{x-3}\\\\\frac{x-3}{(x-3)(x-2)}+ \frac{2(x-2)}{(x-2)(x-3)} \\\\\frac{x-3+2x-4}{(x-3)(x-2)}\\\\\frac{3x-7}{x^2-5x+6}$

3 0

3 years ago