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Wewaii [24]
2 years ago
13

Plsss answer the question in the photo .I rlly need ur help.It s probably easy for most of you guys here .

Mathematics
1 answer:
denis-greek [22]2 years ago
8 0
Box 1: None
Box 2: None
Box 3: A, B, C, E
Box 4: D

Some columns are empty because there cannot be 0 or 1 aute angle in a shape!

I hope this helped! Mark me Brainliest! :) -Raven❤️
You might be interested in
Hey can you please help me posted picture of question
Rudik [331]
Following changes are made in graph of G(x) to obtain F(x)

1) Vertical stretch by a factor of 2
So, G(x) will be changed to 2G(x)

2) Shift towards Right by 2 units
So, 2G(x) will be changed to 2G(x - 2)

3) Upward shift by 2 units.
So, 2G(x - 2) will be changed to 2G(x - 2) + 2

Thus,

F(x) = 2G(x - 2) + 2
F(x) = 2(x - 2)² + 2

Therefore, option C is the correct answer
3 0
3 years ago
Suppose monthly rental prices for a one-bedroom apartment in a large city has a distribution that is skewed to the right with a
omeli [17]

Answer:

a) Nothing, beause the distribution of the monthly rental prices are not normal.

b) 1.43% probability that the sample mean rent price will be greater than $900

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

(a) Suppose a one-bedroom rental listing in this large city is selected at random. What can be said about the probability that the listed rent price will be at least $930?

Nothing, beause the distribution of the monthly rental prices are not normal.

(b) Suppose a random sample 30 one-bedroom rental listing in this large city will be selected, the rent price will be recorded for each listing, and the sample mean rent price will be computed. What can be said about the probability that the sample mean rent price will be greater than $900?

Now we can apply the Central Limit Theorem.

\mu = 880, \sigma = 50, n = 30, s = \frac{50}{\sqrt{30}} = 9.1287

This probability is 1 subtracted by the pvalue of Z when X = 900.

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

By the Central Limit Theorem

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

Z = \frac{900 - 880}{9.1287}

Z = 2.19

Z = 2.19 has a pvalue of 0.9857

1 - 0.9857 = 0.0143

1.43% probability that the sample mean rent price will be greater than $900

8 0
3 years ago
The Area of a square at right is 72 square units Find the value of X, <br>Did I do this right?
Assoli18 [71]
Yes, your work is correct.

The diagram does illustrate x²+6x+9=72.  This is also represented by

(x+3)² = 72

We can take the square root of both sides:
\sqrt{(x+3)^2}=\sqrt{72}&#10;\\&#10;\\x+3=\pm 6 \sqrt2&#10;\\&#10;\\x+3-3=\pm 6 \sqrt{2} -3&#10;\\&#10;\\x=-3\pm 6\sqrt{2}

When we evaluate this, the answers are x = -11.5 or x = 5.5.
5 0
2 years ago
How do I do this ?????
AleksAgata [21]

Answer:

y=3

Step-by-step explanation:

it says that y is the inverse of x so if x=3 and y=8, so if x=8 than y must equal 3

3 0
2 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bx%7D%7B2%7D%20%2B1%20%3D%20%5Csqrt%7B3%7D%20" id="TexFormula1" title=" \frac{x}{2
Tresset [83]
I hope this helps you

3 0
3 years ago
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