All categories

2 years ago

What figures have rotation symmetry? PLEASE HELP I NEED A GOOD GRADE

Mathematics

2 answers:

Darina [25.2K]2 years ago

6 0

if set up like

1 2

3 4

its 1 and 4

1 has 90deg rotationals and 4 has 180 the other two don't.

Ivanshal [37]2 years ago

5 0

The figures on the top left and the bottom right have rotational symmetry.

You might be interested in

For i≥1 , let Xi∼G1/2 be distributed Geometrically with parameter 1/2 . Define Yn=1n−−√∑i=1n(Xi−2) Approximate P(−1≤Yn≤2) with l

murzikaleks [220]

Answer:

The answer is "0.68".

Step-by-step explanation:

Given value:

$X_i \sim \frac{G_1}{2}$

$E(X_i)=2 \\$

$Var (X_i)= \frac{1- \frac{1}{2}}{(\frac{1}{2})^2}\\$

$= \frac{ \frac{2-1}{2}}{\frac{1}{4}}\\\\= \frac{ \frac{1}{2}}{\frac{1}{4}}\\\\= \frac{1}{2} \times \frac{4}{1}\\\\= \frac{4}{2}\\\\=2$

Now we calculate the $\bar X \sim N(2, \sqrt{\frac{2}{n}})\\$

$\to \frac{\bar X - 2}{\sqrt{\frac{2}{n}}} \sim N(0, 1)\\$

$\to \sum^n_{i=1} \frac{X_i - 2}{n} \times\sqrt{\frac{n}{2}}} \sim N(0, 1)\\\\\to \sum^n_{i=1} \frac{X_i - 2}{\sqrt{2n}} \sim N(0, 1)\\$

$\to Z_n = \frac{1}{\sqrt{n}} \sum^n_{i=1} (X_i -2) \sim N(0, 2)\\$

$\to P(-1 \leq X_n \leq 2) = P(Z_n \leq Z) -P(Z_n \leq -1) \\\\$

$= 0.92 -0.24\\\\= 0.68$

6 0

2 years ago

The length of a rectangle is 4 meters longer than the width. if the area is 21 square meters, find the rectangle's dimensions.

brilliants [131]

The length of a rectangle is 4 meters longer than the width. if the area is 21 square meters, find the rectangle's dimensions. round to the nearest tenth of a meter.

7 0

3 years ago

The graph of a system of equations is shown below . What system of equations represents the graph ?

True [87]

Answer:

Step-by-step explanation:

Plug the values of x into each equation

3 0

2 years ago

Suppose the names of the months of the year are placed in a hat and a name is drawn at random. (a) list the sample space for thi

ollegr [7]

(a)

The sample space is a set whose elements are all the possible outcomes for the experiment. Since we will extract one of the months of the years, the sample space is the set composed by all the 12 months:

$\Omega = \{ \text{January},\ \text{February},\ldots,\text{December}\}$

(b)

An event is a subset of the sample space. Events are often defined by their properties. In this example, the event E is the subset of the sample space defined as

$E = \{ x \in \Omega: x \text{ starts with the letter J}\}$

So, we have

$E = \{\text{January},\ \text{June},\ \text{July}\}$

(c)

If all outcomes have equal probability, then the probability of an event is the ratio bewteen its cardinality, and the cardinality of the whole sample space:

$P(E) = \dfrac{n(E)}{n(\Omega)} = \dfrac{3}{12} = \dfrac{1}{4}$

In words, since there are three months beginning with J out of 12 months, we have a probability of 3 over 12 to pick a month starting with J, which simplifies to 1 over 4.

3 0

3 years ago

A right triangle has legs measuring 18 in. and 26 in. What is the length of the hypotenuse? Round to the nearest tenth. A) 18.8

Evgen [1.6K]

Answer:

Step-by-step explanation:

3 0

2 years ago