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3 years ago

What is the perimeter of triangle PQR? Show your work.

Mathematics

1 answer:

zubka84 [21]3 years ago

6 0

You know that the two triangles are congruent because two sides and one angle are congruent.

According to CPCTC, corresponding parts of congruent triangles are congruent, which means that PR and SU are equal.

3y-2=y+4
3y=y+6
2y=6
y=3

3(3)-2=9-2=7

7+4+6=17

The perimeter is 17 feet.

Hope this helps!

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The overhead reach distances of adult females are normally distributed with a mean of 205 cm and a standard deviation of 8.3 cm.

Gre4nikov [31]

a. Find the probability that an individual distance is greater than 214.30 cm

We find for the value of z score using the formula:

z = (x – u) / s

z = (214.30 – 205) / 8.3

z = 1.12

Since we are looking for x > 214.30 cm, we use the right tailed test to find for P at z = 1.12 from the tables:

P = 0.1314

b. Find the probability that the mean for 20 randomly selected distances is greater than 202.80 cm

We find for the value of z score using the formula:

z = (x – u) / s

z = (202.80 – 205) / 8.3

z = -0.265

Since we are looking for x > 202.80 cm, we use the right tailed test to find for P at z = -0.265 from the tables:

P = 0.6045

c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

I believe this is because we are given the population standard deviation sigma rather than the sample standard deviation. So we can use the z test.

7 0

3 years ago

What is the area of this triangle? Enter your answer as a decimal. Round only your final answer to the nearest hundredth.

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See attached picture for solution:

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4 years ago

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Airida [17]

Answer:

A is the right one

Step-by-step explanation:

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Sphinxa [80]

Answer:

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Step-by-step explanation:

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3 years ago

Use left-endpoint approximation to estimate the area under the curve of f(x)=x^3 on [1,3]. Use n = 4. Using the same function us

natulia [17]

Split up the interval [1, 3] into 4 intervals of equal length,

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The left endpoint of the $i$ -th interval is

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The area under the curve is then approximately

$\displaystyle \int_1^3 f(x) \, dx \approx \sum_{i=1}^4 f(\ell_i)\Delta x \\\\ ~~~~~~~~ = \frac12 \sum_{i=1}^4 \left(1 + \frac i2\right)^3 \\\\ ~~~~~~~~ = \frac12 \sum_{i=1}^4 \left(1 + \frac{3i}2 + \frac{3i^2}4 + \frac{i^3}8\right) = \boxed{27}$

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1 year ago