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3 years ago

Topic 4 Pythagorean Theorem Practice 11/18

Mathematics

2 answers:

jasenka [17]3 years ago

7 0

7. ≈6.9
8.≈5.2
9.≈ 7.1
10.≈ 7.6
11. ≈6.7
12.≈ 5.7

omeli [17]3 years ago

4 0

7) 8.9
8) 6.7
9) 12.2
10) 7.6
11) 7.3
12) 6.3

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What is the way to best choose the answers

Sunny_sXe [5.5K]

The B option is the correct answer

7 0

3 years ago

Read 2 more answers

Lesson: 1.08Given this function: f(x) = 4 cos(TTX) + 1Find the following and be sure to show work for period, maximum, and minim

Ber [7]

The given function is

$f(x)=4\cos \text{(}\pi x)+1$

The general form of the cosine function is

$y=a\cos (bx+c)+d$

a is the amplitude

2pi/b is the period

c is the phase shift

d is the vertical shift

By comparing the two functions

a = 4

b = pi

c = 0

d = 1

Then its period is

$\begin{gathered} \text{Period}=\frac{2\pi}{\pi} \\ \text{Period}=2 \end{gathered}$

The equation of the midline is

$y_{ml}=\frac{y_{\max }+y_{\min }}{2}$

Since the maximum is at the greatest value of cos, which is 1, then

$\begin{gathered} y_{\max }=4(1)+1 \\ y_{\max }=5 \end{gathered}$

Since the minimum is at the smallest value of cos, which is -1, then

$\begin{gathered} y_{\min }=4(-1)+1 \\ y_{\min }=-4+1 \\ y_{\min }=-3 \end{gathered}$

Then substitute them in the equation of the midline

$\begin{gathered} y_{ml}=\frac{5+(-3)}{2} \\ y_{ml}=\frac{2}{2} \\ y_{ml}=1 \end{gathered}$

The answers are:

Period = 2

Equation of the midline is y = 1

Maximum = 5

Minimum = -3

3 0

1 year ago

HELPPPPPPPPP ME PLEASE????!!?!?!?!?!

Gnesinka [82]

Answer:

More than half

Step-by-step explanation:

Half is 0.5 this is 0.785 approximately 0.280 more or > 0.5

4 0

3 years ago

Solve the problem. Use the Central Limit Theorem.The annual precipitation amounts in a certain mountain range are normally distr

bazaltina [42]

Answer:

0.8944 = 89.44% probability that the mean annual precipitation during 25 randomly picked years will be less than 112 inches.

Step-by-step explanation:

To solve this question, we use the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 109.0 inches, and a standard deviation of 12 inches.

This means that $\mu = 109, \sigma = 12$

Sample of 25.

This means that $n = 25, s = \frac{12}{\sqrt{25}} = 2.4$

What is the probability that the mean annual precipitation during 25 randomly picked years will be less than 112 inches?

This is the p-value of Z when X = 112. So

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

By the Central Limit Theorem

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

$Z = \frac{112 - 109}{2.4}$

$Z = 1.25$

$Z = 1.25$ has a p-value of 0.8944.

0.8944 = 89.44% probability that the mean annual precipitation during 25 randomly picked years will be less than 112 inches.

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3 years ago

Solve the inequality <br><br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B2x%7D%7Bx%20%2B%203%7D%20%20%5Cleqslant%203" id="Tex

Fofino [41]

$Domain:x\neq-3\\\\\dfrac{2x}{x+3}\leq3\qquad\text{subtract 3 from both sides}\\\\\dfrac{2x}{x+3}-3\leq0\\\\\dfrac{2x}{x+3}-\dfrac{3(x+3)}{x+3}\leq0\\\\\dfrac{2x}{x+3}-\dfrac{3x+9}{x+3}\leq0\\\\\dfrac{2x-(3x+9)}{x+3}\leq0\\\\\dfrac{2x-3x-9}{x+3}\leq0\\\\\dfrac{-x-9}{x+3}\leq0\Rightarrow(x+3)(-x-9)\leq0\\\\x=-3,\ x=-9\\\\x\in(-\infty,\ -9]\ \cup\ (-3,\ \infty)$

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3 years ago