What is the length of the side of a right triangle that has a side length of 20 mm and a hypotenuse of 25 mm?

If 2^3^x/8^y = 1/16, what is the value of x - y?

Ksivusya [100]

Answer: fbhgjybjgk byhhhhnkbm j hun. Hhbnbgn

Step-by-step explanation: Hvjgbkgbjh

8 0

3 years ago

Algebra 2 Standard Deviation

Illusion [34]

Answer:

don't have answer but have how to do them

Step-by-step explanation:

What are z-scores?

A z-score measures exactly how many standard deviations above or below the mean a data point is.

Here's the formula for calculating a z-score:

z=\dfrac{\text{data point}-\text{mean}}{\text{standard deviation}}z=

standard deviation

data point−mean

z, equals, start fraction, start text, d, a, t, a, space, p, o, i, n, t, end text, minus, start text, m, e, a, n, end text, divided by, start text, s, t, a, n, d, a, r, d, space, d, e, v, i, a, t, i, o, n, end text, end fraction

Here's the same formula written with symbols:

z=\dfrac{x-\mu}{\sigma}z=

σ

x−μ

z, equals, start fraction, x, minus, mu, divided by, sigma, end fraction

Here are some important facts about z-scores:

A positive z-score says the data point is above average.

A negative z-score says the data point is below average.

A z-score close to 000 says the data point is close to average.

A data point can be considered unusual if its z-score is above 333 or below -3−3minus, 3. [Really?]

Want to learn more about z-scores? Check out this video.

Example 1

The grades on a history midterm at Almond have a mean of \mu = 85μ=85mu, equals, 85 and a standard deviation of \sigma = 2σ=2sigma, equals, 2.

Michael scored 868686 on the exam.

Find the z-score for Michael's exam grade.

\begin{aligned}z&=\dfrac{\text{his grade}-\text{mean grade}}{\text{standard deviation}}\\ \\ z&=\dfrac{86-85}{2}\\ \\ z&=\dfrac{1}{2}=0.5\end{aligned}

z

=

standard deviation

his grade−mean grade

=

2

86−85

=

2

1

=0.5

Michael's z-score is 0.50.50, point, 5. His grade was half of a standard deviation above the mean.

Example 2

The grades on a geometry midterm at Almond have a mean of \mu = 82μ=82mu, equals, 82 and a standard deviation of \sigma = 4σ=4sigma, equals, 4.

Michael scored 747474 on the exam.

Find the z-score for Michael's exam grade.

\begin{aligned}z&=\dfrac{\text{his grade}-\text{mean grade}}{\text{standard deviation}}\\ \\ z&=\dfrac{74-82}{4}\\ \\ z&=\dfrac{-8}{4}=-2\end{aligned}

z

=

standard deviation

his grade−mean grade

=

4

74−82

=

4

−8

=−2

Michael's z-score is -2−2minus, 2. His grade was two standard deviations below the mean.

https://www.khanacademy.org/math/statistics-probability/modeling-distributions-of-data/z-scores/a/z-scores-review

this should help

6 0

3 years ago

Are 0.6600 and 0.66 equivalent decimals?<br> yes<br> no

malfutka [58]

Answer:

Yes.

Step-by-step explanation:

0.66 is 0.6600.

7 0

2 years ago

Read 2 more answers

When rolling a pair of dice at the Bellagio Casino in Las Vegas, there is 1 chance in 36 that the outcome is a 12.

nasty-shy [4]

Yes, that is correct. likewise, the odds of rolling a 1 are $\frac{1}{36}$

3 0

3 years ago

A researcher compares two compounds (1 and 2) used in the manufacture of car tires that are designed to reduce braking distances

wariber [46]

Answer:

Step-by-step explanation:

Hello!

The objective of this experiment is to compare two compounds, designed to reduce braking distance, used in tire manufacturing to prove if the braking distance of SUV's equipped with tires made with compound 1 is shorter than the braking distance of SUV's equipped with tires made with compound 2.

So you have 2 independent populations, SUV's equipped with tires made using compound 1 and SUV's equipped with tires made using compound 2.

Two samples of 81 braking tests are made and the braking distance was measured each time, the study variables are determined as:

X₁: Braking distance of an SUV equipped with tires made with compound one.

Its sample mean is X[bar]₁= 69 feet

And the Standard deviation S₁= 10.4 feet

X₂: Braking distance of an SUV equipped with tires made with compound two.

Its sample mean is X[bar]₂= 71 feet

And the Standard deviation S₂= 7.6 feet

We don't have any information on the distribution of the study variables, nor the sample data to test it, but since both sample sizes are large enough n₁ and n₂ ≥ 30 we can apply the central limit theorem and approximate the distribution of both variables sample means to normal.

The researcher's hypothesis, as mentioned before, is that the braking distance using compound one is less than the distance obtained using compound 2, symbolically: μ₁ < μ₂

The statistical hypotheses are:

H₀: μ₁ ≥ μ₂

H₁: μ₁ < μ₂

α: 0.05

The statistic to use to compare these two populations is a pooled Z test

$Z= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{\sqrt{\frac{S^2_1}{n_1} +\frac{S^2_2}{n_2} } }$

Z ≈ N(0;1)

$Z_{H_0}= \frac{69-71-0}{\sqrt{\frac{108.16}{81} +\frac{57.76}{81} } }= -1.397$

The rejection region if this hypothesis test is one-tailed to the right, so you'll reject the null hypothesis to small values of the statistic. The critical value for this test is:

$Z_{\alpha } = Z_{0.05}= -1.648$

Decision rule:

If $Z_{H_0}$ > -1.648 , then you do not reject the null hypothesis.

If $Z_{H_0}$ ≤ -1.648 , then you reject the null hypothesis.

Since the statistic value is greater than the critical value, the decision is to not reject the null hypothesis.

At a 5% significance level, you can conclude that the average braking distance of SUV's equipped with tires manufactured used compound 1 is greater than the average braking distance of SUV's equipped with tires manufactured used compound 2.

I hope you have a SUPER day!

4 0

3 years ago